Chapter 6 Linear Inequalities

Given:

The inequality $30x < 200$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$30x < 200$

To solve for $x$, we divide both sides of the inequality by 30. Since 30 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{30x}{30} < \frac{200}{30}$

$x < \frac{\cancel{200}^{20}}{\cancel{30}_{3}}$

$x < \frac{20}{3}$

Converting the fraction to a decimal for easier comparison:

$x < 6.66...$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, ...\}$.

We need to find the natural numbers $x$ that satisfy $x < 6.66...$.

The natural numbers less than $6.66...$ are $1, 2, 3, 4, 5, 6$.

Therefore, the solution set when $x$ is a natural number is $\{1, 2, 3, 4, 5, 6\}$.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 6.66...$.

The integers less than $6.66...$ are all integers less than or equal to 6.

Therefore, the solution set when $x$ is an integer is $\{..., -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.

Given:

The inequality $5x - 3 < 3x + 1$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$5x - 3 < 3x + 1$

Subtract $3x$ from both sides of the inequality:

$5x - 3x - 3 < 3x - 3x + 1$

$2x - 3 < 1$

Add 3 to both sides of the inequality:

$2x - 3 + 3 < 1 + 3$

$2x < 4$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{2x}{2} < \frac{4}{2}$

$x < 2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 2$.

The integers strictly less than 2 are all integers up to 1.

Therefore, the solution set when $x$ is an integer is $\{..., -3, -2, -1, 0, 1\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x < 2$.

This includes all real numbers less than 2.

In interval notation, the solution set is $(-\infty, 2)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x < 2\}$ or $(-\infty, 2)$.

Given:

The inequality $4x + 3 < 6x + 7$.

To Find:

The solution set for the given inequality.

Solution:

We are given the inequality:

$4x + 3 < 6x + 7$

To solve for $x$, we want to gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $4x$ from both sides of the inequality:

$4x - 4x + 3 < 6x - 4x + 7$

$3 < 2x + 7$

Now, subtract 7 from both sides of the inequality:

$3 - 7 < 2x + 7 - 7$

$-4 < 2x$

Finally, divide both sides by 2. Since 2 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{-4}{2} < \frac{2x}{2}$

$-2 < x$

This can also be written as $x > -2$.

Since the type of number $x$ represents (integer, natural, real) is not specified, we assume $x$ is a real number.

The solution set consists of all real numbers greater than -2.

In interval notation, the solution set is $(-2, \infty)$.

Therefore, the solution to the inequality $4x + 3 < 6x + 7$ is $x > -2$ or $(-2, \infty)$.

Given:

The inequality $\frac{5-2x}{3} \leq \frac{x}{6} - 5$.

To Find:

The solution set for the given inequality.

Solution:

We are given the inequality:

$\frac{5-2x}{3} \leq \frac{x}{6} - 5$

To eliminate the fractions, we can find the Least Common Multiple (LCM) of the denominators, which are 3 and 6. The LCM of 3 and 6 is 6.

Multiply both sides of the inequality by the LCM, 6. Since 6 is a positive number, the inequality sign remains unchanged.

$6 \times \left(\frac{5-2x}{3}\right) \leq 6 \times \left(\frac{x}{6} - 5\right)$

$\frac{\cancel{6}^2}{1} \times \frac{(5-2x)}{\cancel{3}_1} \leq \frac{\cancel{6}^1}{1} \times \frac{x}{\cancel{6}_1} - 6 \times 5$

$2(5-2x) \leq x - 30$

Expand the left side by distributing the 2:

$10 - 4x \leq x - 30$

Now, we rearrange the inequality to group the $x$ terms on one side and the constant terms on the other.

Add $4x$ to both sides:

$10 - 4x + 4x \leq x + 4x - 30$

$10 \leq 5x - 30$

Add 30 to both sides:

$10 + 30 \leq 5x - 30 + 30$

$40 \leq 5x$

Finally, divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{40}{5} \leq \frac{5x}{5}$

$8 \leq x$

This is equivalent to $x \geq 8$.

Assuming $x$ is a real number (as not specified otherwise), the solution set includes all real numbers greater than or equal to 8.

In interval notation, the solution set is $[8, \infty)$.

Therefore, the solution to the inequality $\frac{5-2x}{3} \leq \frac{x}{6} - 5$ is $x \geq 8$ or $[8, \infty)$.

Given:

The inequality $7x + 3 < 5x + 9$.

To Find:

The solution set for the given inequality and show the graph of the solutions on a number line.

Solution:

We are given the inequality:

$7x + 3 < 5x + 9$

Subtract $5x$ from both sides of the inequality:

$7x - 5x + 3 < 5x - 5x + 9$

$2x + 3 < 9$

Subtract 3 from both sides of the inequality:

$2x + 3 - 3 < 9 - 3$

$2x < 6$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{2x}{2} < \frac{6}{2}$

$x < 3$

Assuming $x$ is a real number (as not specified otherwise), the solution set consists of all real numbers strictly less than 3.

Therefore, the solution to the inequality $7x + 3 < 5x + 9$ is $x < 3$.

In interval notation, the solution set is $(-\infty, 3)$.

Graph of the solution on the number line:

To represent the solution $x < 3$ on a number line:

1. Draw a horizontal line representing the real numbers.

2. Mark the point representing the number 3 on the line.

3. Since the inequality is strict ($x < 3$), the number 3 is not included in the solution set. We indicate this by drawing an open circle at 3.

4. Shade the portion of the number line to the left of the open circle at 3. This shaded region represents all real numbers less than 3.

5. Draw an arrow pointing to the left from the shaded region to indicate that the solution extends indefinitely towards negative infinity.

Visual Representation Description:

<------------------------------------)-------------------------->

... -2 -1 0 1 2 ○ 4 5 ...

3

The open circle at 3 and the shaded line extending to the left represent the solution set $x < 3$.

Given:

The inequality $\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$.

To Find:

The solution set for the given inequality and show the graph of the solutions on a number line.

Solution:

We are given the inequality:

$\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 2 and 4, which is 4.

Multiply both sides of the inequality by 4. Since 4 is a positive number, the inequality sign remains unchanged.

$4 \times \left(\frac{3x-4}{2}\right) \geq 4 \times \left(\frac{x+1}{4} - 1\right)$

$\frac{\cancel{4}^2}{1} \times \frac{(3x-4)}{\cancel{2}_1} \geq \frac{\cancel{4}^1}{1} \times \frac{(x+1)}{\cancel{4}_1} - 4 \times 1$

$2(3x-4) \geq (x+1) - 4$

Distribute the 2 on the left side and simplify the right side:

$6x - 8 \geq x - 3$

Now, rearrange the inequality to group the $x$ terms on one side and the constant terms on the other.

Subtract $x$ from both sides:

$6x - x - 8 \geq x - x - 3$

$5x - 8 \geq -3$

Add 8 to both sides:

$5x - 8 + 8 \geq -3 + 8$

$5x \geq 5$

Divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{5x}{5} \geq \frac{5}{5}$

$x \geq 1$

Assuming $x$ is a real number (as not specified otherwise), the solution set consists of all real numbers greater than or equal to 1.

Therefore, the solution to the inequality $\frac{3x-4}{2} \geq \frac{x+1}{4} - 1$ is $x \geq 1$.

In interval notation, the solution set is $[1, \infty)$.

Graph of the solution on the number line:

To represent the solution $x \geq 1$ on a number line:

1. Draw a horizontal line representing the real numbers.

2. Mark the point representing the number 1 on the line.

3. Since the inequality is $x \geq 1$ (greater than or equal to), the number 1 is included in the solution set. We indicate this by drawing a closed circle (or solid dot) at 1.

4. Shade the portion of the number line to the right of the closed circle at 1. This shaded region represents all real numbers greater than or equal to 1.

5. Draw an arrow pointing to the right from the shaded region to indicate that the solution extends indefinitely towards positive infinity.

Visual Representation Description:

<--------------------------[------------------------------------>

... -1 0 ● 2 3 4 5 ...

1

The closed circle at 1 and the shaded line extending to the right represent the solution set $x \geq 1$.

Given:

Marks obtained in the first terminal examination = 62.

Marks obtained in the second terminal examination = 48.

The student needs an average of at least 60 marks in three examinations (first terminal, second terminal, and annual).

To Find:

The minimum marks the student should get in the annual examination.

Solution:

Let $x$ be the marks obtained by the student in the annual examination.

The marks in the three examinations are 62, 48, and $x$.

The total marks obtained in the three examinations = $62 + 48 + x = 110 + x$.

The number of examinations is 3.

The average marks for the three examinations is given by:

Average = $\frac{\text{Total marks}}{\text{Number of examinations}} = \frac{110 + x}{3}$

According to the problem, the student must have an average of at least 60 marks. "At least 60" means the average must be greater than or equal to 60.

So, we can set up the inequality:

$\frac{110 + x}{3} \geq 60$

To solve for $x$, we first multiply both sides of the inequality by 3. Since 3 is a positive number, the inequality sign remains the same.

$3 \times \left(\frac{110 + x}{3}\right) \geq 3 \times 60$

$110 + x \geq 180$

Now, subtract 110 from both sides of the inequality:

$110 + x - 110 \geq 180 - 110$

$x \geq 70$

This inequality means that the marks obtained in the annual examination ($x$) must be greater than or equal to 70.

Therefore, the minimum marks the student should get in the annual examination to achieve an average of at least 60 marks is 70.

Given:

We are looking for pairs of consecutive odd natural numbers.

Condition 1: Both numbers in the pair must be larger than 10.

Condition 2: The sum of the two numbers in the pair must be less than 40.

To Find:

All pairs of consecutive odd natural numbers that satisfy the given conditions.

Solution:

Let the smaller of the two consecutive odd natural numbers be $x$.

Since the numbers are consecutive odd natural numbers, the next odd natural number will be $x+2$.

According to the first condition, both numbers must be larger than 10.

$x > 10$

$x+2 > 10$ (This condition is automatically satisfied if $x > 10$)

Since $x$ must be an odd natural number greater than 10, the smallest possible value for $x$ is 11.

According to the second condition, the sum of the two numbers must be less than 40.

$x + (x+2) < 40$

Simplify the inequality:

$2x + 2 < 40$

Subtract 2 from both sides:

$2x < 40 - 2$

$2x < 38$

Divide both sides by 2:

$\frac{2x}{2} < \frac{38}{2}$

$x < 19$

So, we are looking for odd natural numbers $x$ that satisfy both conditions: $x > 10$ and $x < 19$.

Combining the conditions, we need odd natural numbers $x$ such that $10 < x < 19$.

The odd natural numbers between 10 and 19 are 11, 13, 15, and 17.

Now, we find the pairs $(x, x+2)$ for each possible value of $x$:

1. If $x = 11$, the pair is $(11, 11+2) = (11, 13)$.

Check: Both are odd, both > 10. Sum = $11 + 13 = 24 < 40$. This pair satisfies the conditions.

2. If $x = 13$, the pair is $(13, 13+2) = (13, 15)$.

Check: Both are odd, both > 10. Sum = $13 + 15 = 28 < 40$. This pair satisfies the conditions.

3. If $x = 15$, the pair is $(15, 15+2) = (15, 17)$.

Check: Both are odd, both > 10. Sum = $15 + 17 = 32 < 40$. This pair satisfies the conditions.

4. If $x = 17$, the pair is $(17, 17+2) = (17, 19)$.

Check: Both are odd, both > 10. Sum = $17 + 19 = 36 < 40$. This pair satisfies the conditions.

If $x = 19$, the condition $x < 19$ is not satisfied, so we stop here.

The pairs of consecutive odd natural numbers satisfying the given conditions are (11, 13), (13, 15), (15, 17), and (17, 19).

Therefore, the required pairs are (11, 13), (13, 15), (15, 17), (17, 19).

Given:

The inequality $24x < 100$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$24x < 100$

To solve for $x$, we divide both sides of the inequality by 24. Since 24 is a positive number, the direction of the inequality sign remains unchanged.

$\frac{24x}{24} < \frac{100}{24}$

$x < \frac{\cancel{100}^{25}}{\cancel{24}_{6}}$

$x < \frac{25}{6}$

Converting the fraction to a decimal for easier comparison:

$\frac{25}{6} = 4 \frac{1}{6} \approx 4.167$

So the inequality is $x < 4.167$.

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, 5, ...\}$.

We need to find the natural numbers $x$ that satisfy $x < 4.167$.

The natural numbers less than $4.167$ are $1, 2, 3, 4$.

Therefore, the solution set when $x$ is a natural number is $\{1, 2, 3, 4\}$.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 4.167$.

The integers less than $4.167$ are all integers less than or equal to 4.

Therefore, the solution set when $x$ is an integer is $\{..., -2, -1, 0, 1, 2, 3, 4\}$.

Given:

The inequality $-12x > 30$.

To Find:

The solution set for the given inequality when:

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution:

We are given the inequality:

$-12x > 30$

To solve for $x$, we divide both sides of the inequality by -12. Important: When dividing or multiplying both sides of an inequality by a negative number, we must reverse the direction of the inequality sign.

$\frac{-12x}{-12} < \frac{30}{-12}$

$x < -\frac{\cancel{30}^5}{\cancel{12}_2}$

$x < -\frac{5}{2}$

Converting the fraction to a decimal:

$x < -2.5$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is a natural number:

Natural numbers are the set $\{1, 2, 3, 4, ...\}$. These are all positive integers.

We need to find the natural numbers $x$ that satisfy $x < -2.5$.

Since all natural numbers are positive, there are no natural numbers less than -2.5.

Therefore, the solution set when $x$ is a natural number is the empty set, denoted as $\phi$ or { }.

(ii) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < -2.5$.

The integers strictly less than -2.5 are $-3, -4, -5, ...$ and so on, extending towards negative infinity.

Therefore, the solution set when $x$ is an integer is $\{..., -5, -4, -3\}$.

Given:

The inequality $5x - 3 < 7$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$5x - 3 < 7$

Add 3 to both sides of the inequality:

$5x - 3 + 3 < 7 + 3$

$5x < 10$

Divide both sides by 5. Since 5 is a positive number, the inequality sign remains unchanged.

$\frac{5x}{5} < \frac{10}{5}$

$x < 2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x < 2$.

The integers strictly less than 2 are all integers up to 1.

Therefore, the solution set when $x$ is an integer is $\{..., -2, -1, 0, 1\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x < 2$.

This includes all real numbers less than 2.

In interval notation, the solution set is $(-\infty, 2)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x < 2\}$ or $(-\infty, 2)$.

Given:

The inequality $3x + 8 > 2$.

To Find:

The solution set for the given inequality when:

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution:

We are given the inequality:

$3x + 8 > 2$

Subtract 8 from both sides of the inequality:

$3x + 8 - 8 > 2 - 8$

$3x > -6$

Divide both sides by 3. Since 3 is a positive number, the inequality sign remains unchanged.

$\frac{3x}{3} > \frac{-6}{3}$

$x > -2$

Now, we find the solution set based on the given conditions for $x$.

(i) When $x$ is an integer:

Integers are the set $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

We need to find the integers $x$ that satisfy $x > -2$.

The integers strictly greater than -2 are $-1, 0, 1, 2, 3, ...$ and so on, extending towards positive infinity.

Therefore, the solution set when $x$ is an integer is $\{-1, 0, 1, 2, ...\}$.

(ii) When $x$ is a real number:

Real numbers include all rational and irrational numbers.

We need to find the real numbers $x$ that satisfy $x > -2$.

This includes all real numbers greater than -2.

In interval notation, the solution set is $(-2, \infty)$.

Therefore, the solution set when $x$ is a real number is $\{x \in \mathbb{R} \mid x > -2\}$ or $(-2, \infty)$.

Given:

The inequality $4x + 3 < 5x + 7$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$4x + 3 < 5x + 7$

To solve for $x$, we gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $4x$ from both sides:

$4x - 4x + 3 < 5x - 4x + 7$

$3 < x + 7$

Subtract 7 from both sides:

$3 - 7 < x + 7 - 7$

$-4 < x$

This is equivalent to $x > -4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than -4.

Therefore, the solution set for the inequality is $x > -4$.

In interval notation, the solution set is $(-4, \infty)$.

Given:

The inequality $3x - 7 > 5x - 1$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3x - 7 > 5x - 1$

To solve for $x$, we gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $5x$ from both sides:

$3x - 5x - 7 > 5x - 5x - 1$

$-2x - 7 > -1$

Add 7 to both sides:

$-2x - 7 + 7 > -1 + 7$

$-2x > 6$

Divide both sides by -2. Important: Remember to reverse the inequality sign when dividing by a negative number.

$\frac{-2x}{-2} < \frac{6}{-2}$

$x < -3$

Since $x$ must be a real number, the solution set includes all real numbers less than -3.

Therefore, the solution set for the inequality is $x < -3$.

In interval notation, the solution set is $(-\infty, -3)$.

Given:

The inequality $3(x - 1) \leq 2(x - 3)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3(x - 1) \leq 2(x - 3)$

First, distribute the constants on both sides of the inequality:

$3 \times x - 3 \times 1 \leq 2 \times x - 2 \times 3$

$3x - 3 \leq 2x - 6$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $2x$ from both sides:

$3x - 2x - 3 \leq 2x - 2x - 6$

$x - 3 \leq -6$

Add 3 to both sides:

$x - 3 + 3 \leq -6 + 3$

$x \leq -3$

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to -3.

Therefore, the solution set for the inequality is $x \leq -3$.

In interval notation, the solution set is $(-\infty, -3]$.

Given:

The inequality $3(2 - x) \geq 2(1 - x)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$3(2 - x) \geq 2(1 - x)$

First, distribute the constants on both sides of the inequality:

$3 \times 2 - 3 \times x \geq 2 \times 1 - 2 \times x$

$6 - 3x \geq 2 - 2x$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $3x$ to both sides:

$6 - 3x + 3x \geq 2 - 2x + 3x$

$6 \geq 2 + x$

Subtract 2 from both sides:

$6 - 2 \geq 2 - 2 + x$

$4 \geq x$

This is equivalent to $x \leq 4$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 4.

Therefore, the solution set for the inequality is $x \leq 4$.

In interval notation, the solution set is $(-\infty, 4]$.

Given:

The inequality $x + \frac{x}{2} + \frac{x}{3} < 11$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$x + \frac{x}{2} + \frac{x}{3} < 11$

To solve this inequality, we first combine the terms on the left side. The least common multiple (LCM) of the denominators (1, 2, and 3) is 6.

Rewrite the left side with a common denominator of 6:

$\frac{6x}{6} + \frac{3x}{6} + \frac{2x}{6} < 11$

Combine the fractions:

$\frac{6x + 3x + 2x}{6} < 11$

$\frac{11x}{6} < 11$

Now, we want to isolate $x$. Multiply both sides of the inequality by 6. Since 6 is a positive number, the inequality sign remains the same.

$6 \times \left(\frac{11x}{6}\right) < 6 \times 11$

$11x < 66$

Finally, divide both sides by 11. Since 11 is a positive number, the inequality sign remains the same.

$\frac{11x}{11} < \frac{66}{11}$

$x < 6$

Since $x$ must be a real number, the solution set includes all real numbers less than 6.

Therefore, the solution set for the inequality is $x < 6$.

In interval notation, the solution set is $(-\infty, 6)$.

Given:

The inequality $\frac{x}{3} > \frac{x}{2} + 1$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{x}{3} > \frac{x}{2} + 1$

To eliminate the fractions, we can find the Least Common Multiple (LCM) of the denominators, 3 and 2. The LCM is 6.

Multiply both sides of the inequality by 6. Since 6 is a positive number, the inequality sign remains unchanged.

$6 \times \left(\frac{x}{3}\right) > 6 \times \left(\frac{x}{2} + 1\right)$

$\frac{\cancel{6}^2}{1} \times \frac{x}{\cancel{3}_1} > \frac{\cancel{6}^3}{1} \times \frac{x}{\cancel{2}_1} + 6 \times 1$

$2x > 3x + 6$

Now, we rearrange the inequality to isolate $x$. Subtract $3x$ from both sides:

$2x - 3x > 3x - 3x + 6$

$-x > 6$

To solve for $x$, multiply both sides by -1. Important: When multiplying (or dividing) by a negative number, we must reverse the direction of the inequality sign.

$(-1) \times (-x) < (-1) \times 6$

$x < -6$

Since $x$ must be a real number, the solution set includes all real numbers less than -6.

Therefore, the solution set for the inequality is $x < -6$.

In interval notation, the solution set is $(-\infty, -6)$.

Given:

The inequality $\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators, 5 and 3. The LCM is 15.

Multiply both sides of the inequality by 15. Since 15 is a positive number, the inequality sign remains unchanged.

$15 \times \left(\frac{3(x - 2)}{5}\right) \leq 15 \times \left(\frac{5(2 - x)}{3}\right)$

$\frac{\cancel{15}^3}{1} \times \frac{3(x - 2)}{\cancel{5}_1} \leq \frac{\cancel{15}^5}{1} \times \frac{5(2 - x)}{\cancel{3}_1}$

$3 \times 3(x - 2) \leq 5 \times 5(2 - x)$

$9(x - 2) \leq 25(2 - x)$

Distribute the constants on both sides:

$9x - 18 \leq 50 - 25x$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $25x$ to both sides:

$9x + 25x - 18 \leq 50 - 25x + 25x$

$34x - 18 \leq 50$

Add 18 to both sides:

$34x - 18 + 18 \leq 50 + 18$

$34x \leq 68$

Divide both sides by 34. Since 34 is a positive number, the inequality sign remains unchanged.

$\frac{34x}{34} \leq \frac{68}{34}$

$x \leq 2$

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Given:

The inequality $\frac{1}{2}\left( \frac{3x}{5} + 4 \right) \geq \frac{1}{3}(x - 6)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{1}{2}\left( \frac{3x}{5} + 4 \right) \geq \frac{1}{3}(x - 6)$

First, distribute the constants on both sides of the inequality:

$\frac{1}{2} \times \frac{3x}{5} + \frac{1}{2} \times 4 \geq \frac{1}{3} \times x - \frac{1}{3} \times 6$

$\frac{3x}{10} + \frac{4}{2} \geq \frac{x}{3} - \frac{6}{3}$

Simplify the constants:

$\frac{3x}{10} + 2 \geq \frac{x}{3} - 2$

To eliminate the fractions, find the Least Common Multiple (LCM) of the denominators 10 and 3. The LCM is 30.

Multiply both sides of the inequality by 30. Since 30 is a positive number, the inequality sign remains unchanged.

$30 \times \left(\frac{3x}{10} + 2\right) \geq 30 \times \left(\frac{x}{3} - 2\right)$

$30 \times \frac{3x}{10} + 30 \times 2 \geq 30 \times \frac{x}{3} - 30 \times 2$

$\frac{\cancel{30}^3}{1} \times \frac{3x}{\cancel{10}_1} + 60 \geq \frac{\cancel{30}^{10}}{1} \times \frac{x}{\cancel{3}_1} - 60$

$3 \times 3x + 60 \geq 10 \times x - 60$

$9x + 60 \geq 10x - 60$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $9x$ from both sides:

$9x - 9x + 60 \geq 10x - 9x - 60$

$60 \geq x - 60$

Add 60 to both sides:

$60 + 60 \geq x - 60 + 60$

$120 \geq x$

This is equivalent to $x \leq 120$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 120.

Therefore, the solution set for the inequality is $x \leq 120$.

In interval notation, the solution set is $(-\infty, 120]$.

Given:

The inequality $2(2x + 3) - 10 < 6(x - 2)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$2(2x + 3) - 10 < 6(x - 2)$

First, distribute the constants on both sides of the inequality:

$2 \times 2x + 2 \times 3 - 10 < 6 \times x - 6 \times 2$

$4x + 6 - 10 < 6x - 12$

Simplify both sides:

$4x - 4 < 6x - 12$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $4x$ from both sides:

$4x - 4x - 4 < 6x - 4x - 12$

$-4 < 2x - 12$

Add 12 to both sides:

$-4 + 12 < 2x - 12 + 12$

$8 < 2x$

Divide both sides by 2. Since 2 is a positive number, the inequality sign remains unchanged.

$\frac{8}{2} < \frac{2x}{2}$

$4 < x$

This is equivalent to $x > 4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than 4.

Therefore, the solution set for the inequality is $x > 4$.

In interval notation, the solution set is $(4, \infty)$.

Given:

The inequality $37 – (3x + 5) \geq 9x – 8 (x – 3)$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$37 – (3x + 5) \geq 9x – 8 (x – 3)$

First, simplify both sides of the inequality by removing the parentheses.

Left side: $37 - 3x - 5$

Combine constant terms on the left side:

$32 - 3x$

Right side: $9x - 8(x) - 8(-3)$

$9x - 8x + 24$

Combine $x$ terms on the right side:

$x + 24$

So the inequality becomes:

$32 - 3x \geq x + 24$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Add $3x$ to both sides:

$32 - 3x + 3x \geq x + 3x + 24$

$32 \geq 4x + 24$

Subtract 24 from both sides:

$32 - 24 \geq 4x + 24 - 24$

$8 \geq 4x$

Divide both sides by 4. Since 4 is a positive number, the inequality sign remains unchanged.

$\frac{8}{4} \geq \frac{4x}{4}$

$2 \geq x$

This is equivalent to $x \leq 2$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Given:

The inequality $\frac{x}{4} < \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 4, 3, and 5.

LCM(4, 3, 5) = $4 \times 3 \times 5 = 60$.

Multiply both sides of the inequality by 60. Since 60 is a positive number, the inequality sign remains unchanged.

$60 \times \left(\frac{x}{4}\right) < 60 \times \left(\frac{5x - 2}{3} - \frac{7x - 3}{5}\right)$

$60 \times \frac{x}{4} < 60 \times \frac{5x - 2}{3} - 60 \times \frac{7x - 3}{5}$

$\frac{\cancel{60}^{15}}{1} \times \frac{x}{\cancel{4}_1} < \frac{\cancel{60}^{20}}{1} \times \frac{(5x - 2)}{\cancel{3}_1} - \frac{\cancel{60}^{12}}{1} \times \frac{(7x - 3)}{\cancel{5}_1}$

$15x < 20(5x - 2) - 12(7x - 3)$

Distribute the constants on the right side:

$15x < (20 \times 5x - 20 \times 2) - (12 \times 7x - 12 \times 3)$

$15x < (100x - 40) - (84x - 36)$

$15x < 100x - 40 - 84x + 36$

Combine like terms on the right side:

$15x < (100x - 84x) + (-40 + 36)$

$15x < 16x - 4$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $15x$ from both sides:

$15x - 15x < 16x - 15x - 4$

$0 < x - 4$

Add 4 to both sides:

$0 + 4 < x - 4 + 4$

$4 < x$

This is equivalent to $x > 4$.

Since $x$ must be a real number, the solution set includes all real numbers greater than 4.

Therefore, the solution set for the inequality is $x > 4$.

In interval notation, the solution set is $(4, \infty)$.

Given:

The inequality $\frac{(2x - 1)}{3} \geq \frac{(3x - 2)}{4} - \frac{(2 - x)}{5}$.

We need to solve for real $x$.

To Find:

The solution set for the given inequality for real $x$.

Solution:

We are given the inequality:

$\frac{2x - 1}{3} \geq \frac{3x - 2}{4} - \frac{2 - x}{5}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators 3, 4, and 5. The LCM is $3 \times 4 \times 5 = 60$.

Multiply both sides of the inequality by 60. Since 60 is a positive number, the inequality sign remains unchanged.

$60 \times \left(\frac{2x - 1}{3}\right) \geq 60 \times \left(\frac{3x - 2}{4} - \frac{2 - x}{5}\right)$

$60 \times \frac{2x - 1}{3} \geq 60 \times \frac{3x - 2}{4} - 60 \times \frac{2 - x}{5}$

$\frac{\cancel{60}^{20}}{1} \times \frac{(2x - 1)}{\cancel{3}_1} \geq \frac{\cancel{60}^{15}}{1} \times \frac{(3x - 2)}{\cancel{4}_1} - \frac{\cancel{60}^{12}}{1} \times \frac{(2 - x)}{\cancel{5}_1}$

$20(2x - 1) \geq 15(3x - 2) - 12(2 - x)$

Distribute the constants on both sides:

$20 \times 2x - 20 \times 1 \geq (15 \times 3x - 15 \times 2) - (12 \times 2 - 12 \times x)$

$40x - 20 \geq (45x - 30) - (24 - 12x)$

Remove the parentheses on the right side, paying attention to the signs:

$40x - 20 \geq 45x - 30 - 24 + 12x$

Combine like terms on the right side:

$40x - 20 \geq (45x + 12x) + (-30 - 24)$

$40x - 20 \geq 57x - 54$

Now, gather the terms involving $x$ on one side and the constant terms on the other.

Subtract $40x$ from both sides:

$40x - 40x - 20 \geq 57x - 40x - 54$

$-20 \geq 17x - 54$

Add 54 to both sides:

$-20 + 54 \geq 17x - 54 + 54$

$34 \geq 17x$

Divide both sides by 17. Since 17 is a positive number, the inequality sign remains unchanged.

$\frac{34}{17} \geq \frac{17x}{17}$

$2 \geq x$

This is equivalent to $x \leq 2$.

Since $x$ must be a real number, the solution set includes all real numbers less than or equal to 2.

Therefore, the solution set for the inequality is $x \leq 2$.

In interval notation, the solution set is $(-\infty, 2]$.

Solution:

We are given the inequality:

$3x - 2 < 2x + 1$

To solve for $x$, we first move the terms involving $x$ to one side and the constant terms to the other side.

Subtract $2x$ from both sides of the inequality:

$3x - 2 - 2x < 2x + 1 - 2x$

Simplifying both sides, we get:

$x - 2 < 1$

Now, add $2$ to both sides of the inequality:

$x - 2 + 2 < 1 + 2$

Simplifying both sides, we get:

$x < 3$

The solution to the inequality $3x - 2 < 2x + 1$ is $x < 3$.

Graph on Number Line:

The solution set is the set of all real numbers $x$ such that $x$ is strictly less than $3$. This can be represented on a number line as follows:

1. Draw a straight line representing the number line.

2. Locate the number $3$ on the number line.

3. Since the inequality is $x < 3$ (strict inequality), the point $3$ itself is not included in the solution set. This is denoted by drawing an open circle at the point $3$ on the number line.

4. The solution includes all numbers less than $3$. To represent this, shade or darken the part of the number line that is to the left of the open circle at $3$. The shaded region extends infinitely to the left.

The graph will be an open circle at $3$ with the line segment to its left shaded.

Solution:

We are given the inequality:

$5x - 3 \ge 3x - 5$

To solve for $x$, we move the terms involving $x$ to one side and the constant terms to the other side.

Subtract $3x$ from both sides of the inequality:

$5x - 3 - 3x \ge 3x - 5 - 3x$

Simplifying both sides, we get:

$2x - 3 \ge -5$

Now, add $3$ to both sides of the inequality:

$2x - 3 + 3 \ge -5 + 3$

Simplifying both sides, we get:

$2x \ge -2$

Divide both sides by $2$ (which is a positive number, so the inequality sign does not change):

$\frac{2x}{2} \ge \frac{-2}{2}$

Simplifying both sides, we get:

$x \ge -1$

The solution to the inequality $5x - 3 \ge 3x - 5$ is $x \ge -1$.

Graph on Number Line:

The solution set is the set of all real numbers $x$ such that $x$ is greater than or equal to $-1$. This can be represented on a number line as follows:

1. Draw a straight line representing the number line.

2. Locate the number $-1$ on the number line.

3. Since the inequality is $x \ge -1$ (inclusive inequality), the point $-1$ itself is included in the solution set. This is denoted by drawing a closed circle (or filled circle) at the point $-1$ on the number line.

4. The solution includes $-1$ and all numbers greater than $-1$. To represent this, shade or darken the part of the number line that is to the right of the closed circle at $-1$. The shaded region extends infinitely to the right.

The graph will be a closed circle at $-1$ with the line segment to its right shaded.

Solution:

We are given the inequality:

$3(1 - x) < 2(x + 4)$

First, distribute the constants on both sides of the inequality:

$3 \times 1 - 3 \times x < 2 \times x + 2 \times 4$

Simplifying both sides, we get:

$3 - 3x < 2x + 8$

Now, we move the terms involving $x$ to one side and the constant terms to the other side.

Add $3x$ to both sides of the inequality:

$3 - 3x + 3x < 2x + 8 + 3x$

Simplifying both sides, we get:

$3 < 5x + 8$

Subtract $8$ from both sides of the inequality:

$3 - 8 < 5x + 8 - 8$

Simplifying both sides, we get:

$-5 < 5x$

Divide both sides by $5$ (which is a positive number, so the inequality sign does not change):

$\frac{-5}{5} < \frac{5x}{5}$

Simplifying both sides, we get:

$-1 < x$

This is equivalent to $x > -1$.

The solution to the inequality $3(1 - x) < 2(x + 4)$ is $x > -1$.

Graph on Number Line:

The solution set is the set of all real numbers $x$ such that $x$ is strictly greater than $-1$. This can be represented on a number line as follows:

1. Draw a straight line representing the number line.

2. Locate the number $-1$ on the number line.

3. Since the inequality is $x > -1$ (strict inequality), the point $-1$ itself is not included in the solution set. This is denoted by drawing an open circle at the point $-1$ on the number line.

4. The solution includes all numbers greater than $-1$. To represent this, shade or darken the part of the number line that is to the right of the open circle at $-1$. The shaded region extends infinitely to the right.

The graph will be an open circle at $-1$ with the line segment to its right shaded.

Solution:

We are given the inequality:

$\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$

To solve this inequality, we first simplify the right-hand side by finding a common denominator for the fractions.

The least common multiple (LCM) of $3$ and $5$ is $15$.

Rewrite the fractions on the right side with denominator $15$:

$\frac{5x - 2}{3} = \frac{5(5x - 2)}{5 \times 3} = \frac{25x - 10}{15}$

$\frac{7x - 3}{5} = \frac{3(7x - 3)}{3 \times 5} = \frac{21x - 9}{15}$

Substitute these back into the inequality:

$\frac{x}{2} \ge \frac{25x - 10}{15} - \frac{21x - 9}{15}$

Combine the fractions on the right side:

$\frac{x}{2} \ge \frac{(25x - 10) - (21x - 9)}{15}$

$\frac{x}{2} \ge \frac{25x - 10 - 21x + 9}{15}$

$\frac{x}{2} \ge \frac{(25x - 21x) + (-10 + 9)}{15}$

$\frac{x}{2} \ge \frac{4x - 1}{15}$

Now, we eliminate the denominators by multiplying both sides by the LCM of $2$ and $15$, which is $30$. Since $30$ is positive, the inequality direction remains unchanged.

$30 \times \frac{x}{2} \ge 30 \times \frac{4x - 1}{15}$

$15x \ge 2(4x - 1)$

Distribute the $2$ on the right side:

$15x \ge 8x - 2$

Move the terms involving $x$ to the left side by subtracting $8x$ from both sides:

$15x - 8x \ge 8x - 2 - 8x$

$7x \ge -2$

Divide both sides by $7$ (which is a positive number, so the inequality direction remains unchanged):

$\frac{7x}{7} \ge \frac{-2}{7}$

$x \ge -\frac{2}{7}$

The solution to the inequality $\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$ is $x \ge -\frac{2}{7}$.

Graph on Number Line:

The solution set consists of all real numbers $x$ that are greater than or equal to $-\frac{2}{7}$. To represent this on a number line:

1. Draw a horizontal line for the number line.

2. Locate the point corresponding to $-\frac{2}{7}$. This value is slightly greater than $-1/3$ (since $-2/7 \approx -0.28$).

3. Since the inequality is $x \ge -\frac{2}{7}$, the point $-\frac{2}{7}$ is included in the solution. Mark this point with a closed circle (or filled circle).

4. The inequality $x \ge -\frac{2}{7}$ means all numbers to the right of $-\frac{2}{7}$ are part of the solution. Shade or darken the portion of the number line extending infinitely to the right from the closed circle at $-\frac{2}{7}$.

The graph will show a closed circle at $-\frac{2}{7}$ with the region to its right shaded.

Solution:

Let $x$ be the marks Ravi obtains in the third unit test.

The marks obtained in the first two unit tests are 70 and 75.

The average marks of the three tests is calculated by summing the marks and dividing by the number of tests (which is 3).

Average marks $= \frac{\text{Sum of marks}}{\text{Number of tests}}$

Average marks $= \frac{70 + 75 + x}{3}$

We are given that the average of the three tests must be at least 60 marks. "At least 60" means the average must be greater than or equal to 60.

So, we can write the inequality:

$\frac{70 + 75 + x}{3} \ge 60$

Now, we solve this inequality for $x$:

First, simplify the numerator on the left side:

$70 + 75 = 145$

So, the inequality becomes:

$\frac{145 + x}{3} \ge 60$

Multiply both sides of the inequality by $3$ (which is a positive number, so the inequality sign does not change):

$3 \times \frac{145 + x}{3} \ge 60 \times 3$

$145 + x \ge 180$

Subtract $145$ from both sides of the inequality:

$145 + x - 145 \ge 180 - 145$

Simplifying both sides, we get:

$x \ge 35$

The solution $x \ge 35$ means that the marks obtained in the third test must be 35 or more to have an average of at least 60.

Since marks cannot be less than 0, and the minimum requirement is 35, the minimum marks Ravi should get in the third test is 35.

Conclusion:

The minimum marks Ravi should get in the third test to have an average of at least 60 marks is 35.

Solution:

Let the marks Sunita obtains in the fifth examination be $x$.

The marks obtained in the first four examinations are 87, 92, 94, and 95.

The total marks in the five examinations will be the sum of the marks from the first four exams and the marks from the fifth exam:

Total marks $= 87 + 92 + 94 + 95 + x$

Calculate the sum of the marks from the first four exams:

$87 + 92 + 94 + 95 = 368$

So, the total marks $= 368 + x$.

The average marks in the five examinations is calculated by dividing the total marks by the number of examinations (which is 5).

Average marks $= \frac{\text{Total marks}}{\text{Number of examinations}}$

Average marks $= \frac{368 + x}{5}$

To receive Grade ‘A’, Sunita must obtain an average of 90 marks or more. This means the average marks must be greater than or equal to 90.

We can write this as an inequality:

$\frac{368 + x}{5} \ge 90$

Now, we solve this inequality for $x$:

Multiply both sides of the inequality by $5$ (which is a positive number, so the inequality sign does not change):

$5 \times \frac{368 + x}{5} \ge 90 \times 5$

$368 + x \ge 450$

Subtract $368$ from both sides of the inequality:

$368 + x - 368 \ge 450 - 368$

Simplifying both sides, we get:

$x \ge 82$

The inequality $x \ge 82$ tells us that Sunita must obtain 82 marks or more in the fifth examination to achieve an average of 90 or more.

Since the marks are out of 100, a score of 82 or higher is possible.

The minimum marks required is the smallest value of $x$ that satisfies the inequality, which is 82.

Conclusion:

The minimum marks that Sunita must obtain in the fifth examination to get grade ‘A’ in the course is 82.

Solution:

Let the first consecutive odd positive integer be $x$. Since the integers are consecutive and odd, the next consecutive odd positive integer will be $x+2$.

We are given two conditions for these integers:

1. Both integers must be smaller than 10.

This implies $x < 10$ and $x+2 < 10$.

From $x+2 < 10$, we subtract 2 from both sides:

$x < 10 - 2$

$x < 8$

So, we must have $x < 8$.

Since $x$ must be a positive odd integer, the possible positive odd integers smaller than 8 are 1, 3, 5, and 7.

2. Their sum must be more than 11.

The sum of the two integers is $x + (x+2)$.

The condition is:

Combine like terms:

Subtract 2 from both sides:

Divide both sides by 2 (a positive number, so the inequality sign does not change):

We need to find the positive odd integers $x$ that satisfy both $x < 8$ (from condition 1) and $x > 4.5$ (from inequality (iv)).

The positive odd integers less than 8 are {1, 3, 5, 7}. We check which of these are greater than 4.5.

For $x=1$: $1 \ngtr 4.5$

For $x=3$: $3 \ngtr 4.5$

For $x=5$: $5 > 4.5$. This value is valid for the first integer.

For $x=7$: $7 > 4.5$. This value is valid for the first integer.

Therefore, the possible values for the first integer $x$ are 5 and 7.

Now, we find the pairs $(x, x+2)$ based on these values of $x$:

If $x = 5$, the pair is $(5, 5+2) = (5, 7)$.

If $x = 7$, the pair is $(7, 7+2) = (7, 9)$.

Let's verify these pairs:

Pair (5, 7): Both are consecutive odd positive integers. Both 5 and 7 are less than 10. Their sum is $5+7=12$, which is greater than 11. This pair is valid.

Pair (7, 9): Both are consecutive odd positive integers. Both 7 and 9 are less than 10. Their sum is $7+9=16$, which is greater than 11. This pair is valid.

Conclusion:

The pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 are (5, 7) and (7, 9).

Solution:

Let the first consecutive even positive integer be $x$.

Since the integers are consecutive and even, the next consecutive even positive integer will be $x+2$.

We are given two conditions for these integers:

1. Both integers must be larger than 5.

This implies $x > 5$ and $x+2 > 5$.

From $x+2 > 5$, we subtract 2 from both sides:

$x > 5 - 2$

$x > 3$

The condition $x > 5$ is stronger than $x > 3$. So, we must have $x > 5$.

Since $x$ must be an even positive integer, the possible values for $x$ that are greater than 5 are 6, 8, 10, 12, and so on.

2. Their sum must be less than 23.

The sum of the two integers is $x + (x+2)$.

The condition is:

$x + (x+2) < 23$

Combine like terms:

$2x + 2 < 23$

Subtract 2 from both sides:

$2x + 2 - 2 < 23 - 2$

$2x < 21$

Divide both sides by 2 (a positive number, so the inequality sign does not change):

$\frac{2x}{2} < \frac{21}{2}$

$x < 10.5$

We need to find the even positive integers $x$ that satisfy both $x > 5$ and $x < 10.5$.

The even positive integers greater than 5 are {6, 8, 10, 12, ...}.

The even positive integers less than 10.5 are {..., 6, 8, 10}.

Combining these conditions, the even positive integers $x$ that are greater than 5 and less than 10.5 are 6, 8, and 10.

Now, we form the pairs $(x, x+2)$ based on these possible values of $x$:

If $x = 6$, the pair is $(6, 6+2) = (6, 8)$.

If $x = 8$, the pair is $(8, 8+2) = (8, 10)$.

If $x = 10$, the pair is $(10, 10+2) = (10, 12)$.

Let's verify these pairs:

Pair (6, 8): Both are consecutive even positive integers. Both 6 and 8 are larger than 5. Their sum is $6+8=14$, which is less than 23. This pair is valid.

Pair (8, 10): Both are consecutive even positive integers. Both 8 and 10 are larger than 5. Their sum is $8+10=18$, which is less than 23. This pair is valid.

Pair (10, 12): Both are consecutive even positive integers. Both 10 and 12 are larger than 5. Their sum is $10+12=22$, which is less than 23. This pair is valid.

The next possible value for $x$ would be 12, but $12 \nless 10.5$. So, we have found all possible pairs.

Conclusion:

The pairs of consecutive even positive integers both of which are larger than 5 such that their sum is less than 23 are (6, 8), (8, 10), and (10, 12).

Solution:

Let the length of the shortest side of the triangle be $x$ cm.

Based on the problem description:

The longest side is 3 times the shortest side, so its length is $3x$ cm.

The third side is 2 cm shorter than the longest side, so its length is $(3x - 2)$ cm.

For these lengths to form a valid triangle, they must all be positive, and the triangle inequality must hold (the sum of any two sides must be greater than the third side).

Lengths must be positive:

$x > 0$

$3x > 0 \implies x > 0$

$3x - 2 > 0 \implies 3x > 2 \implies x > \frac{2}{3}$

Triangle inequality:

$x + 3x > 3x - 2 \implies 4x > 3x - 2 \implies x > -2$. (Already covered by $x > 2/3$)

$x + (3x - 2) > 3x \implies 4x - 2 > 3x \implies x > 2$.

$3x + (3x - 2) > x \implies 6x - 2 > x \implies 5x > 2 \implies x > \frac{2}{5}$. (Already covered by $x > 2$)

So, for a valid triangle, $x$ must be greater than 2.

The perimeter of the triangle is the sum of the lengths of its three sides:

Perimeter $= x + 3x + (3x - 2)$

Perimeter $= (x + 3x + 3x) - 2$

Perimeter $= 7x - 2$ cm

We are given that the perimeter of the triangle is at least 61 cm. "At least 61 cm" means the perimeter must be greater than or equal to 61 cm.

We can write this as an inequality:

$7x - 2 \ge 61$

Now, we solve this inequality for $x$:

Add 2 to both sides of the inequality:

$7x - 2 + 2 \ge 61 + 2$

$7x \ge 63$

Divide both sides by 7 (which is a positive number, so the inequality sign does not change):

$\frac{7x}{7} \ge \frac{63}{7}$

$x \ge 9$

This inequality $x \ge 9$ tells us that the length of the shortest side must be greater than or equal to 9 cm.

We also established that for a valid triangle, $x > 2$. The condition $x \ge 9$ satisfies $x > 2$.

The question asks for the minimum length of the shortest side.

The minimum value that satisfies $x \ge 9$ is 9.

Conclusion:

The minimum length of the shortest side of the triangle is 9 cm.

Let's verify the side lengths for $x=9$ cm:

Shortest side = 9 cm

Longest side = $3 \times 9 = 27$ cm

Third side = $3 \times 9 - 2 = 27 - 2 = 25$ cm

These lengths (9, 27, 25) form a valid triangle (9+25 > 27, 9+27 > 25, 25+27 > 9).

Perimeter = $9 + 27 + 25 = 61$ cm, which is "at least 61 cm".

Solution:

Let the length of the shortest piece of board be $x$ cm.

Based on the problem statement, the lengths of the three pieces are:

Shortest piece: $x$ cm

Second piece: $3$ cm longer than the shortest, so $(x + 3)$ cm

Third piece: twice as long as the shortest, so $2x$ cm

Since these are lengths, they must be positive:

$x > 0$

$x + 3 > 0 \implies x > -3$

$2x > 0 \implies x > 0$

The condition $x > 0$ satisfies $x > -3$. So, we must have $x > 0$.

We are given two main conditions:

1. The total length of the three pieces cut from a 91 cm board cannot exceed the length of the board. The sum of the lengths must be less than or equal to 91 cm.

$x + (x + 3) + 2x \le 91$

2. The third piece is to be at least 5 cm longer than the second piece.

$2x \ge (x + 3) + 5$

Let's solve the first inequality:

$x + x + 3 + 2x \le 91$

Combine like terms:

$4x + 3 \le 91$

Subtract 3 from both sides:

$4x \le 91 - 3$

$4x \le 88$

Divide both sides by 4 (a positive number):

$\frac{4x}{4} \le \frac{88}{4}$

$x \le 22$

Now let's solve the second inequality:

$2x \ge x + 3 + 5$

Combine constants on the right side:

$2x \ge x + 8$

Subtract $x$ from both sides:

$2x - x \ge 8$

$x \ge 8$

We also have the condition that the length must be positive:

We need to find the values of $x$ that satisfy all three conditions: $x \le 22$, $x \ge 8$, and $x > 0$.

Combining $x \ge 8$ and $x > 0$, we get $x \ge 8$.

Combining $x \ge 8$ and $x \le 22$, we get $8 \le x \le 22$.

The possible lengths of the shortest board are in the interval $[8, 22]$.

Conclusion:

The possible lengths of the shortest board range from 8 cm to 22 cm, inclusive. That is, the length $x$ must satisfy $8 \le x \le 22$.

Solution:

We are asked to solve the linear inequality $3x + 2y > 6$ graphically.

Steps for Graphical Solution:

1. Consider the corresponding linear equation:

$3x + 2y = 6$

This equation represents the boundary line for the inequality.

2. Find at least two points on the line $3x + 2y = 6$ to draw it.

Let $x = 0$: $3(0) + 2y = 6 \implies 2y = 6 \implies y = 3$. The point is $(0, 3)$.

Let $y = 0$: $3x + 2(0) = 6 \implies 3x = 6 \implies x = 2$. The point is $(2, 0)$.

3. Determine the type of boundary line.

The inequality is $3x + 2y > 6$. Since it is a strict inequality ($>$), the points on the line $3x + 2y = 6$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

4. Choose a test point.

A common and convenient test point is the origin $(0, 0)$, provided it does not lie on the boundary line. In this case, $3(0) + 2(0) = 0 \ne 6$, so $(0, 0)$ is not on the line.

5. Substitute the test point $(0, 0)$ into the original inequality $3x + 2y > 6$.

$3(0) + 2(0) > 6$

$0 + 0 > 6$

$0 > 6$

6. Evaluate the result of the test.

The statement $0 > 6$ is false.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a false statement, the solution region is the half-plane that does not contain the origin $(0, 0)$.

The origin $(0, 0)$ is below the line connecting $(0, 3)$ and $(2, 0)$. Therefore, the solution region is the area above the dashed line $3x + 2y = 6$.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 3)$ and $(2, 0)$.

3. Draw a dashed line passing through these two points. This is the boundary line $3x + 2y = 6$.

4. Shade the region above (or to the right of) this dashed line. This shaded region represents all the points $(x, y)$ that satisfy the inequality $3x + 2y > 6$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $3x - 6 \ge 0$ graphically in a two-dimensional plane (which is the Cartesian plane with x and y axes).

The inequality involves only the variable $x$. We can first simplify the inequality:

$3x - 6 \ge 0$

Add 6 to both sides:

$3x \ge 6$

Divide by 3 (a positive number, inequality sign remains the same):

$x \ge 2$

In a two-dimensional plane, the solution to $x \ge 2$ consists of all points $(x, y)$ where the x-coordinate is greater than or equal to 2. The y-coordinate can be any real number.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$x = 2$

This is the equation of a vertical line passing through the point $(2, 0)$ and parallel to the y-axis.

2. Determine the type of boundary line.

The inequality is $x \ge 2$. Since it includes the equals sign ($\ge$), the points on the line $x = 2$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

3. Choose a test point.

A convenient test point is the origin $(0, 0)$, since it does not lie on the line $x=2$ (because the x-coordinate of the origin is 0, and $0 \ne 2$).

4. Substitute the test point $(0, 0)$ into the simplified inequality $x \ge 2$ (or the original $3x - 6 \ge 0$).

Using $x \ge 2$, substitute $x=0$:

$0 \ge 2$

Using $3x - 6 \ge 0$, substitute $(0, 0)$ (only x matters):

$3(0) - 6 \ge 0$

$0 - 6 \ge 0$

$-6 \ge 0$

5. Evaluate the result of the test.

The statement $0 \ge 2$ is false.

The statement $-6 \ge 0$ is also false.

6. Determine the solution region.

Since the test point $(0, 0)$ (which has $x=0$) resulted in a false statement, the solution region is the half-plane that does not contain the origin $(0, 0)$.

The line $x=2$ divides the plane into two half-planes: one where $x < 2$ (which contains the origin) and one where $x > 2$. Since the half-plane containing the origin is not the solution, the solution region is the other half-plane, where $x > 2$. Including the boundary $x=2$ because of the $\ge$ sign, the solution region is where $x \ge 2$.

Graphically, this is the region to the right of the line $x=2$, including the solid line itself.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Locate the point on the x-axis where $x=2$.

3. Draw a solid vertical line passing through $x=2$. This is the boundary line $x = 2$.

4. Shade the region to the right of this solid line. This shaded region represents all the points $(x, y)$ that satisfy the inequality $3x - 6 \ge 0$ (or $x \ge 2$).

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $y < 2$ graphically in a two-dimensional plane (Cartesian plane).

The inequality involves only the variable $y$. It represents all points $(x, y)$ where the y-coordinate is strictly less than 2. The x-coordinate can be any real number.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$y = 2$

This is the equation of a horizontal line where the y-coordinate of every point is 2. It passes through points like $(0, 2), (5, 2), (-3, 2)$, etc. It is parallel to the x-axis.

2. Determine the type of boundary line.

The inequality is $y < 2$. Since it is a strict inequality ($<$), the points on the line $y = 2$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

3. Choose a test point.

The origin $(0, 0)$ is a convenient test point, as it does not lie on the line $y=2$ (since the y-coordinate is 0, and $0 \ne 2$).

4. Substitute the test point $(0, 0)$ into the original inequality $y < 2$.

Substitute $y=0$:

$0 < 2$

5. Evaluate the result of the test.

The statement $0 < 2$ is true.

6. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The horizontal line $y=2$ divides the plane into two half-planes: one where $y < 2$ (which contains the origin) and one where $y > 2$. Since the half-plane containing the origin is the solution, the solution region is where $y < 2$.

Graphically, this is the region below the dashed line $y=2$.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Locate the point on the y-axis where $y=2$.

3. Draw a dashed horizontal line passing through $y=2$. This is the boundary line $y = 2$.

4. Shade the region below this dashed line. This shaded region represents all the points $(x, y)$ that satisfy the inequality $y < 2$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $x + y < 5$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation:

$x + y = 5$

This equation represents the boundary line for the inequality.

2. Find at least two points on the line $x + y = 5$ to draw it.

If $x = 0$, $0 + y = 5 \implies y = 5$. The point is $(0, 5)$.

If $y = 0$, $x + 0 = 5 \implies x = 5$. The point is $(5, 0)$.

3. Determine the type of boundary line.

The inequality is $x + y < 5$. Since it is a strict inequality ($<$), the points on the line $x + y = 5$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $x + y = 5$ (since $0 + 0 = 0 \ne 5$).

5. Substitute the test point $(0, 0)$ into the original inequality $x + y < 5$.

$0 + 0 < 5$

$0 < 5$

6. Evaluate the result of the test.

The statement $0 < 5$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $x + y = 5$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 5)$ and $(5, 0)$.

3. Draw a dashed line passing through these two points. This is the boundary line $x + y = 5$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area below the dashed line $x + y = 5$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $x + y < 5$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $2x + y \ge 6$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$2x + y = 6$

2. Find at least two points on the line $2x + y = 6$ to draw it.

If $x = 0$, $2(0) + y = 6 \implies y = 6$. The point is $(0, 6)$.

If $y = 0$, $2x + 0 = 6 \implies 2x = 6 \implies x = 3$. The point is $(3, 0)$.

3. Determine the type of boundary line.

The inequality is $2x + y \ge 6$. Since it includes the equals sign ($\ge$), the points on the line $2x + y = 6$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $2x + y = 6$ (since $2(0) + 0 = 0 \ne 6$).

5. Substitute the test point $(0, 0)$ into the original inequality $2x + y \ge 6$.

$2(0) + 0 \ge 6$

$0 \ge 6$

6. Evaluate the result of the test.

The statement $0 \ge 6$ is false.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a false statement, the solution region is the half-plane that does not contain the origin $(0, 0)$.

The line $2x + y = 6$ divides the plane into two half-planes. The half-plane that does not contain the origin is the solution region. This region is above or to the right of the line.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 6)$ and $(3, 0)$.

3. Draw a solid line passing through these two points. This is the boundary line $2x + y = 6$.

4. Shade the region that does not contain the origin $(0, 0)$. This will be the area above or to the right of the solid line $2x + y = 6$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $2x + y \ge 6$, including the points on the line.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $3x + 4y \le 12$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$3x + 4y = 12$

2. Find at least two points on the line $3x + 4y = 12$ to draw it.

If $x = 0$: $3(0) + 4y = 12 \implies 4y = 12 \implies y = 3$. Point: $(0, 3)$.

If $y = 0$: $3x + 4(0) = 12 \implies 3x = 12 \implies x = 4$. Point: $(4, 0)$.

3. Determine the type of boundary line.

The inequality is $3x + 4y \le 12$. Since it includes the equals sign ($\le$), the points on the line $3x + 4y = 12$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $3x + 4y = 12$ (since $3(0) + 4(0) = 0 \ne 12$).

5. Substitute the test point $(0, 0)$ into the original inequality $3x + 4y \le 12$.

$3(0) + 4(0) \le 12$

$0 \le 12$

6. Evaluate the result of the test.

The statement $0 \le 12$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $3x + 4y = 12$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 3)$ and $(4, 0)$.

3. Draw a solid line passing through these two points. This is the boundary line $3x + 4y = 12$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area below or to the left of the solid line $3x + 4y = 12$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $3x + 4y \le 12$, including the points on the line.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $y + 8 \ge 2x$ graphically in a two-dimensional plane.

We can rewrite the inequality as $y - 2x \ge -8$ or $2x - y \le 8$. Let's work with $y - 2x \ge -8$.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$y - 2x = -8$

or equivalently,

$y = 2x - 8$

2. Find at least two points on the line $y = 2x - 8$ to draw it.

If $x = 0$, $y = 2(0) - 8 = -8$. The point is $(0, -8)$.

If $y = 0$, $0 = 2x - 8 \implies 2x = 8 \implies x = 4$. The point is $(4, 0)$.

3. Determine the type of boundary line.

The inequality is $y + 8 \ge 2x$, which is $y - 2x \ge -8$. Since it includes the equals sign ($\ge$), the points on the line $y - 2x = -8$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $y - 2x = -8$ (since $0 - 2(0) = 0 \ne -8$).

5. Substitute the test point $(0, 0)$ into the original inequality $y + 8 \ge 2x$.

$0 + 8 \ge 2(0)$

$8 \ge 0$

6. Evaluate the result of the test.

The statement $8 \ge 0$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $y = 2x - 8$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, -8)$ and $(4, 0)$.

3. Draw a solid line passing through these two points. This is the boundary line $y + 8 = 2x$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area above or to the left of the solid line $y + 8 = 2x$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $y + 8 \ge 2x$, including the points on the line.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $x - y \le 2$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$x - y = 2$

2. Find at least two points on the line $x - y = 2$ to draw it.

If $x = 0$, $0 - y = 2 \implies -y = 2 \implies y = -2$. The point is $(0, -2)$.

If $y = 0$, $x - 0 = 2 \implies x = 2$. The point is $(2, 0)$.

3. Determine the type of boundary line.

The inequality is $x - y \le 2$. Since it includes the equals sign ($\le$), the points on the line $x - y = 2$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $x - y = 2$ (since $0 - 0 = 0 \ne 2$).

5. Substitute the test point $(0, 0)$ into the original inequality $x - y \le 2$.

$0 - 0 \le 2$

$0 \le 2$

6. Evaluate the result of the test.

The statement $0 \le 2$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $x - y = 2$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, -2)$ and $(2, 0)$.

3. Draw a solid line passing through these two points. This is the boundary line $x - y = 2$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area above or to the left of the solid line $x - y = 2$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $x - y \le 2$, including the points on the line.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $2x - 3y > 6$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$2x - 3y = 6$

2. Find at least two points on the line $2x - 3y = 6$ to draw it.

If $x = 0$, $2(0) - 3y = 6 \implies -3y = 6 \implies y = -2$. The point is $(0, -2)$.

If $y = 0$, $2x - 3(0) = 6 \implies 2x = 6 \implies x = 3$. The point is $(3, 0)$.

3. Determine the type of boundary line.

The inequality is $2x - 3y > 6$. Since it is a strict inequality ($>$), the points on the line $2x - 3y = 6$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $2x - 3y = 6$ (since $2(0) - 3(0) = 0 \ne 6$).

5. Substitute the test point $(0, 0)$ into the original inequality $2x - 3y > 6$.

$2(0) - 3(0) > 6$

$0 > 6$

6. Evaluate the result of the test.

The statement $0 > 6$ is false.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a false statement, the solution region is the half-plane that does not contain the origin $(0, 0)$.

The line $2x - 3y = 6$ divides the plane into two half-planes. The half-plane that does not contain the origin is the solution region. This region is below the line.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, -2)$ and $(3, 0)$.

3. Draw a dashed line passing through these two points. This is the boundary line $2x - 3y = 6$.

4. Shade the region that does not contain the origin $(0, 0)$. This will be the area below the dashed line $2x - 3y = 6$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $2x - 3y > 6$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $-3x + 2y \ge -6$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$-3x + 2y = -6$

2. Find at least two points on the line $-3x + 2y = -6$ to draw it.

If $x = 0$: $-3(0) + 2y = -6 \implies 2y = -6 \implies y = -3$. The point is $(0, -3)$.

If $y = 0$: $-3x + 2(0) = -6 \implies -3x = -6 \implies x = 2$. The point is $(2, 0)$.

3. Determine the type of boundary line.

The inequality is $-3x + 2y \ge -6$. Since it includes the equals sign ($\ge$), the points on the line $-3x + 2y = -6$ are included in the solution set. Therefore, the boundary line should be drawn as a solid line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $-3x + 2y = -6$ (since $-3(0) + 2(0) = 0 \ne -6$).

5. Substitute the test point $(0, 0)$ into the original inequality $-3x + 2y \ge -6$.

$-3(0) + 2(0) \ge -6$

$0 \ge -6$

6. Evaluate the result of the test.

The statement $0 \ge -6$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $-3x + 2y = -6$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, -3)$ and $(2, 0)$.

3. Draw a solid line passing through these two points. This is the boundary line $-3x + 2y = -6$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area above or to the left of the solid line $-3x + 2y = -6$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $-3x + 2y \ge -6$, including the points on the line.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $3y - 5x < 30$ graphically in a two-dimensional plane.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$3y - 5x = 30$

2. Find at least two points on the line $3y - 5x = 30$ to draw it.

If $x = 0$: $3y - 5(0) = 30 \implies 3y = 30 \implies y = 10$. The point is $(0, 10)$.

If $y = 0$: $3(0) - 5x = 30 \implies -5x = 30 \implies x = -6$. The point is $(-6, 0)$.

3. Determine the type of boundary line.

The inequality is $3y - 5x < 30$. Since it is a strict inequality ($<$), the points on the line $3y - 5x = 30$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

4. Choose a test point.

The origin $(0, 0)$ is a convenient test point because it is not on the line $3y - 5x = 30$ (since $3(0) - 5(0) = 0 \ne 30$).

5. Substitute the test point $(0, 0)$ into the original inequality $3y - 5x < 30$.

$3(0) - 5(0) < 30$

$0 < 30$

6. Evaluate the result of the test.

The statement $0 < 30$ is true.

7. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The line $3y - 5x = 30$ divides the plane into two half-planes. The half-plane containing the origin is the solution region.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 10)$ and $(-6, 0)$.

3. Draw a dashed line passing through these two points. This is the boundary line $3y - 5x = 30$.

4. Shade the region that contains the origin $(0, 0)$. This will be the area below or to the right of the dashed line $3y - 5x = 30$. This shaded region represents all the points $(x, y)$ that satisfy the inequality $3y - 5x < 30$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $y < -2$ graphically in a two-dimensional plane.

The inequality involves only the variable $y$. It represents all points $(x, y)$ where the y-coordinate is strictly less than -2. The x-coordinate can be any real number.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$y = -2$

This is the equation of a horizontal line where the y-coordinate of every point is -2. It passes through points like $(0, -2), (5, -2), (-3, -2)$, etc. It is parallel to the x-axis.

2. Determine the type of boundary line.

The inequality is $y < -2$. Since it is a strict inequality ($<$), the points on the line $y = -2$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

3. Choose a test point.

The origin $(0, 0)$ is a convenient test point, as it does not lie on the line $y=-2$ (since the y-coordinate is 0, and $0 \ne -2$).

4. Substitute the test point $(0, 0)$ into the original inequality $y < -2$.

Substitute $y=0$:

$0 < -2$

5. Evaluate the result of the test.

The statement $0 < -2$ is false.

6. Determine the solution region.

Since the test point $(0, 0)$ resulted in a false statement, the solution region is the half-plane that does not contain the origin $(0, 0)$.

The horizontal line $y=-2$ divides the plane into two half-planes: one where $y < -2$ and one where $y > -2$. The origin $(0,0)$ is in the $y > -2$ region. Since the origin side is false, the solution region is the other side, where $y < -2$.

Graphically, this is the region below the dashed line $y=-2$.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Locate the point on the y-axis where $y=-2$.

3. Draw a dashed horizontal line passing through $y=-2$. This is the boundary line $y = -2$.

4. Shade the region below this dashed line. This shaded region represents all the points $(x, y)$ that satisfy the inequality $y < -2$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the linear inequality $x > -3$ graphically in a two-dimensional plane.

The inequality involves only the variable $x$. It represents all points $(x, y)$ where the x-coordinate is strictly greater than -3. The y-coordinate can be any real number.

Steps for Graphical Solution:

1. Consider the corresponding linear equation, which represents the boundary line:

$x = -3$

This is the equation of a vertical line where the x-coordinate of every point is -3. It passes through points like $(-3, 0), (-3, 5), (-3, -1)$, etc. It is parallel to the y-axis.

2. Determine the type of boundary line.

The inequality is $x > -3$. Since it is a strict inequality ($>$), the points on the line $x = -3$ are not included in the solution set. Therefore, the boundary line should be drawn as a dashed or dotted line.

3. Choose a test point.

The origin $(0, 0)$ is a convenient test point, as it does not lie on the line $x=-3$ (since the x-coordinate is 0, and $0 \ne -3$).

4. Substitute the test point $(0, 0)$ into the original inequality $x > -3$.

Substitute $x=0$:

$0 > -3$

5. Evaluate the result of the test.

The statement $0 > -3$ is true.

6. Determine the solution region.

Since the test point $(0, 0)$ resulted in a true statement, the solution region is the half-plane that contains the origin $(0, 0)$.

The vertical line $x=-3$ divides the plane into two half-planes: one where $x > -3$ (which contains the origin) and one where $x < -3$. Since the half-plane containing the origin is the solution, the solution region is where $x > -3$.

Graphically, this is the region to the right of the dashed line $x=-3$.

Graph:

To graphically represent the solution:

1. Draw the x-axis and y-axis.

2. Locate the point on the x-axis where $x=-3$.

3. Draw a dashed vertical line passing through $x=-3$. This is the boundary line $x = -3$.

4. Shade the region to the right of this dashed line. This shaded region represents all the points $(x, y)$ that satisfy the inequality $x > -3$.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

Solution:

We are asked to solve the following system of linear inequalities graphically:

$x + y \ge 5$

$x - y \le 3$

Steps for Graphical Solution:

We will graph each inequality separately and then find the region where their solutions overlap.

For the first inequality: $x + y \ge 5$

1. Consider the boundary line: $x + y = 5$.

2. Find two points on this line:

If $x = 0$, then $0 + y = 5 \implies y = 5$. Point: $(0, 5)$.

If $y = 0$, then $x + 0 = 5 \implies x = 5$. Point: $(5, 0)$.

3. Determine the line type: Since the inequality is $\ge$, the line is solid.

4. Choose a test point: $(0, 0)$ (not on the line).

5. Test the point $(0, 0)$ in $x + y \ge 5$: $0 + 0 \ge 5 \implies 0 \ge 5$, which is false.

6. Shade the region: The solution region for $x + y \ge 5$ is the half-plane not containing the origin $(0, 0)$, i.e., the region above or to the right of the line $x+y=5$.

For the second inequality: $x - y \le 3$

1. Consider the boundary line: $x - y = 3$.

2. Find two points on this line:

If $x = 0$, then $0 - y = 3 \implies y = -3$. Point: $(0, -3)$.

If $y = 0$, then $x - 0 = 3 \implies x = 3$. Point: $(3, 0)$.

3. Determine the line type: Since the inequality is $\le$, the line is solid.

4. Choose a test point: $(0, 0)$ (not on the line).

5. Test the point $(0, 0)$ in $x - y \le 3$: $0 - 0 \le 3 \implies 0 \le 3$, which is true.

6. Shade the region: The solution region for $x - y \le 3$ is the half-plane containing the origin $(0, 0)$, i.e., the region above or to the left of the line $x-y=3$.

Graph of the System:

To graphically represent the solution of the system:

1. Draw the x-axis and y-axis.

2. Plot the points $(0, 5)$ and $(5, 0)$ and draw a solid line through them for $x + y = 5$.

3. Plot the points $(0, -3)$ and $(3, 0)$ and draw a solid line through them for $x - y = 3$.

4. The solution for $x + y \ge 5$ is the area above/right of the line $x+y=5$.

5. The solution for $x - y \le 3$ is the area above/left of the line $x-y=3$.

6. The solution to the system of inequalities is the region where these two shaded areas overlap. This is the region that is simultaneously above/right of $x+y=5$ and above/left of $x-y=3$.

The intersection point of the two boundary lines can be found by solving the system of equations $x+y=5$ and $x-y=3$. Adding the two equations gives $2x = 8 \implies x = 4$. Substituting $x=4$ into $x+y=5$ gives $4+y=5 \implies y=1$. So the lines intersect at $(4, 1)$. This point is part of the solution region because both lines are solid.

The feasible region is unbounded and lies above the intersection point $(4,1)$, bounded by the two solid lines.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation. The final solution region is the area above the line $x+y=5$ and to the left of the line $x-y=3$, including the boundary lines and their intersection point $(4,1)$.)

Solution:

We are asked to solve the following system of linear inequalities graphically:

Steps for Graphical Solution:

To solve the system graphically, we will graph the solution region for each inequality individually on the same coordinate plane. The solution to the system is the region where all these individual solution regions overlap.

For the first inequality: $5x + 4y \le 40$

1. Consider the boundary line: $5x + 4y = 40$.

2. To draw this line, find two points on it:

If $x = 0$, $5(0) + 4y = 40 \implies 4y = 40 \implies y = 10$. Point: $(0, 10)$.

If $y = 0$, $5x + 4(0) = 40 \implies 5x = 40 \implies x = 8$. Point: $(8, 0)$.

3. The inequality is $\le$, so the boundary line is solid and is included in the solution region.

4. Choose a test point, e.g., the origin $(0, 0)$, which is not on the line $5x+4y=40$.

5. Substitute $(0, 0)$ into $5x + 4y \le 40$: $5(0) + 4(0) \le 40 \implies 0 \le 40$, which is true.

6. The solution region for $5x + 4y \le 40$ is the half-plane that contains the origin $(0, 0)$. This is the region below or to the left of the line $5x + 4y = 40$.

For the second inequality: $x \ge 2$

1. Consider the boundary line: $x = 2$.

2. This is a vertical line passing through $x=2$ on the x-axis. Points on this line are $(2, y)$ for any $y$. Examples: $(2, 0), (2, 3)$.

3. The inequality is $\ge$, so the boundary line is solid and is included in the solution region.

4. Choose a test point, e.g., the origin $(0, 0)$, which is not on the line $x=2$.

5. Substitute $x=0$ into $x \ge 2$: $0 \ge 2$, which is false.

6. The solution region for $x \ge 2$ is the half-plane that does not contain the origin $(0, 0)$. This is the region to the right of the line $x = 2$.

For the third inequality: $y \ge 3$

1. Consider the boundary line: $y = 3$.

2. This is a horizontal line passing through $y=3$ on the y-axis. Points on this line are $(x, 3)$ for any $x$. Examples: $(0, 3), (2, 3)$.

3. The inequality is $\ge$, so the boundary line is solid and is included in the solution region.

4. Choose a test point, e.g., the origin $(0, 0)$, which is not on the line $y=3$.

5. Substitute $y=0$ into $y \ge 3$: $0 \ge 3$, which is false.

6. The solution region for $y \ge 3$ is the half-plane that does not contain the origin $(0, 0)$. This is the region above the line $y = 3$.

Graph of the System:

To graphically represent the solution of the system:

1. Draw the x-axis and y-axis.

2. Draw the solid line $5x + 4y = 40$ through $(0, 10)$ and $(8, 0)$. The solution for this inequality is below/left of this line.

3. Draw the solid vertical line $x = 2$. The solution for this inequality is to the right of this line.

4. Draw the solid horizontal line $y = 3$. The solution for this inequality is above this line.

5. The feasible region for the system is the intersection of these three regions. It is the region that is simultaneously below/left of $5x+4y=40$, to the right of $x=2$, and above $y=3$.

This region is a triangle bounded by the three lines. Its vertices are the intersection points of the boundary lines:

• Intersection of $x=2$ and $y=3$: $(2, 3)$.
• Intersection of $x=2$ and $5x+4y=40$: $5(2) + 4y = 40 \implies 10 + 4y = 40 \implies 4y = 30 \implies y = 7.5$. Point: $(2, 7.5)$.
• Intersection of $y=3$ and $5x+4y=40$: $5x + 4(3) = 40 \implies 5x + 12 = 40 \implies 5x = 28 \implies x = 5.6$. Point: $(5.6, 3)$.

6. Shade the triangular region with vertices $(2, 3)$, $(2, 7.5)$, and $(5.6, 3)$. This shaded region, including its boundaries, represents the solution set for the given system of inequalities.

(Note: A visual graph is required for a complete answer, but cannot be generated here. The description above guides its creation.)

We need to solve the given system of inequalities graphically.

First, consider the linear inequality $8x + 3y \leq 100$. To graph this inequality, we first plot the graph of the corresponding linear equation:

To draw the line (A), we find two points on the line.

Plot the points $(0, \frac{100}{3})$, $(12.5, 0)$ and $(10, \frac{20}{3})$ and draw a solid line passing through them (since the inequality is $\leq$).

Next, we determine the region that satisfies the inequality $8x + 3y \leq 100$. We can use a test point not on the line $8x + 3y = 100$. Let's use the origin $(0, 0)$.

This statement is true. Therefore, the region satisfying $8x + 3y \leq 100$ is the half-plane containing the origin.

Now, consider the inequality $x \geq 0$. This inequality represents the region to the right of the y-axis, including the y-axis itself.

Next, consider the inequality $y \geq 0$. This inequality represents the region above the x-axis, including the x-axis itself.

The inequalities $x \geq 0$ and $y \geq 0$ together represent the first quadrant of the coordinate plane, including the positive x-axis and positive y-axis.

The solution to the system of inequalities is the intersection of the regions represented by inequality (1), inequality (2), and inequality (3).

This intersection is the region in the first quadrant that lies on or below the line $8x + 3y = 100$.

The graphical representation of the solution is the polygonal region bounded by the lines $x=0$, $y=0$, and $8x + 3y = 100$ in the first quadrant. This region is the feasible region.

The vertices of this feasible region are the points of intersection of the boundary lines: $(0, 0)$, the intersection of $y=0$ and $8x+3y=100$ which is $(12.5, 0)$, and the intersection of $x=0$ and $8x+3y=100$ which is $(0, \frac{100}{3})$.

We are asked to solve the following system of inequalities graphically:

Consider the inequality $x + 2y \leq 8$. The corresponding linear equation is $x + 2y = 8$.

To plot the line $x + 2y = 8$, we find two points on the line:

Plot the points $(0, 4)$ and $(8, 0)$ and draw a solid line through them. To determine the region for $x + 2y \leq 8$, test the origin $(0, 0)$: $0 + 2(0) \leq 8 \Rightarrow 0 \leq 8$. This is true, so the region is the half-plane containing the origin.

Consider the inequality $2x + y \leq 8$. The corresponding linear equation is $2x + y = 8$.

To plot the line $2x + y = 8$, we find two points on the line:

Plot the points $(0, 8)$ and $(4, 0)$ and draw a solid line through them. To determine the region for $2x + y \leq 8$, test the origin $(0, 0)$: $2(0) + 0 \leq 8 \Rightarrow 0 \leq 8$. This is true, so the region is the half-plane containing the origin.

Consider the inequalities $x > 0$ and $y > 0$.

$x > 0$ represents the region to the right of the y-axis, excluding the y-axis itself.

$y > 0$ represents the region above the x-axis, excluding the x-axis itself.

Together, $x > 0$ and $y > 0$ represent the interior of the first quadrant (i.e., the first quadrant excluding the positive x-axis and the positive y-axis).

The solution to the system of inequalities is the intersection of the regions represented by (1), (2), (3), and (4). This is the region within the first quadrant (excluding the axes) that is on or below the line $x + 2y = 8$ and on or below the line $2x + y = 8$.

To find the vertices of the feasible region, we find the intersection points of the boundary lines:

1. Intersection of $y=0$ and $2x+y=8$: $2x + 0 = 8 \Rightarrow x=4$. Point $(4,0)$.

2. Intersection of $x=0$ and $x+2y=8$: $0 + 2y = 8 \Rightarrow y=4$. Point $(0,4)$.

3. Intersection of $x+2y=8$ and $2x+y=8$. Multiply the first equation by 2:

Subtract the second equation $2x + y = 8$ from equation (A):

Substitute $y = \frac{8}{3}$ into $x + 2y = 8$:

The intersection point is $(\frac{8}{3}, \frac{8}{3})$.

The feasible region is the open polygonal region in the first quadrant bounded by the lines $x=0$, $y=0$, $x+2y=8$, and $2x+y=8$. The boundary segments on the lines $x+2y=8$ and $2x+y=8$ are included, but the boundary segments on the axes $x=0$ and $y=0$ are excluded due to the strict inequalities $x > 0$ and $y > 0$. The vertices of the region defined by $x+2y \le 8$, $2x+y \le 8$, $x \ge 0$, $y \ge 0$ are $(0,0)$, $(4,0)$, $(\frac{8}{3}, \frac{8}{3})$, and $(0,4)$. The solution region is the interior of this polygon, excluding the parts on the x and y axes.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $x \geq 3$. To graph this, we first consider the corresponding linear equation $x = 3$.

The graph of $x = 3$ is a vertical line parallel to the y-axis, passing through the point $(3, 0)$ on the x-axis.

Since the inequality is $x \geq 3$, the region satisfying this inequality is the half-plane to the right of the line $x = 3$, including the line itself.

Now, consider the second inequality $y \geq 2$. To graph this, we first consider the corresponding linear equation $y = 2$.

The graph of $y = 2$ is a horizontal line parallel to the x-axis, passing through the point $(0, 2)$ on the y-axis.

Since the inequality is $y \geq 2$, the region satisfying this inequality is the half-plane above the line $y = 2$, including the line itself.

The solution to the system of inequalities is the intersection of the regions represented by inequality (1) and inequality (2).

This intersection is the region where $x \geq 3$ and $y \geq 2$ simultaneously.

Graphically, this is the region in the Cartesian plane that is to the right of or on the vertical line $x=3$ and above or on the horizontal line $y=2$.

The feasible region is the unbounded region in the first quadrant bounded by the lines $x=3$ and $y=2$. The lines $x=3$ and $y=2$ are part of the solution region.

The vertex of this feasible region is the point of intersection of the lines $x=3$ and $y=2$, which is $(3, 2)$.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $3x + 2y \leq 12$. The corresponding linear equation is $3x + 2y = 12$.

To plot the line $3x + 2y = 12$, we find two points on the line:

Plot the points $(0, 6)$, $(4, 0)$ and $(2, 3)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $3x + 2y \leq 12$, test the origin $(0, 0)$: $3(0) + 2(0) \leq 12 \Rightarrow 0 \leq 12$. This is true, so the region is the half-plane containing the origin.

Consider the second inequality $x \geq 1$. The corresponding linear equation is $x = 1$.

The graph of $x = 1$ is a vertical line parallel to the y-axis, passing through the point $(1, 0)$. Draw a solid line (since the inequality is $\geq$).

The region satisfying $x \geq 1$ is the half-plane to the right of the line $x = 1$, including the line itself.

Consider the third inequality $y \geq 2$. The corresponding linear equation is $y = 2$.

The graph of $y = 2$ is a horizontal line parallel to the x-axis, passing through the point $(0, 2)$. Draw a solid line (since the inequality is $\geq$).

The region satisfying $y \geq 2$ is the half-plane above the line $y = 2$, including the line itself.

The solution to the system of inequalities is the intersection of the three regions: $3x + 2y \leq 12$, $x \geq 1$, and $y \geq 2$.

This is the region that is:

• On or below the line $3x + 2y = 12$.
• On or to the right of the line $x = 1$.
• On or above the line $y = 2$.

The graphical representation of the solution is the feasible region bounded by the lines $x=1$, $y=2$, and $3x + 2y = 12$. This is a closed polygonal region.

To find the vertices of this feasible region, we find the points of intersection of the boundary lines:

1. Intersection of $x = 1$ and $y = 2$: The point is $(1, 2)$.

2. Intersection of $x = 1$ and $3x + 2y = 12$. Substitute $x = 1$ into the equation:

The intersection point is $(1, 4.5)$ or $(1, \frac{9}{2})$.

3. Intersection of $y = 2$ and $3x + 2y = 12$. Substitute $y = 2$ into the equation:

The intersection point is $(\frac{8}{3}, 2)$.

The vertices of the feasible region are $(1, 2)$, $(1, \frac{9}{2})$, and $(\frac{8}{3}, 2)$.

The feasible region is the triangle with these vertices, including the boundary lines.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $2x + y \geq 6$. The corresponding linear equation is $2x + y = 6$.

To plot the line $2x + y = 6$, we find two points on the line:

Plot the points $(0, 6)$ and $(3, 0)$ (and $(1, 4)$ as a check) and draw a solid line through them (since the inequality is $\geq$). To determine the region for $2x + y \geq 6$, test the origin $(0, 0)$: $2(0) + 0 \geq 6 \Rightarrow 0 \geq 6$. This is false, so the region is the half-plane not containing the origin.

Consider the second inequality $3x + 4y \leq 12$. The corresponding linear equation is $3x + 4y = 12$.

To plot the line $3x + 4y = 12$, we find two points on the line:

Plot the points $(0, 3)$ and $(4, 0)$ (and $(2, 1.5)$ as a check) and draw a solid line through them (since the inequality is $\leq$). To determine the region for $3x + 4y \leq 12$, test the origin $(0, 0)$: $3(0) + 4(0) \leq 12 \Rightarrow 0 \leq 12$. This is true, so the region is the half-plane containing the origin.

The solution to the system of inequalities is the intersection of the two regions: the region that is on or above the line $2x + y = 6$ and the region that is on or below the line $3x + 4y = 12$.

The feasible region is the region that satisfies both inequalities simultaneously. This region is bounded by the lines $2x + y = 6$ and $3x + 4y = 12$.

To find the vertex (or vertices) of this region, we find the point of intersection of the boundary lines $2x + y = 6$ and $3x + 4y = 12$.

From the first equation, $y = 6 - 2x$. Substitute this into the second equation:

Now substitute the value of $x$ back into $y = 6 - 2x$:

The intersection point is $(\frac{12}{5}, \frac{6}{5})$.

The feasible region is the region bounded by the line segments from $(0,6)$ to $(\frac{12}{5}, \frac{6}{5})$ on $2x+y=6$ and from $(\frac{12}{5}, \frac{6}{5})$ to $(4,0)$ on $3x+4y=12$. It includes the boundary lines.

Graphically, this region is the triangle formed by the intersection of the two half-planes, including the segment on $2x+y=6$ above $3x+4y=12$ and the segment on $3x+4y=12$ above $2x+y=6$. The intersection point $(\frac{12}{5}, \frac{6}{5})$ is the common vertex of these segments.

The vertices of the feasible region are the intersection point $(\frac{12}{5}, \frac{6}{5})$ and potentially the points where the boundary lines intersect the axes, but only those points which satisfy the *other* inequality.

Point on $2x+y=6$ intersecting y-axis: $(0,6)$. Check inequality (2): $3(0) + 4(6) = 24$. Is $24 \leq 12$? No. So $(0,6)$ is not part of the feasible region boundary.

Point on $2x+y=6$ intersecting x-axis: $(3,0)$. Check inequality (2): $3(3) + 4(0) = 9$. Is $9 \leq 12$? Yes. So $(3,0)$ is a vertex of the feasible region.

Point on $3x+4y=12$ intersecting y-axis: $(0,3)$. Check inequality (1): $2(0) + 3 = 3$. Is $3 \geq 6$? No. So $(0,3)$ is not part of the feasible region boundary.

Point on $3x+4y=12$ intersecting x-axis: $(4,0)$. Check inequality (1): $2(4) + 0 = 8$. Is $8 \geq 6$? Yes. So $(4,0)$ is a vertex of the feasible region.

The feasible region is the triangular region with vertices $(\frac{12}{5}, \frac{6}{5})$, $(3, 0)$, and $(4, 0)$. Note: The standard vertices are formed by the intersection of boundary lines or boundaries with axes *within* the feasible region. The region is bounded by $2x+y=6$ from above-left, and $3x+4y=12$ from below-right, and the x-axis. The vertices are the intersection of $2x+y=6$ and $3x+4y=12$, and the intersections of these lines with the x-axis ($y=0$) since the region is in the first quadrant area bounded by these lines.

Vertices are $(\frac{12}{5}, \frac{6}{5})$, $(3, 0)$, and $(4, 0)$.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $x + y \geq 4$. The corresponding linear equation is $x + y = 4$.

To plot the line $x + y = 4$, we find two points on the line:

Plot the points $(0, 4)$ and $(4, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $x + y \geq 4$, test the origin $(0, 0)$: $0 + 0 \geq 4 \Rightarrow 0 \geq 4$. This is false, so the region is the half-plane not containing the origin (i.e., above or on the line).

Consider the second inequality $2x - y < 0$. The corresponding linear equation is $2x - y = 0$ or $y = 2x$.

To plot the line $y = 2x$, we find two points on the line:

Plot the points $(0, 0)$ and $(1, 2)$ (and $(-1, -2)$ as a check) and draw a dashed line through them (since the inequality is $<$). To determine the region for $2x - y < 0$, we cannot use the origin as it lies on the line. Let's test the point $(1, 0)$: $2(1) - 0 < 0 \Rightarrow 2 < 0$. This is false. Let's test the point $(-1, 0)$: $2(-1) - 0 < 0 \Rightarrow -2 < 0$. This is true. So the region is the half-plane containing $(-1, 0)$ (i.e., to the left of the line $y=2x$).

The solution to the system of inequalities is the intersection of the two regions: the region that is on or above the line $x + y = 4$ and the region that is strictly to the left of the line $y = 2x$.

The graphical representation of the solution is the feasible region that satisfies both inequalities simultaneously. This region is bounded by the lines $x + y = 4$ and $y = 2x$.

To find the vertex (or vertices) of this region, we find the point of intersection of the boundary lines $x + y = 4$ and $y = 2x$.

Substitute $y = 2x$ from the second equation into the first equation:

Now substitute the value of $x$ back into $y = 2x$:

The intersection point is $(\frac{4}{3}, \frac{8}{3})$.

The feasible region is the unbounded region located above or on the line $x+y=4$ and strictly to the left of the line $y=2x$. The point of intersection $(\frac{4}{3}, \frac{8}{3})$ is on the boundary $x+y=4$ (which is included) but not on the boundary $y=2x$ (which is excluded). Therefore, the vertex $(\frac{4}{3}, \frac{8}{3})$ is part of the boundary of the feasible region.

The boundary of the feasible region consists of the part of the solid line $x+y=4$ that is to the left of the dashed line $y=2x$, and the part of the dashed line $y=2x$ that is above the solid line $x+y=4$.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $2x - y > 1$. The corresponding linear equation is $2x - y = 1$, or $y = 2x - 1$.

To plot the line $y = 2x - 1$, we find two points on the line:

Plot the points $(0, -1)$ and $(1, 1)$ (and $(2, 3)$ as a check) and draw a dashed line through them (since the inequality is $>$). To determine the region for $2x - y > 1$, test the origin $(0, 0)$: $2(0) - 0 > 1 \Rightarrow 0 > 1$. This is false, so the region is the half-plane not containing the origin (i.e., to the right of the line $y=2x-1$).

Consider the second inequality $x - 2y < - 1$. The corresponding linear equation is $x - 2y = -1$, or $2y = x + 1$, so $y = \frac{1}{2}x + \frac{1}{2}$.

To plot the line $y = \frac{1}{2}x + \frac{1}{2}$, we find two points on the line:

Plot the points $(-1, 0)$ and $(1, 1)$ (and $(3, 2)$ as a check) and draw a dashed line through them (since the inequality is $<$). To determine the region for $x - 2y < - 1$, test the origin $(0, 0)$: $0 - 2(0) < - 1 \Rightarrow 0 < -1$. This is false, so the region is the half-plane not containing the origin (i.e., above or to the left of the line $y=\frac{1}{2}x + \frac{1}{2}$).

The solution to the system of inequalities is the intersection of the two regions: the region that is strictly to the right of the line $2x - y = 1$ and the region that is strictly above or to the left of the line $x - 2y = -1$.

The graphical representation of the solution is the feasible region that satisfies both inequalities simultaneously. This region is bounded by the lines $2x - y = 1$ and $x - 2y = -1$.

To find the vertex (or vertices) of this region, we find the point of intersection of the boundary lines $2x - y = 1$ and $x - 2y = -1$.

We have the system of equations:

From equation (A), $y = 2x - 1$. Substitute this into equation (B):

Now substitute the value of $x=1$ back into $y = 2x - 1$:

The intersection point is $(1, 1)$.

The feasible region is the unbounded region bounded by the two dashed lines $2x - y = 1$ and $x - 2y = -1$. The region lies to the right of $2x-y=1$ and to the left of $x-2y=-1$.

The intersection point $(1, 1)$ is on both boundary lines, but since both inequalities are strict, this point is not included in the feasible region.

The feasible region is the region between the two lines, containing points $(x,y)$ such that $y < 2x - 1$ and $y > \frac{1}{2}x + \frac{1}{2}$.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $x + y \leq 6$. The corresponding linear equation is $x + y = 6$.

To plot the line $x + y = 6$, we find two points on the line:

Plot the points $(0, 6)$ and $(6, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $x + y \leq 6$, test the origin $(0, 0)$: $0 + 0 \leq 6 \Rightarrow 0 \leq 6$. This is true, so the region is the half-plane containing the origin (i.e., below or on the line).

Consider the second inequality $x + y \geq 4$. The corresponding linear equation is $x + y = 4$.

To plot the line $x + y = 4$, we find two points on the line:

Plot the points $(0, 4)$ and $(4, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $x + y \geq 4$, test the origin $(0, 0)$: $0 + 0 \geq 4 \Rightarrow 0 \geq 4$. This is false, so the region is the half-plane not containing the origin (i.e., above or on the line).

The solution to the system of inequalities is the intersection of the two regions: the region that is on or below the line $x + y = 6$ and the region that is on or above the line $x + y = 4$.

Since the lines $x + y = 6$ and $x + y = 4$ have the same slope ($-1$), they are parallel lines.

The feasible region is the region between these two parallel lines, including the lines themselves (due to the non-strict inequalities $\leq$ and $\geq$).

This is an unbounded region, extending infinitely in the direction parallel to the lines, but bounded between the lines perpendicular to them.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $2x + y \geq 8$. The corresponding linear equation is $2x + y = 8$.

To plot the line $2x + y = 8$, we find two points on the line:

Plot the points $(0, 8)$ and $(4, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $2x + y \geq 8$, test the origin $(0, 0)$: $2(0) + 0 \geq 8 \Rightarrow 0 \geq 8$. This is false, so the region is the half-plane not containing the origin (i.e., above or on the line).

Consider the second inequality $x + 2y \geq 10$. The corresponding linear equation is $x + 2y = 10$.

To plot the line $x + 2y = 10$, we find two points on the line:

Plot the points $(0, 5)$ and $(10, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $x + 2y \geq 10$, test the origin $(0, 0)$: $0 + 2(0) \geq 10 \Rightarrow 0 \geq 10$. This is false, so the region is the half-plane not containing the origin (i.e., above or on the line).

The solution to the system of inequalities is the intersection of the two regions: the region that is on or above the line $2x + y = 8$ and the region that is on or above the line $x + 2y = 10$.

The graphical representation of the solution is the feasible region that satisfies both inequalities simultaneously. This region is bounded by the lines $2x + y = 8$ and $x + 2y = 10$.

To find the vertex (or vertices) of this region, we find the point of intersection of the boundary lines $2x + y = 8$ and $x + 2y = 10$.

We have the system of equations:

From equation (A), $y = 8 - 2x$. Substitute this into equation (B):

Now substitute the value of $x=2$ back into $y = 8 - 2x$:

The intersection point is $(2, 4)$.

The feasible region is the unbounded region above or on both lines $2x + y = 8$ and $x + 2y = 10$. This region is bounded by the line segment on $2x + y = 8$ starting from its y-intercept $(0,8)$ down to the intersection point $(2,4)$, and the line segment on $x + 2y = 10$ starting from the intersection point $(2,4)$ down to its x-intercept $(10,0)$, extending outwards infinitely from these points in the "above" direction.

The vertices of this unbounded feasible region are the points where the boundary lines meet and form a corner. In this case, the only finite vertex is the intersection point of the two lines.

The vertex of the feasible region is $(2, 4)$. The region extends infinitely upwards and outwards from this vertex.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $x + y \leq 9$. The corresponding linear equation is $x + y = 9$.

To plot the line $x + y = 9$, we find two points on the line:

Plot the points $(0, 9)$ and $(9, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $x + y \leq 9$, test the origin $(0, 0)$: $0 + 0 \leq 9 \Rightarrow 0 \leq 9$. This is true, so the region is the half-plane containing the origin (i.e., on or below the line).

Consider the second inequality $y > x$. The corresponding linear equation is $y = x$.

To plot the line $y = x$, we find two points on the line:

Plot the points $(0, 0)$ and $(5, 5)$ and draw a dashed line through them (since the inequality is $>$). To determine the region for $y > x$, test a point not on the line, e.g., $(1, 0)$: $0 > 1$. This is false. Test $(-1, 0)$: $0 > -1$. This is true. So the region is the half-plane containing $(-1, 0)$ (i.e., strictly above the line $y=x$).

Consider the third inequality $x \geq 0$. The corresponding linear equation is $x = 0$.

The graph of $x = 0$ is the y-axis. Draw a solid line (since the inequality is $\geq$).

The region satisfying $x \geq 0$ is the half-plane to the right of the y-axis, including the y-axis itself.

The solution to the system of inequalities is the intersection of the three regions: the region on or below the line $x + y = 9$, the region strictly above the line $y = x$, and the region on or to the right of the line $x = 0$ (the y-axis).

This intersection is the feasible region.

To find the vertices of this feasible region, we find the points of intersection of the boundary lines:

1. Intersection of $x + y = 9$ and $y = x$. Substitute $y=x$ into the first equation:

Since $y=x$, $y = 4.5$. The intersection point is $(4.5, 4.5)$. This point is on the solid line $x+y=9$ but not on the dashed line $y=x$, so it is part of the boundary but not strictly included in the region.

2. Intersection of $x + y = 9$ and $x = 0$. Substitute $x=0$ into the equation:

The intersection point is $(0, 9)$. This point is on the solid line $x+y=9$ and on the solid line $x=0$. Check if it satisfies $y > x$: $9 > 0$. This is true. So $(0, 9)$ is a vertex of the feasible region.

3. Intersection of $y = x$ and $x = 0$. Substitute $x=0$ into the equation:

The intersection point is $(0, 0)$. This point is on the solid line $x=0$ and on the dashed line $y=x$. Check if it satisfies $y > x$: $0 > 0$. This is false. So $(0, 0)$ is not part of the feasible region (specifically, it is not included due to the strict inequality $y > x$).

The feasible region is the bounded region in the first quadrant bounded by the lines $x=0$ (y-axis), $y=x$, and $x+y=9$. The boundary on $x=0$ and $x+y=9$ is included, but the boundary on $y=x$ is excluded.

The vertices of the feasible region are the points where the included boundaries intersect or where an included boundary intersects an excluded boundary (provided the point is on the included boundary).

The vertices are $(0, 9)$, the intersection of $x+y=9$ and $x=0$.

The other vertex is the intersection of $x+y=9$ and $y=x$, which is $(4.5, 4.5)$. This point is on the solid line $x+y=9$ and is included in the region defined by $x+y \leq 9$ and $x \ge 0$. It is on the boundary $y=x$, which is excluded by $y > x$. So, $(4.5, 4.5)$ is a vertex where an included boundary meets an excluded boundary.

The third potential vertex is the intersection of $y=x$ and $x=0$, which is $(0,0)$. This point satisfies $x \ge 0$ but not $y > x$, so it is not in the feasible region.

The feasible region is the triangular region with vertices $(0, 9)$, $(4.5, 4.5)$, and the point $(0,0)$ is a boundary point for $x \ge 0$ but not a vertex of the feasible region itself due to the strict inequality $y>x$. The feasible region is actually bounded by the y-axis ($x=0$) from the left, the line $x+y=9$ from the top-right, and the line $y=x$ from the bottom-left. The line $y=x$ is a dashed boundary, and the origin $(0,0)$ is excluded.

The vertices of the closed region defined by $x+y \le 9$, $y \ge x$, $x \ge 0$ would be $(0,0)$, $(0,9)$ and $(4.5, 4.5)$. Since $y > x$, the boundary $y=x$ (including $(0,0)$ and $(4.5, 4.5)$) is excluded. However, the region extends rightwards from the y-axis starting at $(0,0)$ along the y-axis up to $(0,9)$, then along $x+y=9$ to $(4.5, 4.5)$, then along $y=x$ (dashed) back towards the origin. The feasible region is the area bounded by the segment from $(0,0)$ to $(0,9)$ on the y-axis, the segment from $(0,9)$ to $(4.5, 4.5)$ on $x+y=9$, and the dashed line segment from $(4.5, 4.5)$ to $(0,0)$ on $y=x$. The point $(0,0)$ is not included as $0 \ngtr 0$. The point $(4.5, 4.5)$ is not included as $4.5 \ngtr 4.5$. The segment from $(0,0)$ to $(0,9)$ on the y-axis is included except for $(0,0)$. The segment from $(0,9)$ to $(4.5, 4.5)$ on $x+y=9$ is included except for $(4.5, 4.5)$.

The vertices of the feasible region are $(0, 9)$ and $(4.5, 4.5)$, but $(4.5, 4.5)$ is not included in the solution set due to the strict inequality $y>x$. The point $(0,0)$ is not a vertex of the *feasible* region either for the same reason. The region is a triangle with vertices $(0,0), (0,9), (4.5, 4.5)$, but the side along $y=x$ is dashed and endpoints on this side are excluded. The side along $x=0$ includes points $(0,y)$ where $0 \le y < 4.5$. The side along $x+y=9$ includes points $(x,y)$ where $4.5 < x \le 9$ (and $y=9-x$).

The vertices of the closed feasible region defined by $x+y \le 9, y \ge x, x \ge 0$ are $(0,0), (0,9), (4.5, 4.5)$. Since $y>x$, the boundary $y=x$ is excluded. So the actual vertices are $(0,9)$ and the boundary point $(4.5, 4.5)$ on $x+y=9$ but not on $y=x$. The region is bounded by $x=0$ for $0 \le y \le 9$, $x+y=9$ for $0 \le x \le 9$, and $y=x$. The points on $y=x$ are excluded. The origin $(0,0)$ is excluded. The vertex $(0,9)$ is included. The vertex $(4.5, 4.5)$ is not included. The boundary from $(0,0)$ to $(0,9)$ on the y-axis is included except for $(0,0)$. The boundary from $(0,9)$ to $(4.5, 4.5)$ on $x+y=9$ is included except for $(4.5, 4.5)$.

The vertices of the feasible region are $(0, 9)$ and approximately $(4.5, 4.5)$ (as an excluded point where boundary lines meet). The region is bounded.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $5x + 4y \leq 20$. The corresponding linear equation is $5x + 4y = 20$.

To plot the line $5x + 4y = 20$, we find two points on the line:

Plot the points $(0, 5)$ and $(4, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $5x + 4y \leq 20$, test the origin $(0, 0)$: $5(0) + 4(0) \leq 20 \Rightarrow 0 \leq 20$. This is true, so the region is the half-plane containing the origin (i.e., on or below the line).

Consider the second inequality $x \geq 1$. The corresponding linear equation is $x = 1$.

The graph of $x = 1$ is a vertical line parallel to the y-axis, passing through the point $(1, 0)$. Draw a solid line (since the inequality is $\geq$).

The region satisfying $x \geq 1$ is the half-plane to the right of the line $x = 1$, including the line itself.

Consider the third inequality $y \geq 2$. The corresponding linear equation is $y = 2$.

The graph of $y = 2$ is a horizontal line parallel to the x-axis, passing through the point $(0, 2)$. Draw a solid line (since the inequality is $\geq$).

The region satisfying $y \geq 2$ is the half-plane above the line $y = 2$, including the line itself.

The solution to the system of inequalities is the intersection of the three regions: the region on or below the line $5x + 4y = 20$, the region on or to the right of the line $x = 1$, and the region on or above the line $y = 2$.

This intersection forms the feasible region.

The graphical representation of the solution is the closed polygonal region bounded by the lines $x=1$, $y=2$, and $5x + 4y = 20$.

To find the vertices of this feasible region, we find the points of intersection of the boundary lines:

1. Intersection of $x = 1$ and $y = 2$: The point is $(1, 2)$.

2. Intersection of $x = 1$ and $5x + 4y = 20$. Substitute $x = 1$ into the equation:

The intersection point is $(1, \frac{15}{4})$.

3. Intersection of $y = 2$ and $5x + 4y = 20$. Substitute $y = 2$ into the equation:

The intersection point is $(\frac{12}{5}, 2)$.

The vertices of the feasible region are $(1, 2)$, $(1, \frac{15}{4})$, and $(\frac{12}{5}, 2)$.

The feasible region is the triangle formed by these three vertices, including the boundary lines.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $3x + 4y \leq 60$. The corresponding linear equation is $3x + 4y = 60$.

To plot the line $3x + 4y = 60$, we find two points on the line:

Plot the points $(0, 15)$ and $(20, 0)$ and draw a solid line through them. To determine the region for $3x + 4y \leq 60$, test the origin $(0, 0)$: $3(0) + 4(0) \leq 60 \Rightarrow 0 \leq 60$. This is true, so the region is the half-plane containing the origin.

Consider the second inequality $x + 3y \leq 30$. The corresponding linear equation is $x + 3y = 30$.

To plot the line $x + 3y = 30$, we find two points on the line:

Plot the points $(0, 10)$ and $(30, 0)$ and draw a solid line through them. To determine the region for $x + 3y \leq 30$, test the origin $(0, 0)$: $0 + 3(0) \leq 30 \Rightarrow 0 \leq 30$. This is true, so the region is the half-plane containing the origin.

Consider the inequalities $x \geq 0$ and $y \geq 0$.

$x \geq 0$ represents the region to the right of the y-axis, including the y-axis.

$y \geq 0$ represents the region above the x-axis, including the x-axis.

Together, $x \geq 0$ and $y \geq 0$ represent the first quadrant of the coordinate plane, including the positive x and y axes.

The solution to the system of inequalities is the intersection of the regions represented by (1), (2), (3), and (4). This is the region in the first quadrant that is on or below the line $3x + 4y = 60$ and on or below the line $x + 3y = 30$.

This intersection forms the feasible region.

The graphical representation of the solution is the closed polygonal region in the first quadrant bounded by the lines $x=0$, $y=0$, $3x + 4y = 60$, and $x + 3y = 30$.

To find the vertices of this feasible region, we find the points of intersection of the boundary lines:

1. Intersection of $x = 0$ and $y = 0$: The point is $(0, 0)$.

2. Intersection of $x = 0$ and $x + 3y = 30$. Substitute $x = 0$: $0 + 3y = 30 \implies y = 10$. The point is $(0, 10)$.

3. Intersection of $y = 0$ and $3x + 4y = 60$. Substitute $y = 0$: $3x + 4(0) = 60 \implies 3x = 60 \implies x = 20$. The point is $(20, 0)$.

4. Intersection of $3x + 4y = 60$ and $x + 3y = 30$. We solve the system of equations:

Multiply equation (B) by 3:

Subtract equation (A) from equation (C):

Substitute $y = 6$ into equation (B):

The intersection point is $(12, 6)$.

The vertices of the feasible region are $(0, 0)$, $(0, 10)$, $(20, 0)$, and $(12, 6)$.

The feasible region is the quadrilateral with these four vertices, including the boundary lines.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $2x + y \geq 4$. The corresponding linear equation is $2x + y = 4$.

To plot the line $2x + y = 4$, we find two points on the line:

Plot the points $(0, 4)$ and $(2, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $2x + y \geq 4$, test the origin $(0, 0)$: $2(0) + 0 \geq 4 \Rightarrow 0 \geq 4$. This is false, so the region is the half-plane not containing the origin (i.e., on or above the line).

Consider the second inequality $x + y \leq 3$. The corresponding linear equation is $x + y = 3$.

To plot the line $x + y = 3$, we find two points on the line:

Plot the points $(0, 3)$ and $(3, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $x + y \leq 3$, test the origin $(0, 0)$: $0 + 0 \leq 3 \Rightarrow 0 \leq 3$. This is true, so the region is the half-plane containing the origin (i.e., on or below the line).

Consider the third inequality $2x - 3y \leq 6$. The corresponding linear equation is $2x - 3y = 6$.

To plot the line $2x - 3y = 6$, we find two points on the line:

Plot the points $(0, -2)$ and $(3, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $2x - 3y \leq 6$, test the origin $(0, 0)$: $2(0) - 3(0) \leq 6 \Rightarrow 0 \leq 6$. This is true, so the region is the half-plane containing the origin (i.e., on or above the line).

The solution to the system of inequalities is the intersection of the three regions: the region on or above the line $2x + y = 4$, the region on or below the line $x + y = 3$, and the region on or above the line $2x - 3y = 6$.

This intersection forms the feasible region.

The graphical representation of the solution is the closed polygonal region bounded by the three lines $2x + y = 4$, $x + y = 3$, and $2x - 3y = 6$.

To find the vertices of this feasible region, we find the points of intersection of the boundary lines:

1. Intersection of $2x + y = 4$ and $x + y = 3$.

Substitute $x=1$ into $x + y = 3$: $1 + y = 3 \implies y = 2$. Intersection point: $(1, 2)$.

2. Intersection of $2x + y = 4$ and $2x - 3y = 6$.

Substitute $y = -\frac{1}{2}$ into $2x + y = 4$: $2x + (-\frac{1}{2}) = 4 \implies 2x = 4 + \frac{1}{2} = \frac{9}{2} \implies x = \frac{9}{4}$. Intersection point: $(\frac{9}{4}, -\frac{1}{2})$.

3. Intersection of $x + y = 3$ and $2x - 3y = 6$.

From $x + y = 3$, we get $x = 3 - y$. Substitute into $2x - 3y = 6$: $2(3 - y) - 3y = 6 \implies 6 - 2y - 3y = 6 \implies 6 - 5y = 6 \implies -5y = 0 \implies y = 0$.

Substitute $y = 0$ into $x + y = 3$: $x + 0 = 3 \implies x = 3$. Intersection point: $(3, 0)$.

The vertices of the feasible region are the intersection points of the boundary lines that satisfy all inequalities. We found three intersection points:

1. $(1, 2)$: $2(1)+2=4 \ge 4$ (T), $1+2=3 \le 3$ (T), $2(1)-3(2)=2-6=-4 \le 6$ (T). Point $(1, 2)$ is a vertex.

2. $(\frac{9}{4}, -\frac{1}{2})$: $2(\frac{9}{4})+(-\frac{1}{2})=\frac{9}{2}-\frac{1}{2}=4 \ge 4$ (T), $\frac{9}{4}+(-\frac{1}{2})=\frac{7}{4} \le 3$ (T), $2(\frac{9}{4})-3(-\frac{1}{2})=\frac{9}{2}+\frac{3}{2}=6 \le 6$ (T). Point $(\frac{9}{4}, -\frac{1}{2})$ is a vertex.

3. $(3, 0)$: $2(3)+0=6 \ge 4$ (T), $3+0=3 \le 3$ (T), $2(3)-3(0)=6 \le 6$ (T). Point $(3, 0)$ is a vertex.

The feasible region is the triangular region bounded by the lines $2x + y = 4$, $x + y = 3$, and $2x - 3y = 6$. All boundary lines are included in the feasible region.

The vertices of the feasible region are $(1, 2)$, $(\frac{9}{4}, -\frac{1}{2})$, and $(3, 0)$.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $x - 2y \leq 3$. The corresponding linear equation is $x - 2y = 3$.

To plot the line $x - 2y = 3$, we find two points on the line:

Plot the points $(-1, -2)$ and $(3, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $x - 2y \leq 3$, test the origin $(0, 0)$: $0 - 2(0) \leq 3 \Rightarrow 0 \leq 3$. This is true, so the region is the half-plane containing the origin (i.e., on or above the line $y = \frac{1}{2}x - \frac{3}{2}$).

Consider the second inequality $3x + 4y \geq 12$. The corresponding linear equation is $3x + 4y = 12$.

To plot the line $3x + 4y = 12$, we find two points on the line:

Plot the points $(0, 3)$ and $(4, 0)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $3x + 4y \geq 12$, test the origin $(0, 0)$: $3(0) + 4(0) \geq 12 \Rightarrow 0 \geq 12$. This is false, so the region is the half-plane not containing the origin (i.e., on or above the line $y = 3 - \frac{3}{4}x$).

Consider the third inequality $x \geq 0$. The corresponding linear equation is $x = 0$ (the y-axis). Draw a solid line.

The region satisfying $x \geq 0$ is the half-plane to the right of or on the y-axis.

Consider the fourth inequality $y \geq 1$. The corresponding linear equation is $y = 1$ (a horizontal line). Draw a solid line.

The region satisfying $y \geq 1$ is the half-plane above or on the line $y = 1$.

The solution to the system of inequalities is the intersection of the four regions:

• On or above the line $x - 2y = 3$.
• On or above the line $3x + 4y = 12$.
• On or to the right of the line $x = 0$.
• On or above the line $y = 1$.

This intersection forms the feasible region.

The feasible region is the set of points $(x,y)$ such that $x \ge 0$, $y \ge 1$, $y \ge \frac{1}{2}x - \frac{3}{2}$, and $y \ge 3 - \frac{3}{4}x$. The region must lie in the part of the first quadrant where $y \ge 1$.

To find the vertices of the feasible region, we find the points of intersection of the boundary lines within the region $x \geq 0, y \geq 1$ that form the corners:

1. Intersection of $x = 0$ and $y = 1$: Point $(0, 1)$. Check other inequalities: $0 - 2(1) = -2 \leq 3$ (T); $3(0) + 4(1) = 4 \geq 12$ (F). $(0, 1)$ is not in the feasible region.

2. Intersection of $x = 0$ and $3x + 4y = 12$: Substitute $x=0 \implies 4y = 12 \implies y = 3$. Point $(0, 3)$. Check other inequalities: $3 \geq 1$ (T); $0 - 2(3) = -6 \leq 3$ (T). Point $(0, 3)$ is a vertex.

3. Intersection of $y = 1$ and $3x + 4y = 12$: Substitute $y=1 \implies 3x + 4(1) = 12 \implies 3x = 8 \implies x = \frac{8}{3}$. Point $(\frac{8}{3}, 1)$. Check other inequalities: $\frac{8}{3} \geq 0$ (T); $\frac{8}{3} - 2(1) = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \leq 3$ (T). Point $(\frac{8}{3}, 1)$ is a vertex.

4. Intersection of $y = 1$ and $x - 2y = 3$: Substitute $y=1 \implies x - 2(1) = 3 \implies x - 2 = 3 \implies x = 5$. Point $(5, 1)$. Check other inequalities: $5 \geq 0$ (T); $3(5) + 4(1) = 15 + 4 = 19 \geq 12$ (T). Point $(5, 1)$ is a vertex.

5. Intersection of $3x + 4y = 12$ and $x - 2y = 3$: Solve the system. Multiply the second equation by 2: $2(x - 2y) = 2(3) \implies 2x - 4y = 6$. Add this to $3x + 4y = 12$: $(3x + 4y) + (2x - 4y) = 12 + 6 \implies 5x = 18 \implies x = \frac{18}{5}$. Substitute $x = \frac{18}{5}$ into $x - 2y = 3$: $\frac{18}{5} - 2y = 3 \implies -2y = 3 - \frac{18}{5} = \frac{15-18}{5} = -\frac{3}{5} \implies y = \frac{3}{10}$. Point $(\frac{18}{5}, \frac{3}{10})$. Check inequality (4): $\frac{3}{10} \geq 1$ (F). This point is not in the feasible region.

The feasible region is an unbounded region. Its vertices are the points where the boundary lines intersect and form a corner within the constraints $x \ge 0$ and $y \ge 1$. The vertices are $(0, 3)$, $(\frac{8}{3}, 1)$, and $(5, 1)$.

The boundary of the feasible region is formed by the line segment from $(0, 3)$ to $(\frac{8}{3}, 1)$ (on $3x+4y=12$), the line segment from $(\frac{8}{3}, 1)$ to $(5, 1)$ (on $y=1$), the ray from $(5, 1)$ along the line $x-2y=3$ (in the direction of increasing $x$), and the ray from $(0, 3)$ along the y-axis ($x=0$) (in the direction of increasing $y$). All boundary lines are included.

We are asked to solve the following system of inequalities graphically:

Consider the inequality $4x + 3y \leq 60$. The corresponding linear equation is $4x + 3y = 60$.

To plot the line $4x + 3y = 60$, we find two points on the line:

Plot the points $(0, 20)$ and $(15, 0)$ and draw a solid line through them (since the inequality is $\leq$). To determine the region for $4x + 3y \leq 60$, test the origin $(0, 0)$: $4(0) + 3(0) \leq 60 \Rightarrow 0 \leq 60$. This is true, so the region is the half-plane containing the origin (i.e., on or below the line).

Consider the inequality $y \geq 2x$. The corresponding linear equation is $y = 2x$.

To plot the line $y = 2x$, we find two points on the line:

Plot the points $(0, 0)$ and $(5, 10)$ and draw a solid line through them (since the inequality is $\geq$). To determine the region for $y \geq 2x$, test a point not on the line, e.g., $(1, 0)$: $0 \geq 2(1) \Rightarrow 0 \geq 2$. This is false. Test $(0, 1)$: $1 \geq 2(0) \Rightarrow 1 \geq 0$. This is true. So the region is the half-plane containing $(0, 1)$ (i.e., on or above the line).

Consider the inequality $x \geq 3$. The corresponding linear equation is $x = 3$.

The graph of $x = 3$ is a vertical line parallel to the y-axis, passing through $(3, 0)$. Draw a solid line (since the inequality is $\geq$).

The region satisfying $x \geq 3$ is the half-plane to the right of the line $x = 3$, including the line itself.

The inequalities $x \geq 0$ and $y \geq 0$ indicate that the solution lies in the first quadrant. However, since we have $x \geq 3$ and $y \geq 2x$ (which for $x \geq 3$ implies $y \geq 6$), the conditions $x \geq 0$ and $y \geq 0$ are automatically satisfied for the feasible region determined by the first three inequalities.

The solution to the system of inequalities is the intersection of the three regions: the region on or below the line $4x + 3y = 60$, the region on or above the line $y = 2x$, and the region on or to the right of the line $x = 3$.

This intersection forms the feasible region.

The graphical representation of the solution is the closed polygonal region bounded by the lines $4x + 3y = 60$, $y = 2x$, and $x = 3$.

To find the vertices of this feasible region, we find the points of intersection of these boundary lines that lie within the region defined by all inequalities:

1. Intersection of $x = 3$ and $y = 2x$. Substitute $x = 3$ into $y = 2x$: $y = 2(3) = 6$. Point $(3, 6)$. Check inequality (1): $4(3) + 3(6) = 12 + 18 = 30 \leq 60$. This is true. Point $(3, 6)$ is a vertex.

2. Intersection of $x = 3$ and $4x + 3y = 60$. Substitute $x = 3$ into $4x + 3y = 60$: $4(3) + 3y = 60 \implies 12 + 3y = 60 \implies 3y = 48 \implies y = 16$. Point $(3, 16)$. Check inequality (2): $16 \geq 2(3) \implies 16 \geq 6$. This is true. Point $(3, 16)$ is a vertex.

3. Intersection of $y = 2x$ and $4x + 3y = 60$. Substitute $y = 2x$ into $4x + 3y = 60$: $4x + 3(2x) = 60 \implies 4x + 6x = 60 \implies 10x = 60 \implies x = 6$. Substitute $x = 6$ into $y = 2x$: $y = 2(6) = 12$. Point $(6, 12)$. Check inequality (3): $6 \geq 3$. This is true. Point $(6, 12)$ is a vertex.

The feasible region is the triangular region with vertices $(3, 6)$, $(3, 16)$, and $(6, 12)$. All boundary lines are included in the feasible region.

We are asked to solve the following system of inequalities graphically:

Consider the first inequality $3x + 2y \leq 150$. The corresponding linear equation is $3x + 2y = 150$.

To plot the line $3x + 2y = 150$, we find two points on the line:

Plot the points $(0, 75)$ and $(50, 0)$ and draw a solid line. The region for $3x + 2y \leq 150$ is the half-plane containing the origin.

Consider the second inequality $x + 4y \leq 80$. The corresponding linear equation is $x + 4y = 80$.

To plot the line $x + 4y = 80$, we find two points on the line:

Plot the points $(0, 20)$ and $(80, 0)$ and draw a solid line. The region for $x + 4y \leq 80$ is the half-plane containing the origin.

Consider the third inequality $x \leq 15$. The corresponding linear equation is $x = 15$.

The graph of $x = 15$ is a vertical line passing through $(15, 0)$. Draw a solid line. The region for $x \leq 15$ is the half-plane to the left of or on the line $x=15$.

The inequalities $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant, including the axes.

The solution to the system of inequalities is the intersection of all five regions. This is the set of points $(x, y)$ in the first quadrant such that $x \leq 15$, $3x + 2y \leq 150$, and $x + 4y \leq 80$.

This intersection forms the feasible region.

The feasible region is a closed polygonal region in the first quadrant. Its vertices are the points of intersection of the boundary lines that satisfy all the given inequalities.

The boundary lines are $x=0$, $y=0$, $x=15$, $3x+2y=150$, and $x+4y=80$.

Let's find the vertices:

1. Intersection of $x=0$ and $y=0$: $(0, 0)$. This satisfies all inequalities: $0 \le 150$, $0 \le 80$, $0 \le 15$, $0 \ge 0$, $0 \ge 0$. So, $(0, 0)$ is a vertex.

2. Intersection of $y=0$ and $x=15$: $(15, 0)$. Check other inequalities: $3(15)+2(0) = 45 \le 150$ (T), $15+4(0) = 15 \le 80$ (T), $15 \ge 0$ (T), $0 \ge 0$ (T). So, $(15, 0)$ is a vertex.

3. Intersection of $x=0$ and $x+4y=80$: Substitute $x=0$ into $x+4y=80 \implies 4y = 80 \implies y = 20$. Point $(0, 20)$. Check other inequalities: $3(0)+2(20) = 40 \le 150$ (T), $0 \le 15$ (T), $20 \ge 0$ (T), $0 \ge 0$ (T). So, $(0, 20)$ is a vertex.

4. Intersection of $x=15$ and $x+4y=80$: Substitute $x=15$ into $x+4y=80 \implies 15+4y=80 \implies 4y=65 \implies y = \frac{65}{4} = 16.25$. Point $(15, 16.25)$. Check other inequalities: $3(15)+2(16.25) = 45+32.5 = 77.5 \le 150$ (T), $15 \ge 0$ (T), $16.25 \ge 0$ (T). So, $(15, 16.25)$ is a vertex.

5. Intersection of $x=0$ and $3x+2y=150$: Substitute $x=0$ into $3x+2y=150 \implies 2y = 150 \implies y = 75$. Point $(0, 75)$. Check inequality (2): $0+4(75) = 300 \not\leq 80$ (F). This point is not a vertex of the feasible region.

6. Intersection of $y=0$ and $3x+2y=150$: Substitute $y=0$ into $3x+2y=150 \implies 3x = 150 \implies x = 50$. Point $(50, 0)$. Check inequality (3): $50 \not\leq 15$ (F). This point is not a vertex of the feasible region.

7. Intersection of $x=15$ and $3x+2y=150$: Substitute $x=15$ into $3x+2y=150 \implies 3(15)+2y=150 \implies 45+2y=150 \implies 2y=105 \implies y = \frac{105}{2} = 52.5$. Point $(15, 52.5)$. Check inequality (2): $15+4(52.5) = 15+210 = 225 \not\leq 80$ (F). This point is not a vertex of the feasible region.

8. Intersection of $3x+2y=150$ and $x+4y=80$: Solve the system. Multiply the second equation by 3: $3(x+4y) = 3(80) \implies 3x + 12y = 240$. Subtract the first equation: $(3x+12y) - (3x+2y) = 240 - 150 \implies 10y = 90 \implies y = 9$. Substitute $y=9$ into $x+4y=80 \implies x+4(9)=80 \implies x+36=80 \implies x=44$. Point $(44, 9)$. Check inequality (3): $44 \not\leq 15$ (F). This point is not a vertex of the feasible region.

The feasible region is the quadrilateral bounded by segments of the lines $x=0$, $y=0$, $x=15$, and $x+4y=80$. The line $3x+2y=150$ passes above the feasible region formed by the other inequalities.

The vertices of the feasible region are $(0, 0)$, $(15, 0)$, $(15, 16.25)$, and $(0, 20)$.

We are asked to solve the following system of inequalities graphically:

Consider $x + 2y \leq 10$. Boundary line: $x + 2y = 10$. Points: $(0, 5), (10, 0)$. Region: contains $(0,0)$ (below or on the line).

Consider $x + y \geq 1$. Boundary line: $x + y = 1$. Points: $(0, 1), (1, 0)$. Region: does not contain $(0,0)$ (above or on the line).

Consider $y \geq x$. Boundary line: $y = x$. Points: $(0, 0), (5, 5)$. Region: contains $(0,1)$ (above or on the line).

The inequalities $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant, including the axes.

The feasible region is the intersection of all these regions: in the first quadrant, on or below $x+2y=10$, on or above $x+y=1$, and on or above $y=x$.

This is a closed polygonal region.

To find the vertices, we find intersection points of the boundary lines within the feasible region:

1. Intersection of $y=x$ and $x+y=1$: Substitute $y=x$ into $x+y=1 \implies x+x=1 \implies 2x=1 \implies x=\frac{1}{2}$. Then $y=\frac{1}{2}$. Point $(\frac{1}{2}, \frac{1}{2})$. Check inequality (1): $\frac{1}{2} + 2(\frac{1}{2}) = \frac{1}{2} + 1 = \frac{3}{2} \le 10$ (T). Check $x \ge 0, y \ge 0$ (T). Vertex: $(\frac{1}{2}, \frac{1}{2})$.

2. Intersection of $y=x$ and $x+2y=10$: Substitute $y=x$ into $x+2y=10 \implies x+2x=10 \implies 3x=10 \implies x=\frac{10}{3}$. Then $y=\frac{10}{3}$. Point $(\frac{10}{3}, \frac{10}{3})$. Check inequality (2): $\frac{10}{3} + \frac{10}{3} = \frac{20}{3} = 6.\overline{6} \ge 1$ (T). Check $x \ge 0, y \ge 0$ (T). Vertex: $(\frac{10}{3}, \frac{10}{3})$.

3. Intersection of $x+y=1$ and $x=0$ (y-axis): Substitute $x=0$ into $x+y=1 \implies y=1$. Point $(0, 1)$. Check inequality (3): $1 \ge 0$ (T). Check $y \ge 0$ (T). Check inequality (1): $0 + 2(1) = 2 \le 10$ (T). Vertex: $(0, 1)$.

4. Intersection of $x+2y=10$ and $x=0$ (y-axis): Substitute $x=0$ into $x+2y=10 \implies 2y=10 \implies y=5$. Point $(0, 5)$. Check inequality (3): $5 \ge 0$ (T). Check inequality (2): $0+5 = 5 \ge 1$ (T). Check $y \ge 0$ (T). Vertex: $(0, 5)$.

Note: Intersection of $x+y=1$ and $y=0$ is $(1,0)$. Check inequality (3): $0 \ge 1$ (F). Not a vertex of the feasible region.

Intersection of $x+2y=10$ and $y=0$ is $(10,0)$. Check inequality (3): $0 \ge 10$ (F). Not a vertex of the feasible region.

The feasible region is the triangular region with vertices $(\frac{1}{2}, \frac{1}{2})$, $(0, 1)$, and $(0, 5)$, and $(\frac{10}{3}, \frac{10}{3})$. No, check the graph. The region is bounded by $y=x$ from below left, $x+y=1$ from below right, $x=0$ from the left, and $x+2y=10$ from above.

Let's re-evaluate the vertices based on the boundary lines: $y=x$, $x+y=1$, $x+2y=10$, $x=0$, $y=0$. We are in the first quadrant ($x \ge 0, y \ge 0$).

Region is $y \ge x$, $x+y \ge 1$, $x+2y \le 10$, $x \ge 0$, $y \ge 0$.

Relevant boundaries: $y=x$, $x+y=1$, $x+2y=10$, $x=0$, $y=0$.

Intersection of $y=x$ and $x+y=1$: $(\frac{1}{2}, \frac{1}{2})$. Satisfies $x+2y \le 10$ (3/2 <= 10). $x \ge 0, y \ge 0$ (T). Vertex: $(\frac{1}{2}, \frac{1}{2})$.

Intersection of $y=x$ and $x+2y=10$: $(\frac{10}{3}, \frac{10}{3})$. Satisfies $x+y \ge 1$ (20/3 >= 1). $x \ge 0, y \ge 0$ (T). Vertex: $(\frac{10}{3}, \frac{10}{3})$.

Intersection of $x+y=1$ and $x=0$ (y-axis): $(0, 1)$. Satisfies $y \ge x$ (1 >= 0). Satisfies $x+2y \le 10$ (2 <= 10). Satisfies $y \ge 0$. Vertex: $(0, 1)$.

Intersection of $x+2y=10$ and $x=0$ (y-axis): $(0, 5)$. Satisfies $y \ge x$ (5 >= 0). Satisfies $x+y \ge 1$ (5 >= 1). Satisfies $y \ge 0$. Vertex: $(0, 5)$.

Intersection of $x+y=1$ and $y=0$ (x-axis): $(1, 0)$. Check $y \ge x$ (0 >= 1) (F). Not a vertex.

Intersection of $x+2y=10$ and $y=0$ (x-axis): $(10, 0)$. Check $y \ge x$ (0 >= 10) (F). Not a vertex.

The vertices are the corners of the feasible region where two or more boundary lines intersect and lie within the allowed region.

The vertices are $(0, 1)$, $(\frac{1}{2}, \frac{1}{2})$, $(\frac{10}{3}, \frac{10}{3})$, and $(0, 5)$.

The feasible region is the quadrilateral with these four vertices, including the boundary lines.

Given:

The inequality is $– 8 \le 5x – 3 < 7$.

Solution:

We need to find the values of $x$ that satisfy the compound inequality.

$– 8 \le 5x – 3 < 7$

$– 5 \le 5x < 10$

$– 1 \le x < 2$

Thus, the solution set consists of all real numbers $x$ such that $x$ is greater than or equal to $-1$ and less than $2$.

In interval notation, the solution set is $[-1, 2)$.

Alternate Solution:

The given inequality $– 8 \le 5x – 3 < 7$ can be treated as two separate inequalities which must hold simultaneously:

1) $– 8 \le 5x – 3$

2) $5x – 3 < 7$

Solving the first inequality:

$– 8 \le 5x – 3$

$– 5 \le 5x$

$– 1 \le x$ or $x \ge – 1$

Solving the second inequality:

$5x – 3 < 7$

$5x < 10$

$x < 2$

Combining the results from both inequalities, we need $x$ such that $x \ge – 1$ and $x < 2$.

Therefore, $– 1 \le x < 2$.

In interval notation, the solution set is $[-1, 2)$.

We are given the inequality:

$-5\leq\frac{5\;-\;3x}{2} \leq8$

To eliminate the denominator, multiply all parts of the inequality by $2$:

Simplifying this gives:

Now, subtract $5$ from all parts of the inequality (ii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $-3$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (vi) in the standard form, with the smaller value on the left:

$-\frac{11}{3} \leq x \leq 5$

This means that $x$ is greater than or equal to $-\frac{11}{3}$ and less than or equal to $5$. In interval notation, the solution is:

$\left[-\frac{11}{3}, 5\right]$

We are asked to solve the system of two linear inequalities:

Let us solve each inequality separately.

Consider inequality (1):

Subtract $x$ from both sides:

Add $7$ to both sides of inequality (iii):

Divide both sides of inequality (iv) by $2$:

Now, consider inequality (2):

Subtract $11$ from both sides:

Divide both sides of inequality (v) by $-5$. Remember to reverse the direction of the inequality sign when dividing by a negative number:

The solution to the system of inequalities is the set of values of $x$ that satisfy both inequality (A) and inequality (B).

From (A), we have $x < 6$.

From (B), we have $x \geq 2$.

Combining these two conditions, we get $x \geq 2$ and $x < 6$.

This can be written as:

$2 \leq x < 6$

In interval notation, the solution set is:

$[2, 6)$

We are given the range of temperature in Celsius:

The conversion formula between Celsius (C) and Fahrenheit (F) is given by:

Substitute the expression for C from (ii) into the inequality (i):

To isolate the term $(F - 32)$, multiply all parts of the inequality (iii) by $\frac{9}{5}$:

Simplify the multiplication:

Now, add $32$ to all parts of the inequality (v) to isolate F:

Simplify the addition:

Thus, the range of temperature in degree Fahrenheit is between $86^\circ$ F and $95^\circ$ F, exclusive of the endpoints.

The range is $86^\circ$ F $< F < 95^\circ$ F.

In interval notation, the range is $(86, 95)$.

Let $x$ be the number of litres of 30% acid solution that must be added.

Initial solution: 600 litres of 12% acid.

Amount of acid in the initial solution = $12\%$ of $600$ litres

Amount of acid = $0.12 \times 600 = 72$ litres.

Added solution: $x$ litres of 30% acid.

Amount of acid in the added solution = $30\%$ of $x$ litres

Amount of acid = $0.30 \times x = 0.3x$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Added volume)

Total amount of acid in the resulting mixture = (Acid in initial solution) + (Acid in added solution)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{72 + 0.3x}{600 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 15% but less than 18%. This gives us a system of two inequalities:

Inequality 1: Acid content > 15%

Inequality 2: Acid content < 18%

Since $x$ represents the amount of solution added, $x \geq 0$. The total volume $600+x$ is always positive. So we can multiply both sides of the inequalities by $600+x$ without changing the direction of the inequality sign.

Solving Inequality (A):

Subtract $0.15x$ from both sides:

Subtract $72$ from both sides:

Divide by $0.15$:

Solving Inequality (B):

Subtract $0.18x$ from both sides:

Subtract $72$ from both sides:

Divide by $0.12$:

For the acid content to be between 15% and 18%, the value of $x$ must satisfy both conditions (C) and (D).

Combining $x > 120$ and $x < 300$, we get:

$120 < x < 300$

Therefore, the manufacturer must add more than 120 litres but less than 300 litres of the 30% acid solution.

The range is $(120, 300)$ litres.

We are given the inequality:

$2 \leq 3x - 4 \leq 5$

To isolate the term with $x$, add $4$ to all parts of the inequality:

Simplifying this gives:

Now, divide all parts of the inequality (ii) by $3$ to solve for $x$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are greater than or equal to $2$ and less than or equal to $3$.

In interval notation, the solution is:

$\mathbf{[2, 3]}$

We are given the inequality:

$6 \leq -3(2x - 4) < 12$

First, distribute the $-3$ on the right side of the first part of the inequality:

So the inequality becomes:

To isolate the term with $x$, subtract $12$ from all parts of the inequality (i):

Simplifying this gives:

Now, divide all parts of the inequality (iii) by $-6$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (v) in the standard form, with the smaller value on the left:

$\mathbf{0 < x \leq 1}$

The solution set consists of all real numbers $x$ that are strictly greater than $0$ and less than or equal to $1$.

In interval notation, the solution is:

$\mathbf{(0, 1]}$

We are given the inequality:

$-3 \leq 4 - \frac{7x}{2} \leq 18$

To isolate the term with $x$, subtract $4$ from all parts of the inequality:

Simplifying this gives:

To eliminate the denominator, multiply all parts of the inequality (i) by $2$. Since $2$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, divide all parts of the inequality (ii) by $-7$. Remember to reverse the direction of the inequality signs when dividing by a negative number:

Simplifying this gives:

We can rewrite the inequality (iii) in the standard form, with the smaller value on the left:

$\mathbf{-4 \leq x \leq 2}$

The solution set consists of all real numbers $x$ that are greater than or equal to $-4$ and less than or equal to $2$.

In interval notation, the solution is:

$\mathbf{[-4, 2]}$

We are given the inequality:

$-15 < \frac{3(x - 2)}{5} \leq 0$

To eliminate the denominator, multiply all parts of the inequality by $5$. Since $5$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, distribute the $3$ on the right side of the left part of inequality (ii):

To isolate the term with $x$, add $6$ to all parts of the inequality (iii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are strictly greater than $-23$ and less than or equal to $2$.

In interval notation, the solution is:

$\mathbf{(-23, 2]}$

We are given the inequality:

$-12 < 4 - \frac{3x}{-5} \leq 2$

First, simplify the term $-\frac{3x}{-5} = -(-\frac{3x}{5}) = \frac{3x}{5}$. The inequality becomes:

To isolate the term with $x$, subtract $4$ from all parts of the inequality (i):

Simplifying this gives:

To eliminate the denominator, multiply all parts of the inequality (iii) by $5$. Since $5$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Finally, divide all parts of the inequality (v) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are strictly greater than $-\frac{80}{3}$ and less than or equal to $-\frac{10}{3}$.

In interval notation, the solution is:

$\mathbf{\left(-\frac{80}{3}, -\frac{10}{3}\right]}$

We are given the inequality:

$7 \leq \frac{3x + 11}{2} \leq 11$

To eliminate the denominator, multiply all parts of the inequality by $2$. Since $2$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

To isolate the term with $x$, subtract $11$ from all parts of the inequality (ii):

Simplifying this gives:

Finally, divide all parts of the inequality (iv) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

The solution set consists of all real numbers $x$ that are greater than or equal to $1$ and less than or equal to $\frac{11}{3}$.

In interval notation, the solution is:

$\mathbf{\left[1, \frac{11}{3}\right]}$

We are given a system of two linear inequalities:

Let's solve inequality (1):

Subtract $1$ from both sides:

Divide both sides by $5$ (positive number, so inequality direction does not change):

Now, let's solve inequality (2):

Add $1$ to both sides:

Divide both sides by $5$ (positive number, so inequality direction does not change):

The solution to the system of inequalities is the set of values of $x$ that satisfy both inequality (A) and inequality (B).

From (A), we have $x > -5$.

From (B), we have $x < 5$.

Combining these two conditions, we get $x$ must be greater than $-5$ and less than $5$.

The combined inequality is:

$\mathbf{-5 < x < 5}$

In interval notation, the solution set is:

$\mathbf{(-5, 5)}$

Graphical Representation:

The solution set on the number line is the open interval between $-5$ and $5$. This is represented by drawing a number line, marking the points $-5$ and $5$, and drawing an open circle at each of these points (to indicate that $-5$ and $5$ are not included in the solution). Then, a line segment is drawn between these two open circles.

A visual representation (conceptual description as drawing is not feasible in text):

Number line:
... -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 ...

Mark open circles at $-5$ and $5$. Shade the region between $-5$ and $5$.

We are given a system of two linear inequalities:

Let's solve inequality (1):

Distribute the 2 on the left side:

Subtract $x$ from both sides of inequality (iii):

Add $2$ to both sides:

Now, let's solve inequality (2):

Distribute the 3 on the left side:

Add $x$ to both sides of inequality (iv):

Subtract $6$ from both sides:

Divide both sides by $4$ (positive number, so inequality direction does not change):

The solution to the system of inequalities is the set of values of $x$ that satisfy both inequality (A) and inequality (B).

From (A), we have $x < 7$.

From (B), we have $x > -1$.

Combining these two conditions, we get $x$ must be greater than $-1$ and less than $7$.

The combined inequality is:

$\mathbf{-1 < x < 7}$

In interval notation, the solution set is:

$\mathbf{(-1, 7)}$

Graphical Representation:

The solution set on the number line is the open interval between $-1$ and $7$. This is represented by drawing a number line, marking the points $-1$ and $7$, and drawing an open circle at each of these points (to indicate that $-1$ and $7$ are not included in the solution). Then, a line segment is drawn between these two open circles.

A visual representation (conceptual description as drawing is not feasible in text):

Number line:
... -2 -1 0 1 2 3 4 5 6 7 8 ...

Mark open circles at $-1$ and $7$. Shade the region between $-1$ and $7$.

We are given a system of two linear inequalities:

Let's solve inequality (1):

Distribute the 2 on the right side:

Subtract $2x$ from both sides of inequality (iii):

Add $7$ to both sides:

Now, let's solve inequality (2):

Add $2x$ to both sides:

Subtract $6$ from both sides:

The solution to the system of inequalities is the set of values of $x$ that satisfy both inequality (A) and inequality (B).

From (A), we have $x > -5$.

From (B), we have $x > 5$.

For $x$ to satisfy both conditions, it must be greater than the larger of the two lower bounds. Since $5 > -5$, the condition $x > 5$ implies $x > -5$.

Thus, the combined inequality is:

$\mathbf{x > 5}$

In interval notation, the solution set is:

$\mathbf{(5, \infty)}$

Graphical Representation:

The solution set on the number line is the ray extending to the right from $5$, not including $5$. This is represented by drawing a number line, marking the point $5$, and drawing an open circle at $5$ (to indicate that $5$ is not included in the solution). Then, a line segment is drawn to the right of $5$ with an arrow indicating it extends infinitely.

A visual representation (conceptual description as drawing is not feasible in text):

Number line:
... 3 4 5 6 7 8 9 ...

Mark an open circle at $5$. Shade the region to the right of $5$ with an arrow pointing right.

We are given a system of two linear inequalities:

Let's solve inequality (1):

Distribute the numbers:

Combine like terms:

Add $44$ to both sides:

Divide both sides by $4$ (positive number, so inequality direction does not change):

Now, let's solve inequality (2):

Subtract $2x$ from both sides:

Subtract $47$ from both sides:

Divide both sides by $4$ (positive number, so inequality direction does not change):

The solution to the system of inequalities is the set of values of $x$ that satisfy both inequality (A) and inequality (B).

From (A), we have $x \leq 11$.

From (B), we have $x \geq -7$.

Combining these two conditions, we get $x$ must be greater than or equal to $-7$ and less than or equal to $11$.

The combined inequality is:

$\mathbf{-7 \leq x \leq 11}$

In interval notation, the solution set is:

$\mathbf{[-7, 11]}$

Graphical Representation:

The solution set on the number line is the closed interval between $-7$ and $11$. This is represented by drawing a number line, marking the points $-7$ and $11$, and drawing a closed circle (or a filled dot) at each of these points (to indicate that $-7$ and $11$ are included in the solution). Then, a line segment is drawn between these two closed circles.

A visual representation (conceptual description as drawing is not feasible in text):

Number line:
... -8 -7 -6 ... 0 ... 10 11 12 ...

Mark closed circles at $-7$ and $11$. Shade the region between $-7$ and $11$.

We are given the range of temperature in Fahrenheit:

The conversion formula between Fahrenheit (F) and Celsius (C) is given by:

We need to find the range of temperature in Celsius. Substitute the expression for F from (ii) into the inequality (i):

To isolate the term with C, subtract $32$ from all parts of the inequality (iii):

Simplifying this gives:

Now, multiply all parts of the inequality (v) by $\frac{5}{9}$ to isolate C. Since $\frac{5}{9}$ is positive, the direction of the inequality signs remains unchanged:

Simplify the multiplication:

Thus, the range of temperature in degree Celsius is between $20^\circ$ C and $25^\circ$ C, exclusive of the endpoints.

The range is $\mathbf{20^\circ \text{C} < C < 25^\circ \text{C}}$.

In interval notation, the range is $\mathbf{(20, 25)}$.

Let $x$ be the number of litres of 2% boric acid solution that must be added.

Since we are adding a volume of solution, $x \geq 0$.

Initial solution: 640 litres of 8% boric acid.

Amount of boric acid in the initial solution = $8\%$ of $640$ litres

Amount of boric acid = $0.08 \times 640 = 51.2$ litres.

Added solution: $x$ litres of 2% boric acid.

Amount of boric acid in the added solution = $2\%$ of $x$ litres

Amount of boric acid = $0.02 \times x = 0.02x$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Added volume)

Total amount of boric acid in the resulting mixture = (Acid in initial solution) + (Acid in added solution)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{51.2 + 0.02x}{640 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 4% but less than 6%. This gives us a compound inequality:

Divide the inequality (iii) by 100:

Since $x \geq 0$, the total volume $640 + x$ is always positive. We can multiply all parts of the inequality (iv) by $(640 + x)$ without changing the direction of the inequality signs.

This compound inequality can be split into two separate inequalities:

Inequality 1: $0.04(640 + x) < 51.2 + 0.02x$

Distribute the 0.04:

Subtract $0.02x$ from both sides:

Subtract $25.6$ from both sides:

Divide by $0.02$ (positive number):

Inequality 2: $51.2 + 0.02x < 0.06(640 + x)$

Distribute the 0.06:

Subtract $0.02x$ from both sides:

Subtract $38.4$ from both sides:

Divide by $0.04$ (positive number):

To satisfy the conditions of the problem, $x$ must satisfy both inequality (A) and inequality (B).

From (A), $x < 1280$.

From (B), $x > 320$.

Combining these two conditions, we get:

$\mathbf{320 < x < 1280}$

Therefore, the manufacturer must add more than 320 litres but less than 1280 litres of the 2% boric acid solution.

The range of the amount of 2% solution to be added is $\mathbf{(320, 1280)}$ litres.

Let $x$ be the number of litres of water added to the 1125 litres of 45% acid solution.

Since $x$ is a volume of water added, $x \geq 0$.

Initial solution: 1125 litres of 45% acid.

Amount of acid in the initial solution = $45\%$ of $1125$ litres

Let's calculate $0.45 \times 1125$:

$\begin{array}{cc} & 1 & 1 & 2 & 5 \\ \times & & & 4 & 5 \\ \hline & & 5 & 6 & 2 & 5 \\ 4 & 5 & 0 & 0 & \times \\ \hline 5 & 0 & 6 & 2 & 5 \\ \hline \end{array}$

$0.45 \times 1125 = 506.25$.

So, the initial amount of acid is $506.25$ litres.

Added solution: $x$ litres of water (0% acid).

Amount of acid in the added water = $0\%$ of $x$ litres = $0$ litres.

Resulting mixture:

Total volume of the mixture = (Initial volume) + (Volume of water added)

Total amount of acid in the resulting mixture = (Acid in initial solution) + (Acid in added water)

The acid content in the resulting mixture is given by:

Percentage of acid = $\frac{\text{Total amount of acid}}{\text{Total volume}} \times 100$

Percentage of acid = $\frac{506.25}{1125 + x} \times 100$

According to the problem, the acid content in the resulting mixture must be more than 25% but less than 30%. This gives us the compound inequality:

Divide the inequality (iv) by 100:

Since $x \geq 0$, the total volume $1125 + x$ is always positive. We can multiply all parts of the inequality (v) by $(1125 + x)$ without changing the direction of the inequality signs.

This compound inequality can be split into two separate inequalities:

Inequality 1: $0.25(1125 + x) < 506.25$

Distribute the 0.25:

Calculate $0.25 \times 1125 = \frac{1}{4} \times 1125 = 281.25$:

Subtract $281.25$ from both sides:

Divide by $0.25$ (positive number):

Inequality 2: $506.25 < 0.30(1125 + x)$

Distribute the 0.30:

Calculate $0.30 \times 1125 = 337.5$:

Subtract $337.5$ from both sides:

Divide by $0.30$ (positive number):

To satisfy the conditions of the problem, $x$ must satisfy both inequality (A) and inequality (B).

From (A), $x < 900$.

From (B), $x > 562.5$.

Combining these two conditions, we get:

$\mathbf{562.5 < x < 900}$

Therefore, the number of litres of water that must be added is more than 562.5 litres but less than 900 litres.

The range of water to be added is $\mathbf{(562.5, 900)}$ litres.

The formula for IQ is given by:

where MA is the mental age and CA is the chronological age.

We are given that the children are 12 years old, so their chronological age is $\text{CA} = 12$ years.

We are also given the range of IQ for this group of children:

Substitute the value of CA = 12 into the IQ formula (i):

Now, substitute the expression for IQ from (iii) into the inequality (ii):

Simplify the term with MA:

So the inequality (iv) becomes:

To isolate MA, first multiply all parts of the inequality (v) by $3$. Since $3$ is positive, the direction of the inequality signs remains unchanged:

Simplifying this gives:

Now, divide all parts of the inequality (vii) by $25$. Since $25$ is positive, the direction of the inequality signs remains unchanged:

Simplify the fractions:

So the inequality (viii) becomes:

Therefore, the range of mental age (MA) for this group of 12-year-old children is between 9.6 years and 16.8 years, inclusive of the endpoints.

The range of their mental age is $\mathbf{9.6 \leq \text{MA} \leq 16.8}$ years.

In interval notation, the range is $\mathbf{[9.6, 16.8]}$.