Chapter 3 Pair Of Linear Equations In Two Variables

The given pair of linear equations is:

$5x - 15y = 8$

$3x - 9y = \frac{24}{5}$

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We can write the equations in the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$.

From the first equation, $a_1 = 5$, $b_1 = -15$, $c_1 = 8$.

From the second equation, $a_2 = 3$, $b_2 = -9$, $c_2 = \frac{24}{5}$.

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Now we compare the ratios of the coefficients:

Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{5}{3}$

Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{-15}{-9} = \frac{15}{9} = \frac{5}{3}$

Ratio of constant terms: $\frac{c_1}{c_2} = \frac{8}{\frac{24}{5}} = 8 \times \frac{5}{24} = \frac{40}{24} = \frac{5}{3}$

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We observe that the ratios are equal:

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This condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ indicates that the pair of linear equations represents coincident lines.

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Coincident lines overlap completely, meaning every point on one line is also a point on the other line.

Therefore, a system of coincident lines has infinitely many solutions.

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Hence, the correct answer is (C) infinitely many solutions.

Let the two-digit number be represented as $10x + y$, where $x$ is the digit in the tens place and $y$ is the digit in the units place.

The number obtained by reversing the digits is $10y + x$.

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According to the first condition, the sum of the digits is 9.

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According to the second condition, if 27 is added to the number, the digits get reversed.

Rearranging the terms:

$10x - x + y - 10y = -27$

$9x - 9y = -27$

Dividing the entire equation by 9:

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Now we have a system of two linear equations with two variables:

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Adding equation (i) and equation (ii):

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Substitute the value of $x = 3$ into equation (i):

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The digits are $x=3$ and $y=6$.

The original two-digit number is $10x + y = 10(3) + 6 = 30 + 6 = 36$.

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Let's check the conditions:

Sum of digits: $3 + 6 = 9$ (Correct)

Number + 27: $36 + 27 = 63$. The reversed number is $10(6) + 3 = 60 + 3 = 63$ (Correct).

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The number is 36.

The correct answer is (D) 36.

The given pair of linear equations is:

$6x - 3y + 10 = 0$

$2x - y + 9 = 0$

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We compare these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

From the first equation:

From the second equation:

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Now, we find the ratios of the coefficients:

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Comparing the ratios, we see that:

But,

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Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and non-coincident.

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Graphically, a pair of linear equations representing parallel lines that are not coincident will never intersect, meaning they have no solution.

Therefore, the pair of equations represents two lines which are parallel.

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The correct answer is (D) parallel.

The given pair of linear equations is:

$x + 2y + 5 = 0$

$-3x - 6y + 1 = 0$

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We compare these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

From the first equation:

From the second equation:

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Now, we find the ratios of the coefficients:

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Comparing the ratios, we observe that:

But,

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Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of linear equations represents parallel lines.

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Parallel lines do not intersect at any point.

Therefore, a system of parallel lines has no solution.

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The correct answer is (D) no solution.

A pair of linear equations is said to be consistent if it has at least one solution.

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There are two cases for a pair of linear equations to have solutions:

1. Unique Solution: This occurs when the lines intersect at exactly one point.

The condition for a unique solution for the pair $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

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2. Infinitely Many Solutions: This occurs when the lines are coincident (one line lies exactly on top of the other).

The condition for infinitely many solutions is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

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If a pair of linear equations is inconsistent, it has no solution. This occurs when the lines are parallel and non-coincident. The condition for no solution is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

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Since a consistent system has at least one solution, the lines must either intersect (unique solution) or be coincident (infinitely many solutions).

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Therefore, if a pair of linear equations is consistent, the lines will be intersecting or coincident.

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The correct answer is (C) intersecting or coincident.

The given pair of equations is:

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The equation $y = 0$ represents the x-axis in the Cartesian coordinate system.

The equation $y = -7$ represents a horizontal line parallel to the x-axis, located 7 units below the x-axis.

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For a pair of linear equations to have a solution, the lines represented by the equations must intersect.

The x-axis ($y=0$) and the line $y = -7$ are two distinct horizontal lines.

Parallel lines that are not coincident never intersect.

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Since these two lines are parallel and distinct, they do not have any point in common.

Therefore, there is no solution that satisfies both equations simultaneously.

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The pair of equations has no solution.

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The correct answer is (D) no solution.

The given pair of equations is:

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The equation $x = a$ represents a vertical line in the Cartesian coordinate system. This line is parallel to the y-axis and passes through the point $(a, 0)$.

The equation $y = b$ represents a horizontal line in the Cartesian coordinate system. This line is parallel to the x-axis and passes through the point $(0, b)$.

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A vertical line and a horizontal line in a 2-dimensional plane will always intersect at exactly one point, unless one of the constants $a$ or $b$ makes the equations invalid (which is not the case here as $a$ and $b$ are constants representing specific values).

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The point where the two lines intersect must satisfy both equations simultaneously.

From the first equation, the x-coordinate of any point on the line is $a$.

From the second equation, the y-coordinate of any point on the line is $b$.

Therefore, the point of intersection is $(a, b)$.

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Graphically, the pair of equations $x = a$ and $y = b$ represents two lines which are intersecting at (a, b).

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The correct answer is (D) intersecting at (a, b).

The given pair of linear equations is:

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We can rewrite the second equation in the standard form $a_2x + b_2y + c_2 = 0$:

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Comparing the given equations with the standard forms $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we get:

From the first equation ($3x - y + 8 = 0$):

From the second equation ($6x - ky + 16 = 0$):

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For the lines to be coincident, the ratios of the corresponding coefficients must be equal:

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Let's set up the ratios:

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For coincident lines, we must have $\frac{a_1}{a_2} = \frac{b_1}{b_2}$.

Cross-multiplying gives:

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We should also check if $\frac{b_1}{b_2} = \frac{c_1}{c_2}$ for $k=2$.

Since $\frac{1}{2} = \frac{1}{2}$, the condition $\frac{b_1}{b_2} = \frac{c_1}{c_2}$ is also satisfied when $k=2$.

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Thus, for the equations to represent coincident lines, the value of k must be 2.

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The correct answer is (C) 2.

The given pair of linear equations is:

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We can rewrite the first equation in the standard form $a_1x + b_1y + c_1 = 0$:

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Comparing the equations with the standard forms $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we get:

From the first equation ($3x + 2ky - 2 = 0$):

From the second equation ($2x + 5y + 1 = 0$):

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For two linear equations to represent parallel lines, the condition is:

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We use the equality part of the condition to find the value of k:

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Cross-multiply to solve for k:

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We should also verify the inequality $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ for $k = \frac{15}{4}$.

Since $\frac{3}{2} \neq -2$, the condition for parallel lines is satisfied for $k = \frac{15}{4}$.

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The value of k for which the given lines are parallel is $\frac{15}{4}$.

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The correct answer is (C) $\frac{15}{4}$.

The given pair of linear equations is:

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We rewrite these equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Comparing the coefficients with the standard forms, we get:

From the first equation ($cx - y - 2 = 0$):

From the second equation ($6x - 2y - 3 = 0$):

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For a pair of linear equations to have infinitely many solutions, the condition is that the lines must be coincident. This happens when the ratios of the corresponding coefficients are equal:

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Let's calculate the ratios:

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For infinitely many solutions, we must have all three ratios equal:

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Let's check the equality of the last two ratios:

To check if this is true, we can cross-multiply:

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This is false. The ratio $\frac{b_1}{b_2}$ is $\frac{1}{2}$ and the ratio $\frac{c_1}{c_2}$ is $\frac{2}{3}$, and $\frac{1}{2} \neq \frac{2}{3}$.

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Since the condition $\frac{b_1}{b_2} = \frac{c_1}{c_2}$ is not satisfied, the lines can never be coincident, regardless of the value of c.

This pair of equations actually represents parallel lines because $\frac{a_1}{a_2} = \frac{c}{6}$, $\frac{b_1}{b_2} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{2}{3}$. If we set $\frac{c}{6} = \frac{1}{2}$, we get $c=3$. In that case, the ratios would be $\frac{3}{6} = \frac{1}{2}$, $\frac{-1}{-2} = \frac{1}{2}$, and $\frac{-2}{-3} = \frac{2}{3}$. This satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, which is the condition for parallel lines (no solution).

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Therefore, there is no value of c for which the pair of equations will have infinitely many solutions.

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The correct answer is (D) no value.

A pair of linear equations is dependent if they represent coincident lines. This means that the lines overlap completely, and there are infinitely many solutions.

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For two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be dependent (coincident), the ratios of their corresponding coefficients must be equal:

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The first given equation is $-5x + 7y = 2$. We can rewrite it in the standard form as $-5x + 7y - 2 = 0$.

From this equation, we have:

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Now, let's examine each option for the second equation and check the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

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Option (A): $10x + 14y + 4 = 0$

$a_2 = 10$, $b_2 = 14$, $c_2 = 4$

Here, $\frac{a_1}{a_2} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = -\frac{1}{2}$. The ratios are not equal.

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Option (B): –10x – 14y + 4 = 0

$a_2 = -10$, $b_2 = -14$, $c_2 = 4$

Here, $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = -\frac{1}{2}$, $\frac{c_1}{c_2} = -\frac{1}{2}$. The ratios are not equal.

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Option (C): –10x + 14y + 4 = 0

$a_2 = -10$, $b_2 = 14$, $c_2 = 4$

Here, $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = -\frac{1}{2}$. The first two ratios are equal, but the third is different. This represents parallel lines (inconsistent system).

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Option (D): 10x – 14y = –4

We rewrite this in the standard form as $10x - 14y + 4 = 0$.

$a_2 = 10$, $b_2 = -14$, $c_2 = 4$

Here, $\frac{a_1}{a_2} = -\frac{1}{2}$, $\frac{b_1}{b_2} = -\frac{1}{2}$, $\frac{c_1}{c_2} = -\frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$, the condition for coincident lines is satisfied.

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Therefore, the second equation that can form a pair of dependent linear equations with $-5x + 7y = 2$ is $10x – 14y = –4$.

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The correct answer is (D) 10x – 14y = –4.

The given unique solution is $x = 2$ and $y = -3$.

For a pair of linear equations to have this unique solution, the point $(2, -3)$ must satisfy both equations in the pair.

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Let's check each option by substituting $x = 2$ and $y = -3$ into the equations.

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Option (A): $x + y = -1$ and $2x - 3y = -5$

Equation 1: $x + y = -1$

Substitute $x=2, y=-3$: $2 + (-3) = 2 - 3 = -1$. This is true.

Equation 2: $2x - 3y = -5$

Substitute $x=2, y=-3$: $2(2) - 3(-3) = 4 - (-9) = 4 + 9 = 13$. This is not equal to $-5$.

Since the second equation is not satisfied, option (A) is incorrect.

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Option (B): $2x + 5y = -11$ and $4x + 10y = -22$

Equation 1: $2x + 5y = -11$

Substitute $x=2, y=-3$: $2(2) + 5(-3) = 4 - 15 = -11$. This is true.

Equation 2: $4x + 10y = -22$

Substitute $x=2, y=-3$: $4(2) + 10(-3) = 8 - 30 = -22$. This is true.

Both equations are satisfied by the point $(2, -3)$. However, let's check the nature of the solution for this pair.

The equations are $2x + 5y + 11 = 0$ and $4x + 10y + 22 = 0$.

Comparing coefficients: $a_1=2, b_1=5, c_1=11$ and $a_2=4, b_2=10, c_2=22$.

Ratios: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{5}{10} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{11}{22} = \frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, and the system has infinitely many solutions, not a unique solution.

Option (B) is incorrect.

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Option (C): $2x - y = 1$ and $3x + 2y = 0$

Equation 1: $2x - y = 1$

Substitute $x=2, y=-3$: $2(2) - (-3) = 4 + 3 = 7$. This is not equal to 1.

Since the first equation is not satisfied, option (C) is incorrect.

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Option (D): $x - 4y - 14 = 0$ and $5x - y - 13 = 0$

Equation 1: $x - 4y - 14 = 0$

Substitute $x=2, y=-3$: $2 - 4(-3) - 14 = 2 + 12 - 14 = 14 - 14 = 0$. This is true.

Equation 2: $5x - y - 13 = 0$

Substitute $x=2, y=-3$: $5(2) - (-3) - 13 = 10 + 3 - 13 = 13 - 13 = 0$. This is true.

Both equations are satisfied by the point $(2, -3)$.

Let's check the nature of the solution for this pair to ensure it's unique.

The equations are $x - 4y - 14 = 0$ and $5x - y - 13 = 0$.

Comparing coefficients: $a_1=1, b_1=-4, c_1=-14$ and $a_2=5, b_2=-1, c_2=-13$.

Ratios: $\frac{a_1}{a_2} = \frac{1}{5}$, $\frac{b_1}{b_2} = \frac{-4}{-1} = 4$.

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ ($\frac{1}{5} \neq 4$), the lines intersect at exactly one point, meaning the system has a unique solution.

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Since both equations in option (D) are satisfied by $x=2, y=-3$, and the system has a unique solution, option (D) is the correct answer.

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The correct answer is (D) x – 4y – 14 = 0, 5x – y – 13 = 0.

The given pair of linear equations is:

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We can solve this system using the elimination method.

Adding equation (i) and equation (ii):

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Now, substitute the value of $x = 3$ into equation (ii):

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The solution to the system of equations is $x = 3$ and $y = 1$.

We are given that the solution is $x = a$ and $y = b$.

Therefore, we have $a = 3$ and $b = 1$.

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The values of a and b are 3 and 1, respectively.

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The correct answer is (C) 3 and 1.

Let the number of Re 1 coins be $x$.

Let the number of Rs 2 coins be $y$.

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According to the first condition, the total number of coins is 50.

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According to the second condition, the total amount of money is $\textsf{₹}75$.

The value of $x$ Re 1 coins is $1 \times x = x$ rupees.

The value of $y$ Rs 2 coins is $2 \times y = 2y$ rupees.

The total value is the sum of the values of Re 1 and Rs 2 coins.

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We now have a system of two linear equations:

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Subtract equation (i) from equation (ii):

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Substitute the value of $y = 25$ into equation (i):

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So, the number of Re 1 coins is $x = 25$ and the number of Rs 2 coins is $y = 25$.

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The number of Re 1 and Rs 2 coins are 25 and 25, respectively.

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The correct answer is (D) 25 and 25.

Given:

1. The father's current age is six times his son's current age.

2. Four years from now, the father's age will be four times his son's age at that time.

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To Find:

The present ages of the son and the father.

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Solution:

Let the present age of the son be $s$ years.

Let the present age of the father be $f$ years.

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According to the first condition:

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Four years hence:

Son's age will be $(s+4)$ years.

Father's age will be $(f+4)$ years.

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According to the second condition:

Expand the right side:

Rearrange the equation:

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Now we have a system of two linear equations:

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Substitute the value of $f$ from equation (i) into equation (ii):

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Now substitute the value of $s = 6$ into equation (i) to find $f$:

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The present age of the son is 6 years, and the present age of the father is 36 years.

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Let's verify the answer with the options given.

Option (C) gives son's age = 6 and father's age = 36.

Check condition 1: $36 = 6 \times 6$. This is true.

Check condition 2: In 4 years, son's age will be $6+4 = 10$. In 4 years, father's age will be $36+4 = 40$. Is $40 = 4 \times 10$? Yes, this is true.

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The present ages of the son and the father are 6 years and 36 years, respectively.

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The correct answer is (C) 6 and 36.

Given:

The pair of linear equations:

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To Check:

If the pair of equations has a unique solution.

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Justification:

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, it has a unique solution if and only if the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$. That is, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

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Comparing the given equations with the standard forms:

From the first equation ($-x + 2y + 2 = 0$):

From the second equation ($\frac{1}{2}x - \frac{1}{4}y - 1 = 0$):

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Now, let's calculate the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$:

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Comparing the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$:

So, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

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The condition for a unique solution is satisfied.

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Therefore, the pair of equations has a unique solution.

So, it is true to say that the pair of equations has a unique solution.

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Answer:

True.

Justification:

The given equations are $-x + 2y + 2 = 0$ and $\frac{1}{2}x - \frac{1}{4}y - 1 = 0$.

Comparing coefficients, we have $a_1=-1, b_1=2, a_2=\frac{1}{2}, b_2=-\frac{1}{4}$.

The ratio of x-coefficients is $\frac{a_1}{a_2} = \frac{-1}{1/2} = -2$.

The ratio of y-coefficients is $\frac{b_1}{b_2} = \frac{2}{-1/4} = -8$.

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (as $-2 \neq -8$), the lines represented by the equations intersect at a single point, which means the pair of equations has a unique solution.

Given:

The pair of linear equations:

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To Check:

If the equations represent a pair of coincident lines.

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Justification:

First, let's rewrite the given equations in the standard form $ax + by + c = 0$.

Equation 1: $4x + 3y - 1 = 5$

Equation 2: $12x + 9y = 15$

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For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to represent coincident lines, the ratios of their corresponding coefficients must be equal:

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From the given equations, we identify the coefficients:

For $4x + 3y - 6 = 0$:

For $12x + 9y - 15 = 0$:

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Now, we calculate the ratios of the corresponding coefficients:

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Comparing the ratios, we see that:

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We have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{3}$, but $\frac{c_1}{c_2} = \frac{2}{5}$.

So, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

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The condition for coincident lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$) is not satisfied.

The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ represents a pair of parallel lines, which have no solution.

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Therefore, the given equations do not represent a pair of coincident lines.

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Answer:

No.

Justification:

The given equations are $4x + 3y - 6 = 0$ and $12x + 9y - 15 = 0$.

Comparing coefficients, we get $a_1=4, b_1=3, c_1=-6$ and $a_2=12, b_2=9, c_2=-15$.

We calculate the ratios: $\frac{a_1}{a_2} = \frac{4}{12} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$, and $\frac{c_1}{c_2} = \frac{-6}{-15} = \frac{2}{5}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines represented by the equations are parallel and distinct, not coincident. Thus, they do not represent a pair of coincident lines.

Given:

The pair of linear equations:

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To Check:

If the pair of equations is consistent.

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Justification:

A pair of linear equations is consistent if it has at least one solution. This occurs when the lines are either intersecting (unique solution) or coincident (infinitely many solutions).

The pair is inconsistent if it has no solution, which occurs when the lines are parallel and distinct.

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Let's write the second equation in the standard form $ax + by + c = 0$:

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Now, we compare the coefficients of the given equations with the standard forms $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

From the first equation ($x + 2y - 3 = 0$):

From the second equation ($3x + 6y - 9 = 0$):

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We find the ratios of the corresponding coefficients:

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Comparing the ratios, we observe that:

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This condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ means that the lines represented by the equations are coincident.

Coincident lines overlap completely, so they have infinitely many solutions.

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Since the system has infinitely many solutions, it has at least one solution.

Therefore, the pair of equations is consistent.

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Answer:

Yes, the pair of equations is consistent.

Justification:

The given equations are $x + 2y - 3 = 0$ and $3x + 6y - 9 = 0$.

Comparing coefficients, we have $a_1=1, b_1=2, c_1=-3$ and $a_2=3, b_2=6, c_2=-9$.

The ratios of the corresponding coefficients are $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}$, and $\frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident. A system with coincident lines has infinitely many solutions, which means it is consistent (as it has at least one solution).

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has no solution if the lines they represent are parallel and distinct. This condition is given by:

We will examine each pair of equations.

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(i) 2x + 4y = 3 , 12y + 6x = 6

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Compare the ratios:

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct. Therefore, the pair of equations has no solution.

Answer for (i): Yes

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(ii) x = 2y , y = 2x

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Compare the ratios:

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines are intersecting. Therefore, the pair of equations has a unique solution (which is $x=0, y=0$ in this case).

Answer for (ii): No

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(iii) 3x + y – 3 = 0 , 2x + $\frac{2}{3}$y = 2

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Compare the ratios:

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident. Therefore, the pair of equations has infinitely many solutions.

Answer for (iii): No

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ represents coincident lines if the ratios of the corresponding coefficients are equal:

We examine each pair of equations.

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(i) 3x + $\frac{1}{7}$y = 3 , 7x + 3y = 7

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Comparing the ratios, we see that $\frac{a_1}{a_2} = \frac{3}{7}$, $\frac{b_1}{b_2} = \frac{1}{21}$, and $\frac{c_1}{c_2} = \frac{3}{7}$.

Since $\frac{3}{7} \neq \frac{1}{21}$, the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ is not satisfied.

Therefore, the equations do not represent a pair of coincident lines.

Answer for (i): No

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(ii) –2x – 3y = 1 , 6y + 4x = – 2

Rewrite the equations in the standard form $ax + by + c = 0$:

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Comparing the ratios, we see that $\frac{a_1}{a_2} = -\frac{1}{2}$, $\frac{b_1}{b_2} = -\frac{1}{2}$, and $\frac{c_1}{c_2} = -\frac{1}{2}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the condition for coincident lines is satisfied.

Therefore, the equations represent a pair of coincident lines.

Answer for (ii): Yes

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(iii) $\frac{x}{2}$ + y + $\frac{2}{5}$ = 0 , 4x + 8y + $\frac{5}{16}$ = 0

The equations are already in the standard form $ax + by + c = 0$.

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Calculate the ratios:

Comparing the ratios, we see that $\frac{a_1}{a_2} = \frac{1}{8}$, $\frac{b_1}{b_2} = \frac{1}{8}$, and $\frac{c_1}{c_2} = \frac{32}{25}$.

Since $\frac{1}{8} = \frac{1}{8}$, but $\frac{1}{8} \neq \frac{32}{25}$, we have $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

This condition represents parallel and distinct lines.

Therefore, the equations do not represent a pair of coincident lines.

Answer for (iii): No

Question 1:

A pair of equations has no solution if the lines are parallel: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

(i) $2x + 4y - 3 = 0$, $6x + 12y - 6 = 0$.

$a_1=2, b_1=4, c_1=-3$, $a_2=6, b_2=12, c_2=-6$.

Ratios: $\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{4}{12} = \frac{1}{3}$, $\frac{c_1}{c_2} = \frac{-3}{-6} = \frac{1}{2}$.

Comparison: $\frac{1}{3} = \frac{1}{3} \neq \frac{1}{2}$. The lines are parallel.

Answer for (i): Yes.

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(ii) $x - 2y = 0$, $2x - y = 0$.

$a_1=1, b_1=-2, c_1=0$, $a_2=2, b_2=-1, c_2=0$.

Ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-2}{-1} = 2$.

Comparison: $\frac{1}{2} \neq 2$. The lines are intersecting (unique solution).

Answer for (ii): No.

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(iii) $3x + y - 3 = 0$, $2x + \frac{2}{3}y - 2 = 0$.

$a_1=3, b_1=1, c_1=-3$, $a_2=2, b_2=\frac{2}{3}, c_2=-2$.

Ratios: $\frac{a_1}{a_2} = \frac{3}{2}$, $\frac{b_1}{b_2} = \frac{1}{2/3} = \frac{3}{2}$, $\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$.

Comparison: $\frac{3}{2} = \frac{3}{2} = \frac{3}{2}$. The lines are coincident (infinitely many solutions).

Answer for (iii): No.

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Question 2:

A pair of equations represents coincident lines if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

(i) $3x + \frac{1}{7}y - 3 = 0$, $7x + 3y - 7 = 0$.

$a_1=3, b_1=\frac{1}{7}, c_1=-3$, $a_2=7, b_2=3, c_2=-7$.

Ratios: $\frac{a_1}{a_2} = \frac{3}{7}$, $\frac{b_1}{b_2} = \frac{1/7}{3} = \frac{1}{21}$, $\frac{c_1}{c_2} = \frac{-3}{-7} = \frac{3}{7}$.

Comparison: $\frac{3}{7} \neq \frac{1}{21}$. Not coincident.

Answer for (i): No.

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(ii) $-2x - 3y - 1 = 0$, $4x + 6y + 2 = 0$.

$a_1=-2, b_1=-3, c_1=-1$, $a_2=4, b_2=6, c_2=2$.

Ratios: $\frac{a_1}{a_2} = \frac{-2}{4} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-1}{2}$.

Comparison: $-\frac{1}{2} = -\frac{1}{2} = -\frac{1}{2}$. Coincident.

Answer for (ii): Yes.

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(iii) $\frac{1}{2}x + y + \frac{2}{5} = 0$, $4x + 8y + \frac{5}{16} = 0$.

$a_1=\frac{1}{2}, b_1=1, c_1=\frac{2}{5}$, $a_2=4, b_2=8, c_2=\frac{5}{16}$.

Ratios: $\frac{a_1}{a_2} = \frac{1/2}{4} = \frac{1}{8}$, $\frac{b_1}{b_2} = \frac{1}{8}$, $\frac{c_1}{c_2} = \frac{2/5}{5/16} = \frac{32}{25}$.

Comparison: $\frac{1}{8} = \frac{1}{8} \neq \frac{32}{25}$. Parallel.

Answer for (iii): No.

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Question 3:

A pair of equations is consistent if it has at least one solution (intersecting or coincident). This is true if $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ or $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. It is inconsistent if parallel: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

(i) $-3x - 4y - 12 = 0$, $3x + 4y - 12 = 0$.

$a_1=-3, b_1=-4, c_1=-12$, $a_2=3, b_2=4, c_2=-12$.

Ratios: $\frac{a_1}{a_2} = -1$, $\frac{b_1}{b_2} = -1$, $\frac{c_1}{c_2} = 1$.

Comparison: $-1 = -1 \neq 1$. Parallel (inconsistent).

Answer for (i): No.

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(ii) $\frac{3}{5}x - y - \frac{1}{2} = 0$, $\frac{1}{5}x - 3y - \frac{1}{6} = 0$.

$a_1=\frac{3}{5}, b_1=-1, c_1=-\frac{1}{2}$, $a_2=\frac{1}{5}, b_2=-3, c_2=-\frac{1}{6}$.

Ratios: $\frac{a_1}{a_2} = 3$, $\frac{b_1}{b_2} = \frac{1}{3}$.

Comparison: $3 \neq \frac{1}{3}$. Intersecting (consistent).

Answer for (ii): Yes.

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(iii) $2ax + by - a = 0$, $4ax + 2by - 2a = 0$ ($a,b \neq 0$).

$a_1=2a, b_1=b, c_1=-a$, $a_2=4a, b_2=2b, c_2=-2a$.

Ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{1}{2}$.

Comparison: $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$. Coincident (consistent).

Answer for (iii): Yes.

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(iv) $x + 3y - 11 = 0$, $4x + 12y - 22 = 0$.

$a_1=1, b_1=3, c_1=-11$, $a_2=4, b_2=12, c_2=-22$.

Ratios: $\frac{a_1}{a_2} = \frac{1}{4}$, $\frac{b_1}{b_2} = \frac{3}{12} = \frac{1}{4}$, $\frac{c_1}{c_2} = \frac{-11}{-22} = \frac{1}{2}$.

Comparison: $\frac{1}{4} = \frac{1}{4} \neq \frac{1}{2}$. Parallel (inconsistent).

Answer for (iv): No.

Given:

The pair of linear equations:

Statement: For infinitely many solutions, $\lambda = 1$.

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To Check:

If the statement is true.

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Reasoning:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Identify the coefficients:

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For a pair of linear equations to have infinitely many solutions, the lines must be coincident. The condition for coincident lines is:

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Calculate the ratios of the coefficients:

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For infinitely many solutions, we must have all three ratios equal:

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Let's examine the equality $\frac{1}{2} = -\frac{1}{2}$. This equality is false.

Since the ratio of the y-coefficients is not equal to the ratio of the constant terms ($\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$), the condition for coincident lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$) can never be satisfied for any value of $\lambda$.

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If we check the condition for parallel lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$), we set the first two ratios equal:

If $\lambda = 1$, the ratios become $\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}$. This satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, which is the condition for no solution (parallel lines).

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Therefore, for $\lambda = 1$, the pair of equations has no solution, not infinitely many solutions.

The statement is false.

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Answer:

False.

Reasons:

The given equations are $\lambda x + 3y + 7 = 0$ and $2x + 6y - 14 = 0$.

The coefficients are $a_1=\lambda, b_1=3, c_1=7$ and $a_2=2, b_2=6, c_2=-14$.

For infinitely many solutions, we need $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Calculating the ratios: $\frac{a_1}{a_2} = \frac{\lambda}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{7}{-14} = -\frac{1}{2}$.

For infinitely many solutions, we would need $\frac{1}{2} = -\frac{1}{2}$, which is not true.

Hence, there is no value of $\lambda$ for which the equations have infinitely many solutions.

Specifically, when $\lambda = 1$, the ratios are $\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}$, satisfying $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, which means the lines are parallel and have no solution.

Given:

The pair of linear equations:

Statement: For all real values of c, this pair has a unique solution.

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To Justify:

Whether the given statement is true or false.

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Reasoning:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Identify the coefficients:

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For a pair of linear equations to have a unique solution, the condition is:

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Let's calculate the ratios of the coefficients of x and y:

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Comparing these two ratios, we find that:

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Since the ratio of the x-coefficients is equal to the ratio of the y-coefficients ($\frac{a_1}{a_2} = \frac{b_1}{b_2}$), the condition for a unique solution ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$) is never satisfied for this pair of equations, regardless of the value of c.

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This means the lines are either parallel (no solution) or coincident (infinitely many solutions).

The condition depends on the ratio of the constant terms, $\frac{c_1}{c_2} = \frac{-8}{-c} = \frac{8}{c}$ (assuming $c \neq 0$).

• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (parallel lines, no solution): $\frac{1}{5} \neq \frac{8}{c} \implies c \neq 40$.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (coincident lines, infinitely many solutions): $\frac{1}{5} = \frac{8}{c} \implies c = 40$.

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In either case (c=40 or c $\neq$ 40), the system does not have a unique solution.

Therefore, the statement "For all real values of c, the pair of equations have a unique solution" is false.

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Answer:

False.

Reasons:

The given equations are $x - 2y - 8 = 0$ and $5x - 10y - c = 0$.

Comparing coefficients, we get $a_1=1, b_1=-2, c_1=-8$ and $a_2=5, b_2=-10, c_2=-c$.

We calculate the ratios: $\frac{a_1}{a_2} = \frac{1}{5}$ and $\frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5}$.

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ for all values of c, the lines are either parallel or coincident. They are never intersecting at a single point to have a unique solution.

Specifically, if c = 40, $\frac{c_1}{c_2} = \frac{-8}{-40} = \frac{1}{5}$, so $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{5}$, meaning infinitely many solutions.

If c $\neq$ 40, $\frac{c_1}{c_2} \neq \frac{1}{5}$, so $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, meaning no solution.

Thus, the equations never have a unique solution, and the statement is false.

Given:

The statement: The line represented by $x = 7$ is parallel to the x-axis.

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To Justify:

Whether the statement is true or false.

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Reasoning:

The equation of a line in the Cartesian coordinate system specifies the relationship between the x and y coordinates of all points on that line.

The equation $x = 7$ represents a line where the x-coordinate of every point is always 7, regardless of the y-coordinate.

Graphically, this is a line that passes through the point $(7, 0)$ and is perpendicular to the x-axis. It is a vertical line.

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The x-axis is the horizontal line defined by the equation $y = 0$.

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A vertical line and a horizontal line in a 2-dimensional plane are perpendicular to each other, unless one is the y-axis and the other is the x-axis, in which case they are also perpendicular.

Parallel lines have the same slope and never intersect (or are coincident).

The line $x = 7$ is a vertical line and the x-axis ($y=0$) is a horizontal line. These lines intersect at the point $(7, 0)$ at a 90-degree angle.

Since the line $x = 7$ is perpendicular to the x-axis, it cannot be parallel to the x-axis.

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Therefore, the statement "The line represented by x = 7 is parallel to the x–axis" is false.

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Answer:

False.

Reasons:

The equation $x = 7$ represents a vertical line which is perpendicular to the x-axis. A line parallel to the x-axis is a horizontal line of the form $y = k$. Since a vertical line is not parallel to a horizontal line (unless it is the y-axis and the horizontal line is not the x-axis, which is not the case here), the statement is false.

Given:

The pair of linear equations:

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To Find:

Values of p and q for which the equations have infinitely many solutions.

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Solution:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Identify the coefficients:

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For the pair of linear equations to have infinitely many solutions, the condition is that the lines are coincident. This means the ratios of the corresponding coefficients must be equal:

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Substitute the coefficients into the condition:

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We can equate the first two ratios to form an equation:

Cross-multiply:

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Now, we equate the second and third ratios to form another equation:

Cross-multiply:

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Substitute the value of q from equation (i) into equation (ii):

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Substitute the value of $p = -1$ back into equation (i):

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Thus, for the pair of linear equations to have infinitely many solutions, the values of p and q must be $p = -1$ and $q = 2$.

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Final Answer:

The values are p = -1 and q = 2.

Given:

The pair of linear equations:

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To Find:

The values of $x$ and $y$ that satisfy both equations.

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Solution:

We can solve this system by adding and subtracting the two equations, as the coefficients of $x$ and $y$ are interchanged.

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Add equation (i) and equation (ii):

Divide both sides by 68:

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Subtract equation (i) from equation (ii):

Divide both sides by 26:

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Now we have a simpler system of equations (iii) and (iv):

Add equation (iii) and equation (iv):

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Substitute the value of $x = 3$ into equation (iii):

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The solution to the given pair of linear equations is $x = 3$ and $y = 1$.

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Verification:

Substitute $x=3$ and $y=1$ into equation (i):

Substitute $x=3$ and $y=1$ into equation (ii):

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Final Answer:

The solution is x = 3, y = 1.

Given:

The pair of linear equations:

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To Find:

1. Draw the graphs of the given equations.

2. Calculate the area of the triangle formed by the lines and the x-axis.

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Solution:

To draw the graph of a linear equation, we find at least two points that satisfy the equation.

For Equation (1): $x - y + 2 = 0 \implies y = x + 2$

Let's find some points:

If $x = 0$, $y = 0 + 2 = 2$. Point: $(0, 2)$.

If $y = 0$, $0 = x + 2 \implies x = -2$. Point: $(-2, 0)$.

If $x = 1$, $y = 1 + 2 = 3$. Point: $(1, 3)$.

If $x = -1$, $y = -1 + 2 = 1$. Point: $(-1, 1)$.

Points for plotting Line 1: $(-2, 0), (0, 2), (1, 3), (-1, 1)$.

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For Equation (2): $4x - y - 4 = 0 \implies y = 4x - 4$

Let's find some points:

If $x = 0$, $y = 4(0) - 4 = -4$. Point: $(0, -4)$.

If $y = 0$, $0 = 4x - 4 \implies 4x = 4 \implies x = 1$. Point: $(1, 0)$.

If $x = 2$, $y = 4(2) - 4 = 8 - 4 = 4$. Point: $(2, 4)$.

Points for plotting Line 2: $(0, -4), (1, 0), (2, 4)$.

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Graph:

Draw a Cartesian coordinate system. Plot the points found for each equation and draw a straight line passing through them. Label the lines.

Line 1 passes through $(-2, 0), (0, 2)$, etc.

Line 2 passes through $(1, 0), (0, -4)$, etc.

(Note: Actual drawing cannot be provided in this format, but the points and description guide the user to draw it.)

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Vertices of the triangle:

The triangle is formed by the two lines and the x-axis ($y=0$).

The vertices are the intersection points of these three lines.

Vertex 1: Intersection of Line 1 with the x-axis ($y=0$). This is the x-intercept of Line 1, which is $(-2, 0)$.

Vertex 2: Intersection of Line 2 with the x-axis ($y=0$). This is the x-intercept of Line 2, which is $(1, 0)$.

Vertex 3: Intersection of Line 1 and Line 2.

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To find the intersection of Line 1 ($y = x+2$) and Line 2 ($y = 4x-4$), we can set the expressions for y equal:

Solve for x:

Substitute $x=2$ into either equation to find y. Using $y = x+2$:

The intersection point is $(2, 4)$.

Vertex 3: $(2, 4)$.

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The vertices of the triangle are (-2, 0), (1, 0), and (2, 4).

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Area of the triangle:

The base of the triangle lies on the x-axis, connecting the points $(-2, 0)$ and $(1, 0)$.

Length of the base $= |1 - (-2)| = |1 + 2| = 3$ units.

The height of the triangle is the perpendicular distance from the third vertex $(2, 4)$ to the base (the x-axis). This distance is the absolute value of the y-coordinate of the vertex $(2, 4)$, which is $|4| = 4$ units.

Area of a triangle $= \frac{1}{2} \times \text{base} \times \text{height}$.

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Final Answer:

The vertices of the triangle formed by the lines and the x-axis are $(-2, 0)$, $(1, 0)$, and $(2, 4)$.

The area of the triangle is 6 square units.

Given:

The pair of linear equations:

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To Find:

The values of $\lambda$ for which the system has no solution, infinitely many solutions, or a unique solution.

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Solution:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Comparing with the standard forms, we identify the coefficients:

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The nature of solutions for a pair of linear equations depends on the ratios of their coefficients:

Unique solution: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

No solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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Calculate the ratios of the coefficients:

We need to consider the case where $\lambda = 0$ separately, as $\frac{1}{\lambda}$ and $\lambda^2$ would be undefined or zero.

If $\lambda = 0$, the equations become $y=0$ and $x=1$. This is a system with a unique solution $(x,y)=(1,0)$. Let's see if our general conditions cover this later.

Assuming $\lambda \neq 0$, let's evaluate the conditions based on the ratios.

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Equate the first two ratios:

Multiplying by $\lambda$ (since $\lambda \neq 0$):

This equality holds when $\lambda = 1$ or $\lambda = -1$.

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Equate the second and third ratios:

Multiplying by $\lambda$ (since $\lambda \neq 0$):

The real solution for this equation is $\lambda = 1$. The quadratic factor $\lambda^2 + \lambda + 1$ has no real roots as its discriminant is negative ($1^2 - 4(1)(1) = -3$).

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Now let's answer the specific questions:

(i) No solution: The condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

We need $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, which occurs when $\lambda = 1$ or $\lambda = -1$.

We also need $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, which occurs when $\frac{1}{\lambda} \neq \lambda^2$, i.e., when $\lambda^3 \neq 1$, which means $\lambda \neq 1$ (for real $\lambda$).

Combining these, we need ($\lambda = 1$ or $\lambda = -1$) and ($\lambda \neq 1$).

This is satisfied only when $\lambda = -1$.

Let's check $\lambda = -1$: $\frac{a_1}{a_2} = \frac{-1}{1} = -1$, $\frac{b_1}{b_2} = \frac{1}{-1} = -1$, $\frac{c_1}{c_2} = \frac{-(-1)^2}{-1} = \frac{-1}{-1} = 1$. Ratios are $-1, -1, 1$. We have $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. So, $\lambda = -1$ gives no solution.

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(ii) Infinitely many solutions: The condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

We need $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ ($\lambda = 1$ or $\lambda = -1$) and $\frac{b_1}{b_2} = \frac{c_1}{c_2}$ ($\lambda = 1$).

Both conditions are satisfied only when $\lambda = 1$.

Let's check $\lambda = 1$: $\frac{a_1}{a_2} = \frac{1}{1} = 1$, $\frac{b_1}{b_2} = \frac{1}{1} = 1$, $\frac{c_1}{c_2} = \frac{-(1)^2}{-1} = \frac{-1}{-1} = 1$. Ratios are $1, 1, 1$. We have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. So, $\lambda = 1$ gives infinitely many solutions.

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(iii) A unique solution: The condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

This condition is met when $\lambda \neq 1$ and $\lambda \neq -1$ (based on our analysis of $\frac{a_1}{a_2} = \frac{b_1}{b_2}$).

We should also consider the case $\lambda=0$ which we initially excluded. If $\lambda=0$, the equations are $y=0$ and $x=1$, which has a unique solution $(1,0)$. Since $0 \neq 1$ and $0 \neq -1$, this confirms that $\lambda=0$ is included in the range for unique solution.

Thus, the system has a unique solution for all real values of $\lambda$ except 1 and -1.

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Summary of Results:

(i) The pair of equations has no solution when $\lambda = -1$.

(ii) The pair of equations has infinitely many solutions when $\lambda = 1$.

(iii) The pair of equations has a unique solution when $\lambda \in \mathbb{R} \setminus \{-1, 1\}$, i.e., for all real values of $\lambda$ such that $\lambda \neq 1$ and $\lambda \neq -1$.

Given:

The pair of linear equations:

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To Find:

The value(s) of k for which the system has no solution.

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Solution:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Comparing with the standard forms, we identify the coefficients:

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For a pair of linear equations to have no solution, the lines they represent must be parallel and distinct. The condition for this is:

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Calculate the ratios of the coefficients:

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For no solution, we must satisfy two conditions simultaneously:

Condition 1: $\frac{a_1}{a_2} = \frac{b_1}{b_2}$

Cross-multiply:

Taking the square root of both sides:

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Condition 2: $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Since we are considering cases where $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, which implies $k \neq 0$ (otherwise the ratio $\frac{3}{k}$ is undefined), we can multiply both sides by k:

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For no solution, k must satisfy both conditions:

(k = 6 or k = -6) AND (k $\neq$ 6).

The value of k that satisfies both is $k = -6$.

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Let's verify for $k=-6$:

The ratios are $-\frac{1}{2}, -\frac{1}{2}, \frac{3}{2}$. Since $-\frac{1}{2} = -\frac{1}{2} \neq \frac{3}{2}$, the condition for no solution is satisfied for $k=-6$.

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Now, let's consider the case when $k=6$ from Condition 1. If $k=6$, Condition 2 gives $6 \neq 6$, which is false. So $k=6$ does not give no solution. For $k=6$, the ratios are $\frac{6}{12} = \frac{1}{2}$, $\frac{3}{6} = \frac{1}{2}$, $\frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}$. Ratios are $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$. This satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, which means infinitely many solutions.

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Therefore, the pair of equations has no solution only when $k = -6$.

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Final Answer:

The value of k for which the pair of equations has no solution is k = -6.

Given:

The pair of linear equations:

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To Find:

Values of a and b for which the equations have infinitely many solutions.

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Solution:

Rewrite the given equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Comparing with the standard forms, we identify the coefficients:

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For the pair of linear equations to have infinitely many solutions, the lines must be coincident. The condition for this is:

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Substitute the coefficients into the condition:

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Equate the first two ratios:

Cross-multiply:

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Equate the second and third ratios:

Cross-multiply:

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Now we have a system of two linear equations for a and b:

Substitute the value of a from equation (i) into equation (ii):

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Substitute the value of $b = 1$ back into equation (i):

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Thus, for the pair of linear equations to have infinitely many solutions, the values of a and b must be $a = 3$ and $b = 1$.

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Final Answer:

The values are a = 3 and b = 1.

We analyze each pair of equations based on the conditions for different types of solutions.

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(i) Equations: $3x - y - 5 = 0$ and $6x - 2y - p = 0$.

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

For parallel lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Ratios: $\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-5}{-p} = \frac{5}{p}$ (assuming $p \neq 0$).

The condition becomes $\frac{1}{2} = \frac{1}{2} \neq \frac{5}{p}$.

The equality $\frac{1}{2} = \frac{1}{2}$ is always true, indicating the lines always have the same slope.

The inequality $\frac{1}{2} \neq \frac{5}{p}$ requires $p \neq 10$.

If $p=0$, the second equation is $6x - 2y = 0$, which is $3x - y = 0$. $\frac{c_1}{c_2} = \frac{-5}{0}$ (undefined). The first eq is $3x-y=5$. $3x-y=5$ and $3x-y=0$ are parallel lines ($\frac{3}{3}=\frac{-1}{-1} \neq \frac{5}{0}$). So $p=0$ works.

The condition for parallel lines is satisfied for all real values of p except when $\frac{5}{p} = \frac{1}{2}$, which is when $p=10$. If $p=10$, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, giving coincident lines.

Thus, the lines are parallel when $p \neq 10$.

Value(s) of p: p $\neq$ 10.

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(ii) Equations: $-x + py = 1$ and $px - y = 1$.

Standard form: $-x + py - 1 = 0$ and $px - y - 1 = 0$.

Comparing coefficients: $a_1 = -1, b_1 = p, c_1 = -1$; $a_2 = p, b_2 = -1, c_2 = -1$.

For no solution, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Ratios: $\frac{a_1}{a_2} = \frac{-1}{p}$, $\frac{b_1}{b_2} = \frac{p}{-1} = -p$, $\frac{c_1}{c_2} = \frac{-1}{-1} = 1$.

The condition becomes $\frac{-1}{p} = -p \neq 1$. (Assume $p \neq 0$)

Equality part: $\frac{-1}{p} = -p \implies -1 = -p^2 \implies p^2 = 1 \implies p = \pm 1$.

Inequality part: $-p \neq 1 \implies p \neq -1$.

Combining the conditions ($p=1$ or $p=-1$) and ($p \neq -1$), the value of p is $p=1$.

Let's check $p=0$: Eq1: $-x=1 \implies x=-1$. Eq2: $-y=1 \implies y=-1$. Unique solution $(-1,-1)$. So $p=0$ does not give no solution.

Thus, the pair of equations has no solution when $p = 1$.

Value(s) of p: p = 1.

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(iii) Equations: $-3x + 5y = 7$ and $2px - 3y = 1$.

Standard form: $-3x + 5y - 7 = 0$ and $2px - 3y - 1 = 0$.

Comparing coefficients: $a_1 = -3, b_1 = 5, c_1 = -7$; $a_2 = 2p, b_2 = -3, c_2 = -1$.

For intersecting lines at a unique point, the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Ratios: $\frac{a_1}{a_2} = \frac{-3}{2p}$ (assuming $p \neq 0$), $\frac{b_1}{b_2} = \frac{5}{-3} = -\frac{5}{3}$.

The condition becomes $\frac{-3}{2p} \neq -\frac{5}{3}$.

Cross-multiply: $(-3)(3) \neq (2p)(-5) \implies -9 \neq -10p$.

Dividing by -10 (and reversing inequality direction because we divide by a negative number): $\frac{-9}{-10} \neq p \implies p \neq \frac{9}{10}$.

If $p=0$, the equations are $-3x+5y=7$ and $-3y=1 \implies y=-1/3$. Substituting $y=-1/3$ into the first eq: $-3x + 5(-1/3) = 7 \implies -3x - 5/3 = 7 \implies -3x = 7 + 5/3 = 26/3 \implies x = -26/9$. Unique solution $(-26/9, -1/3)$. So $p=0$ gives a unique solution, and $0 \neq 9/10$.

Thus, the lines are intersecting at a unique point for all values of p except $\frac{9}{10}$.

Value(s) of p: p $\neq \frac{9}{10}$.

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(iv) Equations: $2x + 3y – 5 = 0$ and $px – 6y – 8 = 0$.

Standard form: $2x + 3y - 5 = 0$ and $px - 6y - 8 = 0$.

Comparing coefficients: $a_1 = 2, b_1 = 3, c_1 = -5$; $a_2 = p, b_2 = -6, c_2 = -8$.

For a unique solution, the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Ratios: $\frac{a_1}{a_2} = \frac{2}{p}$ (assuming $p \neq 0$), $\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}$.

The condition becomes $\frac{2}{p} \neq -\frac{1}{2}$.

Cross-multiply: $2 \times 2 \neq p \times (-1) \implies 4 \neq -p$.

Multiplying by -1: $-4 \neq p$.

If $p=0$, the equations are $2x+3y=5$ and $-6y=8$. The second equation gives $y=-8/6=-4/3$. Substituting into the first: $2x + 3(-4/3) = 5 \implies 2x - 4 = 5 \implies 2x = 9 \implies x = 9/2$. Unique solution $(9/2, -4/3)$. So $p=0$ gives a unique solution, and $0 \neq -4$.

Thus, the pair of equations has a unique solution for all values of p except -4.

Value(s) of p: p $\neq$ -4.

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(v) Equations: $2x + 3y = 7$ and $2px + py = 28 – qy$.

Standard form:

Rewrite the second equation: $2px + py + qy - 28 = 0 \implies 2px + (p+q)y - 28 = 0$.

Comparing coefficients: $a_1 = 2, b_1 = 3, c_1 = -7$; $a_2 = 2p, b_2 = p+q, c_2 = -28$.

For infinitely many solutions, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Ratios: $\frac{a_1}{a_2} = \frac{2}{2p} = \frac{1}{p}$ (assuming $p \neq 0$), $\frac{b_1}{b_2} = \frac{3}{p+q}$ (assuming $p+q \neq 0$), $\frac{c_1}{c_2} = \frac{-7}{-28} = \frac{1}{4}$.

The condition becomes $\frac{1}{p} = \frac{3}{p+q} = \frac{1}{4}$.

From $\frac{1}{p} = \frac{1}{4}$, we get $p = 4$.

From $\frac{3}{p+q} = \frac{1}{4}$, we get $3 \times 4 = 1 \times (p+q) \implies 12 = p+q$.

Substitute $p=4$ into $12 = p+q$: $12 = 4 + q \implies q = 12 - 4 = 8$.

The values are $p=4$ and $q=8$. Let's check $p \neq 0$ and $p+q \neq 0$. $4 \neq 0$ and $4+8=12 \neq 0$. The assumptions are valid.

Thus, the pair of equations has infinitely many solutions when $p=4$ and $q=8$.

Values of p and q: p = 4, q = 8.

Given:

The equations representing two straight paths:

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To Check:

Whether the paths cross each other (i.e., whether the lines intersect).

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Solution:

Two straight lines cross each other if and only if they intersect at a unique point. In the context of a pair of linear equations, this corresponds to having a unique solution.

If the lines are parallel and distinct (no solution) or coincident (infinitely many solutions), they do not cross at a single, distinct point.

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To determine if the lines intersect, we examine the ratios of the coefficients of the given linear equations.

Rewrite the equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

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Comparing with the standard forms, we identify the coefficients:

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Calculate the ratios of the coefficients:

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Compare the ratios:

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We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, but $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (since $-\frac{1}{2} \neq \frac{2}{5}$).

The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ indicates that the lines represented by the equations are parallel and distinct.

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Parallel and distinct lines never intersect. Therefore, the two straight paths do not cross each other.

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Final Answer:

No, the paths do not cross each other.

Reason:

Comparing the coefficients of the equations $x - 3y - 2 = 0$ and $-2x + 6y - 5 = 0$, we find the ratios $\frac{a_1}{a_2} = \frac{1}{-2} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct, meaning they do not intersect and hence the paths do not cross each other.

Given:

A unique solution $x = -1$, $y = 3$.

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To Find:

1. A pair of linear equations that has this unique solution.

2. The number of such pairs that can be written.

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Solution:

To find a pair of linear equations that has $x = -1$ and $y = 3$ as its unique solution, we can construct two linear equations such that the point $(-1, 3)$ satisfies both equations, and the lines represented by these equations are intersecting (not parallel or coincident).

A general form of a linear equation is $Ax + By = C$. If the point $(-1, 3)$ lies on this line, it must satisfy the equation:

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Let's choose simple values for A and B to create two distinct equations.

Equation 1: Let $A = 1$ and $B = 1$.

So, the first equation is $1x + 1y = 2$, or $x + y = 2$.

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Equation 2: We need to choose values for A and B (let's call them $A'$ and $B'$) such that the resulting equation is different from the first and the lines are not parallel. The condition for non-parallel lines is $\frac{A}{A'} \neq \frac{B}{B'}$. For the first equation, $A=1, B=1$. So we need $\frac{1}{A'} \neq \frac{1}{B'}$, i.e., $A' \neq B'$.

Let $A' = 2$ and $B' = 1$. Note that $A' \neq B'$ (2 $\neq$ 1).

So, the second equation is $2x + 1y = 1$, or $2x + y = 1$.

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The pair of linear equations is:

Let's check if $(x, y) = (-1, 3)$ is a solution:

Eq 1: $-1 + 3 = 2$ (True)

Eq 2: $2(-1) + 3 = -2 + 3 = 1$ (True)

To confirm it's a unique solution, check the ratios of coefficients: $a_1=1, b_1=1, a_2=2, b_2=1$. $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{1}{1} = 1$. Since $\frac{1}{2} \neq 1$, the lines are intersecting and have a unique solution.

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Number of such pairs:

A unique solution $(x_0, y_0)$ represents a single point in the Cartesian plane. A pair of linear equations has $(x_0, y_0)$ as its unique solution if and only if the lines represented by the equations intersect at exactly that point.

Geometrically, infinitely many distinct lines can pass through a single point in a plane. Any two distinct lines passing through the point $(-1, 3)$ will intersect at that point and only that point, thus forming a pair of linear equations with $(-1, 3)$ as the unique solution.

Since we can write infinitely many distinct linear equations passing through $(-1, 3)$ (by choosing different values for A and B in $Ax + By = -A + 3B$, as long as A and B are not both zero), and any two of these equations whose lines are not parallel will form such a pair, there are infinitely many such pairs of linear equations.

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Final Answer:

One pair of linear equations with the unique solution $x=-1, y=3$ is:

$x + y = 2$

$2x + y = 1$

You can write infinitely many such pairs of linear equations.

Given:

The pair of linear equations:

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To Find:

The values of $5y - 2x$ and $\frac{y}{x} - 2$.

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Solution:

First, we need to solve the given system of equations to find the values of x and y.

We can use the elimination method. Notice that the coefficients of y in the two equations are opposite (1 and -1).

Add equation (i) and equation (ii):

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Now, substitute the value of $x = 7$ into equation (i):

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So, the solution to the system is $x = 7$ and $y = 9$.

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Now, we evaluate the given expressions using $x = 7$ and $y = 9$.

Value of $5y - 2x$:

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Value of $\frac{y}{x} - 2$:

To subtract, find a common denominator:

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Final Answer:

The value of $5y - 2x$ is 31.

The value of $\frac{y}{x} - 2$ is $-\frac{5}{7}$.

Given:

A rectangle with side lengths as shown in the figure. The lengths of the sides are expressed in terms of $x$ and $y$.

The side lengths are given as $x + 3y$, $13$, $3x + y$, and $7$.

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To Find:

The values of $x$ and $y$.

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Solution:

In a rectangle, opposite sides are equal in length.

From the figure, we can equate the lengths of the opposite sides to form a pair of linear equations.

Equating the longer sides:

Equating the shorter sides:

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We now have a system of two linear equations:

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We can solve this system using the elimination method. Multiply equation (ii) by 3:

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Subtract equation (i) from equation (iii):

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Substitute the value of $x = 1$ into equation (i):

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So, the values are $x = 1$ and $y = 4$.

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Let's verify the side lengths with these values:

$x + 3y = 1 + 3(4) = 1 + 12 = 13$ (Correct)

$3x + y = 3(1) + 4 = 3 + 4 = 7$ (Correct)

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Final Answer:

The values are x = 1 and y = 4.

(i) Solve: $x + y = 3.3$ and $\frac{0.6}{3x - 2y} = -1$

Given equations:

From the second equation, since $3x - 2y \neq 0$, we can multiply by $(3x - 2y)$: $0.6 = -1(3x - 2y)$

Multiply equation (1) by 2:

Add equation (3) and equation (2):

Substitute $x = 1.2$ into equation (1):

Solution: x = 1.2, y = 2.1

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(ii) Solve: $\frac{x}{3} + \frac{y}{4} = 4$ and $\frac{5x}{6} - \frac{y}{8} = 4$

Given equations:

Multiply the first equation by the LCM of 3 and 4 (which is 12):

Multiply the second equation by the LCM of 6 and 8 (which is 24):

Add equation (1) and equation (2):

Substitute $x = 6$ into equation (1):

Solution: x = 6, y = 8

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(iii) Solve: $4x + \frac{6}{y} = 15$ and $6x - \frac{8}{y} = 14$, y $\neq$ 0

Given equations:

These equations are not linear. Let $v = \frac{1}{y}$. The equations become linear in terms of x and v:

Multiply equation (3) by 4 and equation (4) by 3 to eliminate v:

Add equation (5) and equation (6):

Substitute $x = 3$ into equation (3):

Now substitute back $v = \frac{1}{y}$:

The condition $y \neq 0$ is satisfied.

Solution: x = 3, y = 2

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(iv) Solve: $\frac{1}{2x} - \frac{1}{y} = -1$ and $\frac{1}{x} + \frac{1}{2y} = 8$, x, y $\neq$ 0

Given equations:

These equations are not linear. Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become linear in terms of u and v:

Multiply both equations by 2 to clear fractions:

Multiply equation (4) by 2:

Add equation (3) and equation (5):

Substitute $u = 6$ into equation (4):

Now substitute back $u = \frac{1}{x}$ and $v = \frac{1}{y}$:

The conditions $x \neq 0, y \neq 0$ are satisfied.

Solution: x = $\frac{1}{6}$, y = $\frac{1}{4}$

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(v) Solve: $43x + 67y = – 24$ and $67x + 43y = 24$

Given equations:

Add equation (1) and equation (2):

Subtract equation (1) from equation (2):

Add equation (3) and equation (4):

Substitute $x = 1$ into equation (3):

Solution: x = 1, y = -1

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(vi) Solve: $\frac{x}{a} + \frac{y}{b} = a + b$ and $\frac{x}{a^2} + \frac{y}{b^2} = 2$, a, b $\neq$ 0

Given equations:

Multiply equation (1) by $\frac{1}{a}$:

Subtract equation (3) from equation (2):

Assuming $a \neq b$, we can divide both sides by $(a-b)$:

Substitute $y = b^2$ into equation (1):

If $a=b \neq 0$, the original equations become $\frac{x}{a} + \frac{y}{a} = 2a \implies x+y=2a^2$ and $\frac{x}{a^2} + \frac{y}{a^2} = 2 \implies x+y=2a^2$. These are coincident lines, having infinitely many solutions. The solution $x=a^2, y=b^2$ gives $x=a^2, y=a^2$, which satisfies $x+y = a^2+a^2 = 2a^2$. So, $x=a^2, y=b^2$ is a valid solution even when $a=b$.

Solution: x = $a^2$, y = $b^2$

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(vii) Solve: $\frac{2xy}{x + y} = \frac{3}{2}$ and $\frac{xy}{2x - y} = \frac{-3}{10}$, x + y $\neq$ 0, 2x – y $\neq$ 0

Given equations:

Assume $x \neq 0$ and $y \neq 0$. Invert the first equation:

Multiply by 6:

Invert the second equation:

Multiply by 3:

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become linear in terms of u and v:

Add equation (3) and equation (4):

Substitute $v = -\frac{2}{3}$ into equation (3):

Now substitute back $u = \frac{1}{x}$ and $v = \frac{1}{y}$:

The conditions $x+y \neq 0$ and $2x-y \neq 0$ are satisfied for $x = 1/2, y = -3/2$ as $x+y = 1/2 - 3/2 = -1$ and $2x-y = 2(1/2) - (-3/2) = 1 + 3/2 = 5/2$. The conditions $x, y \neq 0$ for inversion are also satisfied.

Solution: x = $\frac{1}{2}$, y = $-\frac{3}{2}$

Given:

The pair of linear equations:

Also, $y = \lambda x + 5$ at the solution.

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To Find:

1. The solution (x, y) of the given pair of equations.

2. The value of $\lambda$ using the solution and the equation $y = \lambda x + 5$.

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Solution:

Rewrite the given equations to eliminate fractions and put them in standard form.

Equation 1: $\frac{x}{10} + \frac{y}{5} - 1 = 0$. Multiply by the LCM of 10 and 5 (which is 10):

Equation 2: $\frac{x}{8} + \frac{y}{6} = 15$. Multiply by the LCM of 8 and 6 (which is 24):

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Now we have a system of two linear equations:

Multiply equation (1) by 2 to eliminate y:

Subtract equation (3) from equation (2):

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Substitute the value of $x = 340$ into equation (1):

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The solution of the pair of equations is $x = 340$ and $y = -165$.

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Now we use the relation $y = \lambda x + 5$ and the found values of x and y to find $\lambda$.

Substitute $x = 340$ and $y = -165$ into $y = \lambda x + 5$:

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Final Answer:

The solution of the pair of equations is x = 340, y = -165.

The value of $\lambda$ is $-\frac{1}{2}$.

To solve a pair of linear equations graphically, we plot the graph of each equation on the same coordinate plane. The nature of the solution depends on how the lines intersect:

• Intersecting lines: Unique solution. The system is consistent.
• Parallel lines: No solution. The system is inconsistent.
• Coincident lines: Infinitely many solutions. The system is consistent.

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(i) 3x + y + 4 = 0 , 6x – 2y + 4 = 0

Equation 1: $3x + y + 4 = 0 \implies y = -3x - 4$

Points for Line 1:

x	y = -3x - 4	Point (x, y)
0	-4	(0, -4)
-1	-1	(-1, -1)
-2	2	(-2, 2)

Equation 2: $6x - 2y + 4 = 0 \implies 2y = 6x + 4 \implies y = 3x + 2$

Points for Line 2:

x	y = 3x + 2	Point (x, y)
0	2	(0, 2)
1	5	(1, 5)
-1	-1	(-1, -1)

Draw the x and y axes on a graph paper. Plot the points for each equation and draw a straight line passing through them. Observe the graph.

Graphical Observation: The lines intersect at a single point.

Consistency: Since the lines intersect at a single point, the pair of equations is consistent.

Solution: The point of intersection is the unique solution. From the table, we see that the point (-1, -1) lies on both lines. Graphically, the intersection point is (-1, -1).

Solution: x = -1, y = -1.

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(ii) x – 2y = 6 , 3x – 6y = 0

Equation 1: $x - 2y = 6 \implies 2y = x - 6 \implies y = \frac{1}{2}x - 3$

Points for Line 1:

x	y = $\frac{1}{2}$x - 3	Point (x, y)
0	-3	(0, -3)
6	0	(6, 0)
2	-2	(2, -2)

Equation 2: $3x - 6y = 0 \implies 6y = 3x \implies y = \frac{3}{6}x = \frac{1}{2}x$

Points for Line 2:

x	y = $\frac{1}{2}$x	Point (x, y)
0	0	(0, 0)
2	1	(2, 1)
4	2	(4, 2)

Draw the x and y axes on a graph paper. Plot the points for each equation and draw a straight line passing through them. Observe the graph.

Graphical Observation: The lines have the same slope ($\frac{1}{2}$) but different y-intercepts (-3 and 0). They appear to be parallel and distinct.

Consistency: Since the lines are parallel and distinct, they do not intersect. Therefore, the pair of equations is inconsistent.

Solution: No solution exists.

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(iii) x + y = 3 , 3x + 3y = 9

Equation 1: $x + y = 3 \implies y = -x + 3$

Points for Line 1:

x	y = -x + 3	Point (x, y)
0	3	(0, 3)
3	0	(3, 0)
1	2	(1, 2)

Equation 2: $3x + 3y = 9$. Divide the entire equation by 3: $x + y = 3 \implies y = -x + 3$.

This is the same equation as Line 1.

Points for Line 2 (same as Line 1):

x	y = -x + 3	Point (x, y)
0	3	(0, 3)
3	0	(3, 0)
1	2	(1, 2)

Draw the x and y axes on a graph paper. Plot the points for each equation. Observe the graph.

Graphical Observation: The points for both equations lie on the same line. The lines are coincident.

Consistency: Since the lines are coincident, they have infinitely many points in common. Therefore, the pair of equations is consistent.

Solution: There are infinitely many solutions. Any point (x, y) that satisfies the equation $x + y = 3$ is a solution.

Solution: Infinitely many solutions.

Given:

The pair of linear equations:

---

To Find:

1. Draw the graphs of the given equations.

2. Write the vertices of the triangle formed by the lines and the y-axis.

3. Find the area of this triangle.

---

Solution:

To draw the graph of each linear equation, we find at least two points that satisfy the equation.

For Equation (1): $2x + y = 4 \implies y = 4 - 2x$

Points for plotting Line 1:

x	y = 4 - 2x	Point (x, y)
0	4 - 2(0) = 4	(0, 4)
2	4 - 2(2) = 0	(2, 0)
1	4 - 2(1) = 2	(1, 2)

---

For Equation (2): $2x - y = 4 \implies y = 2x - 4$

Points for plotting Line 2:

x	y = 2x - 4	Point (x, y)
0	2(0) - 4 = -4	(0, -4)
2	2(2) - 4 = 0	(2, 0)
1	2(1) - 4 = -2	(1, -2)

---

Graph:

Draw a Cartesian coordinate system. Plot the points found for each equation and draw a straight line passing through them. Label the lines.

Line 1 passes through $(0, 4), (2, 0)$, etc.

Line 2 passes through $(0, -4), (2, 0)$, etc.

(Note: Actual drawing cannot be provided in this format, but the points and description guide the user to draw it.)

---

Vertices of the triangle:

The triangle is formed by the two lines ($2x + y = 4$ and $2x - y = 4$) and the y-axis (which is the line $x=0$).

The vertices are the intersection points of these three lines.

Vertex 1: Intersection of Line 1 with the y-axis ($x=0$). From the table for Line 1, when $x=0$, $y=4$. This point is $(0, 4)$.

Vertex 2: Intersection of Line 2 with the y-axis ($x=0$). From the table for Line 2, when $x=0$, $y=-4$. This point is $(0, -4)$.

Vertex 3: Intersection of Line 1 and Line 2.

To find the intersection of $2x + y = 4$ and $2x - y = 4$, we can solve the system:

Add equation (1) and equation (2):

Substitute $x=2$ into equation (1):

The intersection point is $(2, 0)$.

Vertex 3: $(2, 0)$.

The vertices of the triangle are (0, 4), (0, -4), and (2, 0).

---

Area of the triangle:

The base of the triangle lies on the y-axis, connecting the two points where the lines intersect the y-axis: $(0, 4)$ and $(0, -4)$.

The length of the base $= |4 - (-4)| = |4 + 4| = 8$ units.

The height of the triangle is the perpendicular distance from the third vertex $(2, 0)$ to the base (the y-axis, $x=0$). This distance is the absolute value of the x-coordinate of the vertex $(2, 0)$.

Height $= |2| = 2$ units.

The area of a triangle is given by the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

---

Final Answer:

The vertices of the triangle formed by the lines and the y-axis are (0, 4), (0, -4), and (2, 0).

The area of the triangle is 8 square units.

Given:

The pair of linear equations:

---

To Find:

1. An equation of a line passing through the solution of the given pair of equations.

2. The number of such lines.

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Solution:

First, we need to find the solution of the given pair of linear equations. This solution is the point where the two lines intersect.

Add equation (1) and equation (2):

---

Substitute the value of $x = 1$ into equation (1):

---

The solution of the given pair of equations is the point $(1, 1)$.

---

We need to write an equation of a line passing through the point $(1, 1)$.

A linear equation representing a line passing through $(1, 1)$ can be written in the form $Ax + By = C$, where $A(1) + B(1) = C$, i.e., $A + B = C$.

We can choose any values for A and B (not both zero) and calculate C to get an equation of a line passing through $(1, 1)$.

For example, let $A = 1$ and $B = 0$. Then $C = 1 + 0 = 1$. The equation is $1x + 0y = 1$, or $x = 1$. This is a vertical line passing through $(1, 1)$.

Another example, let $A = 0$ and $B = 1$. Then $C = 0 + 1 = 1$. The equation is $0x + 1y = 1$, or $y = 1$. This is a horizontal line passing through $(1, 1)$.

Another example, let $A = 1$ and $B = 1$. Then $C = 1 + 1 = 2$. The equation is $x + y = 2$. Note that this is one of the original equations, which is valid as it passes through its own solution point.

Another example, let $A = 2$ and $B = -1$. Then $C = 2 + (-1) = 1$. The equation is $2x - y = 1$. This is the other original equation, also valid.

Another example, let $A = 1$ and $B = -1$. Then $C = 1 + (-1) = 0$. The equation is $x - y = 0$, or $x = y$. This line passes through $(1, 1)$.

---

So, an equation of a line passing through the point representing the solution is $x + y = 2$ (or $2x - y = 1$, or $x = 1$, or $y = 1$, or $x - y = 0$, etc.). Let's provide one such equation.

An example equation is x + y = 2.

---

The point $(1, 1)$ is a single point in the coordinate plane. Infinitely many straight lines can pass through a single point in a plane.

Therefore, we can find infinitely many such lines.

---

Final Answer:

An equation of a line passing through the solution is x + y = 2 (or any other valid line equation like $x=1$, $y=1$, $x-y=0$, etc.).

We can find infinitely many such lines.

Given:

1. $(x + 1)$ is a factor of the polynomial $P(x) = 2x^3 + ax^2 + 2bx + 1$.

2. The linear equation $2a - 3b = 4$.

---

To Find:

The values of $a$ and $b$.

---

Solution:

According to the Factor Theorem, if $(x - k)$ is a factor of a polynomial $P(x)$, then $P(k) = 0$.

In this case, the factor is $(x + 1)$, which can be written as $(x - (-1))$. So, $k = -1$.

Therefore, since $(x + 1)$ is a factor of $P(x) = 2x^3 + ax^2 + 2bx + 1$, we must have $P(-1) = 0$.

Substitute $x = -1$ into the polynomial:

Set $P(-1)$ equal to 0:

---

We are also given a second linear equation in terms of a and b:

---

Now we have a system of two linear equations with two variables a and b:

Multiply equation (i) by 2 to eliminate a:

Subtract equation (iii) from equation (ii):

---

Substitute the value of $b = 2$ into equation (i):

---

The values of a and b are 5 and 2, respectively.

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Final Answer:

The values are a = 5 and b = 2.

Given:

The three angles of a triangle are $x$, $y$, and $40^\circ$.

The difference between the angles $x$ and $y$ is $30^\circ$. This means $|x - y| = 30^\circ$.

---

To Find:

The values of $x$ and $y$.

---

Solution:

The sum of the interior angles in any triangle is always $180^\circ$.

We are given that the difference between $x$ and $y$ is $30^\circ$. This leads to two possible cases:

Case 1: The difference is $x - y = 30^\circ$.

We now have a system of two linear equations:

Adding equation (i) and equation (ii):

Substitute the value of $x$ into equation (i):

So, in Case 1, $x = 85^\circ$ and $y = 55^\circ$. The angles are $85^\circ, 55^\circ, 40^\circ$. Their sum is $85+55+40 = 180^\circ$, and $85-55 = 30^\circ$. This is a valid solution.

---

Case 2: The difference is $y - x = 30^\circ$.

We now have a system of two linear equations:

Adding equation (i) and equation (iii):

Substitute the value of $y$ into equation (i):

So, in Case 2, $x = 55^\circ$ and $y = 85^\circ$. The angles are $55^\circ, 85^\circ, 40^\circ$. Their sum is $55+85+40 = 180^\circ$, and $85-55 = 30^\circ$. This is also a valid solution.

---

Therefore, there are two possible pairs of values for $x$ and $y$.

The possible values are:

• $x = 85^\circ$ and $y = 55^\circ$
• $x = 55^\circ$ and $y = 85^\circ$

Given:

Condition 1: Two years ago, Salim's age was thrice his daughter's age.

Condition 2: Six years later, Salim's age will be four years older than twice his daughter's age.

---

To Find:

Their present ages.

---

Solution:

Let Salim's present age be $S$ years.

Let his daughter's present age be $D$ years.

According to the first condition:

Salim's age two years ago $= S - 2$

Daughter's age two years ago $= D - 2$

According to the second condition:

Salim's age six years later $= S + 6$

Daughter's age six years later $= D + 6$

We now have a system of two linear equations with two variables $S$ and $D$:

Subtract equation (i) from equation (ii):

Substitute the value of $D$ into equation (ii):

Thus, Salim's present age is $38$ years and his daughter's present age is $14$ years.

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Verification:

Two years ago: Salim was $38 - 2 = 36$ years old. Daughter was $14 - 2 = 12$ years old.

Six years later: Salim will be $38 + 6 = 44$ years old. Daughter will be $14 + 6 = 20$ years old.

---

Answer:

Salim's present age is $38$ years.

His daughter's present age is $14$ years.

Given:

Condition 1: The father's present age is twice the sum of the present ages of his two children.

Condition 2: After 20 years, the father's age will be equal to the sum of the ages of his children at that time.

---

To Find:

The present age of the father.

---

Solution:

Let the present age of the father be $F$ years.

Let the sum of the present ages of his two children be $C$ years.

According to the first condition:

After 20 years:

Father's age will be $F + 20$ years.

Since there are two children, each child's age will increase by 20 years. So, the sum of their ages will increase by $20 + 20 = 40$ years.

The sum of the children's ages after 20 years will be $C + 40$ years.

According to the second condition:

Now we have a system of two linear equations:

Substitute the value of $F$ from equation (i) into equation (ii):

Now substitute the value of $C$ back into equation (i) to find $F$:

The present age of the father is 40 years.

---

Verification:

Present ages: Father = 40 years, Sum of children's ages = 20 years.

After 20 years: Father = $40 + 20 = 60$ years. Sum of children's ages = $20 + 40 = 60$ years.

---

Answer:

The present age of the father is $40$ years.

Given:

The ratio of two numbers is $5 : 6$.

When 8 is subtracted from each number, their ratio becomes $4 : 5$.

---

To Find:

The two numbers.

---

Solution:

Let the two numbers be $5x$ and $6x$, where $x$ is a common multiplier.

According to the second condition, when 8 is subtracted from each number, their new values are $5x - 8$ and $6x - 8$. The ratio of these new numbers is $4 : 5$.

We can write this as an equation:

Now, we cross-multiply to solve for $x$:

Rearrange the terms to one side to solve for $x$:

Now substitute the value of $x$ back into the expressions for the two numbers:

First number $= 5x = 5 \times 8 = 40$

Second number $= 6x = 6 \times 8 = 48$

The two numbers are $40$ and $48$.

---

Verification:

The ratio of the numbers is $40 : 48 = \frac{40}{48} = \frac{\cancel{8} \times 5}{\cancel{8} \times 6} = \frac{5}{6}$. This matches the first condition.

Subtract 8 from each number: $40 - 8 = 32$ and $48 - 8 = 40$.

The new ratio is $32 : 40 = \frac{32}{40} = \frac{\cancel{8} \times 4}{\cancel{8} \times 5} = \frac{4}{5}$. This matches the second condition.

---

Answer:

The two numbers are $40$ and $48$.

Given:

Condition 1: If 10 students are sent from hall A to hall B, the number of students in both halls becomes equal.

Condition 2: If 20 students are sent from hall B to hall A, the number of students in hall A becomes double the number of students in hall B.

---

To Find:

The initial number of students in hall A and hall B.

---

Solution:

Let the initial number of students in hall A be $N_A$.

Let the initial number of students in hall B be $N_B$.

According to the first condition, when 10 students are sent from A to B:

Number of students in A becomes $N_A - 10$.

Number of students in B becomes $N_B + 10$.

Since the numbers are equal:

According to the second condition, when 20 students are sent from B to A:

Number of students in A becomes $N_A + 20$.

Number of students in B becomes $N_B - 20$.

The number of students in A becomes double the number in B:

Now we have a system of two linear equations:

Subtract equation (ii) from equation (i):

Substitute the value of $N_B$ into equation (i):

The initial number of students in hall A is $100$ and in hall B is $80$.

---

Verification:

Initial students: A=100, B=80.

Condition 1: 10 from A to B. New counts: A = $100 - 10 = 90$, B = $80 + 10 = 90$. $90 = 90$ (True).

Condition 2: 20 from B to A. New counts: A = $100 + 20 = 120$, B = $80 - 20 = 60$. Is A double B? $120 = 2 \times 60$ (True).

---

Answer:

The number of students in hall A is $100$.

The number of students in hall B is $80$.

Given:

A fixed charge for the first two days and an additional charge for each day thereafter.

Latika paid $\textsf{₹}22$ for keeping a book for six days.

Anand paid $\textsf{₹}16$ for keeping a book for four days.

---

To Find:

The fixed charge and the additional charge for each extra day.

---

Solution:

Let the fixed charge for the first two days be $\textsf{₹}F$.

Let the additional charge for each extra day be $\textsf{₹}E$.

For Latika, the book was kept for 6 days. This is 2 fixed days plus $6 - 2 = 4$ extra days.

The total charge paid by Latika is the fixed charge plus the charge for 4 extra days.

For Anand, the book was kept for 4 days. This is 2 fixed days plus $4 - 2 = 2$ extra days.

The total charge paid by Anand is the fixed charge plus the charge for 2 extra days.

We have a system of two linear equations:

Subtract equation (ii) from equation (i):

Substitute the value of $E$ into equation (ii):

The fixed charge for the first two days is $\textsf{₹}10$ and the additional charge for each extra day is $\textsf{₹}3$.

---

Verification:

For Latika (6 days): Fixed charge ($\textsf{₹}10$) + 4 extra days ($\textsf{₹}3$ per day) = $10 + 4 \times 3 = 10 + 12 = \textsf{₹}22$. This matches the given information.

For Anand (4 days): Fixed charge ($\textsf{₹}10$) + 2 extra days ($\textsf{₹}3$ per day) = $10 + 2 \times 3 = 10 + 6 = \textsf{₹}16$. This matches the given information.

---

Answer:

The fixed charge is $\textsf{₹}10$.

The charge for each extra day is $\textsf{₹}3$.

Given:

Total number of questions answered = 120.

Marks awarded for each correct answer = 1.

Marks deducted for each wrong answer = $\frac{1}{2}$.

Total marks obtained = 90.

---

To Find:

The number of questions answered correctly.

---

Solution:

Let the number of questions answered correctly be $c$.

Let the number of questions answered incorrectly (wrong) be $w$.

The total number of questions answered is 120, so:

The total marks obtained is the sum of marks for correct answers minus the marks deducted for wrong answers.

Marks from correct answers = $1 \times c = c$

Marks deducted for wrong answers = $\frac{1}{2} \times w = \frac{w}{2}$

Total marks = $c - \frac{w}{2}$

We are given that the total marks obtained is 90, so:

We now have a system of two linear equations:

From equation (i), we can express $w$ in terms of $c$:

Substitute this expression for $w$ into equation (ii):

Multiply the entire equation by 2 to eliminate the fraction:

Simplify and solve for $c$:

The number of questions answered correctly is 100.

We can also find the number of wrong answers:

---

Verification:

Correct answers = 100, Wrong answers = 20.

Total questions = $100 + 20 = 120$ (Correct).

Marks obtained = (Correct answers $\times$ 1) - (Wrong answers $\times$ $\frac{1}{2}$)

---

Answer:

Jayanti answered 100 questions correctly.

Given:

ABCD is a cyclic quadrilateral.

The angles are given as:

$\angle A = (6x + 10)^\circ$

$\angle B = (5x)^\circ$

$\angle C = (x + y)^\circ$

$\angle D = (3y – 10)^\circ$

---

To Find:

The values of $x$ and $y$, and the measure of each angle $\angle A$, $\angle B$, $\angle C$, and $\angle D$.

---

Solution:

In a cyclic quadrilateral, the sum of opposite angles is $180^\circ$ (supplementary).

Therefore, we have the following equations:

$\angle A + \angle C = 180^\circ$

Simplify Equation (i):

$6x + x + y + 10 = 180$

Also,

$\angle B + \angle D = 180^\circ$

Simplify Equation (iii):

$5x + 3y - 10 = 180$

Now we have a system of two linear equations with two variables $x$ and $y$:

$7x + y = 170 \quad (1)$

$5x + 3y = 190 \quad (2)$

From Equation (1), we can express $y$ in terms of $x$:

Substitute this expression for $y$ into Equation (2):

$5x + 3(170 - 7x) = 190$

$5x + 510 - 21x = 190$

$5x - 21x = 190 - 510$

$-16x = -320$

Now substitute the value of $x = 20$ into Equation (iv) to find the value of $y$:

$y = 170 - 7(20)$

$y = 170 - 140$

So, the values of $x$ and $y$ are $x=20$ and $y=30$.

---

Now, we find the measure of each angle using the values $x=20$ and $y=30$:

$\angle A = (6x + 10)^\circ = (6(20) + 10)^\circ = (120 + 10)^\circ = \mathbf{130^\circ}$

$\angle B = (5x)^\circ = (5(20))^\circ = \mathbf{100^\circ}$

$\angle C = (x + y)^\circ = (20 + 30)^\circ = \mathbf{50^\circ}$

$\angle D = (3y – 10)^\circ = (3(30) – 10)^\circ = (90 – 10)^\circ = \mathbf{80^\circ}$

---

Verification:

Check if opposite angles are supplementary:

$\angle A + \angle C = 130^\circ + 50^\circ = 180^\circ$

$\angle B + \angle D = 100^\circ + 80^\circ = 180^\circ$

The conditions for a cyclic quadrilateral are satisfied.

---

Final Answer:

The value of $x$ is 20 and the value of $y$ is 30.

The measures of the angles are:

$\angle A = \mathbf{130^\circ}$

$\angle B = \mathbf{100^\circ}$

$\angle C = \mathbf{50^\circ}$

$\angle D = \mathbf{80^\circ}$

Given:

Lines: $x = -2$, $y = 3$, the $x$-axis, and the $y$-axis.

---

To Find:

1. The vertices of the figure formed by these lines.

2. The area of the figure.

---

Solution:

The four lines are:

1. $x = -2$ (a vertical line parallel to the $y$-axis, passing through $x = -2$ on the $x$-axis)

2. $y = 3$ (a horizontal line parallel to the $x$-axis, passing through $y = 3$ on the $y$-axis)

3. The $x$-axis, which is the line $y = 0$.

4. The $y$-axis, which is the line $x = 0$.

---

To find the vertices of the figure formed by these lines, we find the points of intersection of these lines.

Intersection of $x = -2$ and $y = 3$: The point is $(-2, 3)$.

Intersection of $x = -2$ and $y = 0$ (x-axis): Substitute $y=0$ into $x=-2$. The point is $(-2, 0)$.

Intersection of $y = 3$ and $x = 0$ (y-axis): Substitute $x=0$ into $y=3$. The point is $(0, 3)$.

Intersection of $x = 0$ (y-axis) and $y = 0$ (x-axis): This is the origin, the point $(0, 0)$.

These four points form the vertices of the figure.

The vertices are: $(-2, 3)$, $(-2, 0)$, $(0, 3)$, and $(0, 0)$.

The figure formed by these lines is a rectangle.

---

To find the area of the rectangle, we need its length and width.

The length of the side along the $x$-axis (or parallel to it) is the distance between the points $(-2, 0)$ and $(0, 0)$ or $(-2, 3)$ and $(0, 3)$. This distance is $|0 - (-2)| = |2| = 2$ units.

The width of the side along the $y$-axis (or parallel to it) is the distance between the points $(0, 0)$ and $(0, 3)$ or $(-2, 0)$ and $(-2, 3)$. This distance is $|3 - 0| = |3| = 3$ units.

The area of a rectangle is given by Length $\times$ Width.

Area $= 2 \times 3 = 6$ square units.

---

Final Answer:

The vertices of the figure formed by the lines are $(0, 0)$, $(-2, 0)$, $(-2, 3)$, and $(0, 3)$.

The area of the figure is 6 square units.

Given:

The equations of the three lines are:

$5x - y = 5$

$x + 2y = 1$

$6x + y = 17$

---

To Find:

The vertices of the triangle formed by the intersection of these lines.

---

Solution:

The vertices of the triangle are the points where any two of the given lines intersect. We need to find the intersection points of the three pairs of lines.

---

Intersection of Line 1 ($5x - y = 5$) and Line 2 ($x + 2y = 1$):

Equation (1): $5x - y = 5$

Equation (2): $x + 2y = 1$

From Equation (1), we can express $y$ in terms of $x$:

$y = 5x - 5$

Substitute this expression for $y$ into Equation (2):

$x + 2(5x - 5) = 1$

$x + 10x - 10 = 1$

$11x - 10 = 1$

$11x = 1 + 10$

$11x = 11$

Substitute the value of $x = 1$ back into the expression for $y$:

$y = 5(1) - 5$

$y = 5 - 5$

The first vertex (intersection of Line 1 and Line 2) is $(1, 0)$.

---

Intersection of Line 1 ($5x - y = 5$) and Line 3 ($6x + y = 17$):

Equation (1): $5x - y = 5$

Equation (3): $6x + y = 17$

Add Equation (1) and Equation (3) to eliminate $y$:

$(5x - y) + (6x + y) = 5 + 17$

$5x + 6x - y + y = 22$

$11x = 22$

Substitute the value of $x = 2$ back into Equation (3):

$6(2) + y = 17$

$12 + y = 17$

$y = 17 - 12$

The second vertex (intersection of Line 1 and Line 3) is $(2, 5)$.

---

Intersection of Line 2 ($x + 2y = 1$) and Line 3 ($6x + y = 17$):

Equation (2): $x + 2y = 1$

Equation (3): $6x + y = 17$

From Equation (3), we can express $y$ in terms of $x$:

$y = 17 - 6x$

Substitute this expression for $y$ into Equation (2):

$x + 2(17 - 6x) = 1$

$x + 34 - 12x = 1$

$x - 12x = 1 - 34$

$-11x = -33$

Substitute the value of $x = 3$ back into the expression for $y$ from Equation (3):

$y = 17 - 6(3)$

$y = 17 - 18$

The third vertex (intersection of Line 2 and Line 3) is $(3, -1)$.

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Final Answer:

The vertices of the triangle formed by the given lines are $(1, 0)$, $(2, 5)$, and $(3, -1)$.

Given:

In the first case, the selling price of the table (with 10% profit) and the chair (with 25% profit) is $\textsf{₹} 1050$.

In the second case, the selling price of the table (with 25% profit) and the chair (with 10% profit) is $\textsf{₹} 1065$.

---

To Find:

The cost price of the table and the cost price of the chair.

---

Solution:

Let the cost price of the table be $\textsf{₹} x$.

Let the cost price of the chair be $\textsf{₹} y$.

According to the first condition:

Selling price of table with 10% profit = $x + 10\%\text{ of } x = x + \frac{10}{100}x = x + 0.1x = 1.1x$.

Selling price of chair with 25% profit = $y + 25\%\text{ of } y = y + \frac{25}{100}y = y + 0.25y = 1.25y$.

Total selling price in the first case = $1.1x + 1.25y$.

Multiply by 100 to remove decimals:

$110x + 125y = 105000$

Dividing by 5, we get:

According to the second condition:

Selling price of table with 25% profit = $x + 25\%\text{ of } x = x + \frac{25}{100}x = x + 0.25x = 1.25x$.

Selling price of chair with 10% profit = $y + 10\%\text{ of } y = y + \frac{10}{100}y = y + 0.1y = 1.1y$.

Total selling price in the second case = $1.25x + 1.1y$.

Multiply by 100 to remove decimals:

$125x + 110y = 106500$

Dividing by 5, we get:

Now we have a system of two linear equations:

$22x + 25y = 21000$ (i)

$25x + 22y = 21300$ (ii)

Adding equation (i) and equation (ii):

Dividing both sides by 47:

Subtracting equation (i) from equation (ii):

Dividing both sides by 3:

Now we solve the system of equations (iii) and (iv):

$x + y = 900$ (iii)

$x - y = 100$ (iv)

Adding equation (iii) and equation (iv):

Substitute the value of $x$ in equation (iii):

Thus, the cost price of the table is $\textsf{₹} 500$ and the cost price of the chair is $\textsf{₹} 400$.

Given:

Time taken to fill the pool by two pipes together = 12 hours.

Time taken by the larger pipe for partial filling = 4 hours.

Time taken by the smaller pipe for partial filling = 9 hours.

Fraction of the pool filled by partial use = $\frac{1}{2}$.

---

To Find:

Time taken by each pipe to fill the pool separately.

---

Solution:

Let the time taken by the larger pipe to fill the pool separately be $x$ hours.

Let the time taken by the smaller pipe to fill the pool separately be $y$ hours.

Work done by the larger pipe in 1 hour = $\frac{1}{x}$.

Work done by the smaller pipe in 1 hour = $\frac{1}{y}$.

According to the first condition, both pipes together fill the pool in 12 hours.

Work done by both pipes in 1 hour = $\frac{1}{12}$.

According to the second condition, the larger pipe works for 4 hours and the smaller pipe works for 9 hours to fill half the pool.

Work done by the larger pipe in 4 hours = $4 \times \frac{1}{x} = \frac{4}{x}$.

Work done by the smaller pipe in 9 hours = $9 \times \frac{1}{y} = \frac{9}{y}$.

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Substituting these values into the equations (i) and (ii), we get:

From equation (iii), multiply by 4:

Subtract equation (v) from equation (iv):

Substitute the value of $v$ in equation (iii):

Find the LCM of 12 and 30. LCM(12, 30) = 60.

Now, we find $x$ and $y$ using $u = \frac{1}{x}$ and $v = \frac{1}{y}$.

$u = \frac{1}{x} \implies \frac{1}{20} = \frac{1}{x} \implies x = 20$.

$v = \frac{1}{y} \implies \frac{1}{30} = \frac{1}{y} \implies y = 30$.

Therefore, the larger pipe takes 20 hours to fill the pool separately, and the smaller pipe takes 30 hours to fill the pool separately.

Given:

The pair of linear equations:

$2x + y = 6$

$2x - y + 2 = 0$

---

To Find:

The graphical solution of the given pair of equations.

The ratio of the areas of the two triangles formed by the lines with the x-axis and with the y-axis.

---

Solution:

To solve the equations graphically, we find at least two points on each line and plot them on a graph paper. Then we draw the lines passing through these points. The point of intersection of the two lines is the graphical solution.

For the first equation: $2x + y = 6 \implies y = 6 - 2x$

Let's find two points:

If $x = 0$, $y = 6 - 2(0) = 6$. Point: $(0, 6)$

If $y = 0$, $2x = 6 \implies x = 3$. Point: $(3, 0)$

We can also find another point for better accuracy:

If $x = 1$, $y = 6 - 2(1) = 4$. Point: $(1, 4)$

---

For the second equation: $2x - y + 2 = 0 \implies y = 2x + 2$

Let's find two points:

If $x = 0$, $y = 2(0) + 2 = 2$. Point: $(0, 2)$

If $y = 0$, $2x + 2 = 0 \implies 2x = -2 \implies x = -1$. Point: $(-1, 0)$

We can also find another point:

If $x = 1$, $y = 2(1) + 2 = 4$. Point: $(1, 4)$

---

When we plot these points and draw the lines, we observe that the lines intersect at the point $(1, 4)$.

Thus, the graphical solution is $x = 1$ and $y = 4$.

---

Now, we find the two triangles formed by these lines and the coordinate axes.

Triangle formed with the x-axis:

The lines intersect the x-axis ($y=0$) at $(3, 0)$ (from $2x+y=6$) and $(-1, 0)$ (from $2x-y+2=0$).

The two lines intersect each other at $(1, 4)$.

The vertices of the triangle formed by the lines and the x-axis are $(3, 0)$, $(-1, 0)$, and $(1, 4)$.

The base of this triangle lies on the x-axis, from $x = -1$ to $x = 3$.

Length of the base = $|3 - (-1)| = 3 + 1 = 4$ units.

The height of the triangle is the perpendicular distance from the intersection point $(1, 4)$ to the x-axis, which is the absolute value of the y-coordinate.

Height = $|4| = 4$ units.

Area of the triangle with the x-axis ($A_1$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$ square units.

---

Triangle formed with the y-axis:

The lines intersect the y-axis ($x=0$) at $(0, 6)$ (from $2x+y=6$) and $(0, 2)$ (from $2x-y+2=0$).

The two lines intersect each other at $(1, 4)$.

The vertices of the triangle formed by the lines and the y-axis are $(0, 6)$, $(0, 2)$, and $(1, 4)$.

The base of this triangle lies on the y-axis, from $y = 2$ to $y = 6$.

Length of the base = $|6 - 2| = 4$ units.

The height of the triangle is the perpendicular distance from the intersection point $(1, 4)$ to the y-axis, which is the absolute value of the x-coordinate.

Height = $|1| = 1$ unit.

Area of the triangle with the y-axis ($A_2$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 1 = 2$ square units.

---

Now, we find the ratio of the areas of the two triangles ($A_1 : A_2$).

Ratio = $\frac{A_1}{A_2} = \frac{8}{2} = \frac{4}{1}$.

The ratio of the areas of the two triangles formed by the lines with the x-axis and the lines with the y-axis is $4:1$.

Given:

The three linear equations:

$y = x$

$3y = x$

$x + y = 8$

---

To Find:

The vertices of the triangle formed by the given lines graphically.

---

Solution:

To determine the vertices graphically, we need to plot each line on a graph paper. The points where any two lines intersect are the vertices of the triangle.

For the first equation: $y = x$

We find two points on this line:

If $x = 0$, $y = 0$. Point: $(0, 0)$

If $x = 4$, $y = 4$. Point: $(4, 4)$

---

For the second equation: $3y = x$ or $y = \frac{x}{3}$

We find two points on this line:

If $x = 0$, $y = \frac{0}{3} = 0$. Point: $(0, 0)$

If $x = 6$, $y = \frac{6}{3} = 2$. Point: $(6, 2)$

---

For the third equation: $x + y = 8$ or $y = 8 - x$

We find two points on this line:

If $x = 0$, $y = 8 - 0 = 8$. Point: $(0, 8)$

If $x = 8$, $y = 8 - 8 = 0$. Point: $(8, 0)$

---

Plot these points on a graph paper and draw the lines passing through them. The points where any two lines intersect are the vertices of the triangle.

Intersection of $y = x$ and $3y = x$:

Substitute $y=x$ into $3y=x$: $3x = x \implies 2x = 0 \implies x = 0$. Since $y=x$, $y=0$. Intersection point: $(0, 0)$.

Intersection of $y = x$ and $x + y = 8$:

Substitute $y=x$ into $x+y=8$: $x + x = 8 \implies 2x = 8 \implies x = 4$. Since $y=x$, $y=4$. Intersection point: $(4, 4)$.

Intersection of $3y = x$ and $x + y = 8$:

Substitute $x=3y$ into $x+y=8$: $3y + y = 8 \implies 4y = 8 \implies y = 2$. Since $x=3y$, $x = 3(2) = 6$. Intersection point: $(6, 2)$.

---

The lines intersect at the points $(0, 0)$, $(4, 4)$, and $(6, 2)$. These points are the vertices of the triangle formed by the given lines.

The vertices of the triangle are (0, 0), (4, 4), and (6, 2).

Given:

The equations of the lines are $x = 3$, $x = 5$, and $2x - y - 4 = 0$. The fourth line is the x-axis, which has the equation $y = 0$.

---

To Find:

The graphs of the given lines.

The area of the quadrilateral formed by these lines and the x-axis.

---

Solution:

To draw the graphs, we find points on each line.

For the line $x = 3$, all points have an x-coordinate of 3. We can take points like (3, 0), (3, 2), (3, -1), etc. This is a vertical line parallel to the y-axis, passing through $x=3$.

For the line $x = 5$, all points have an x-coordinate of 5. We can take points like (5, 0), (5, 2), (5, -1), etc. This is a vertical line parallel to the y-axis, passing through $x=5$.

For the line $2x - y - 4 = 0$, we can rewrite it as $y = 2x - 4$. We find a few points:

• If $x = 0$, $y = 2(0) - 4 = -4$. Point: $(0, -4)$.
• If $x = 2$, $y = 2(2) - 4 = 4 - 4 = 0$. Point: $(2, 0)$.
• If $x = 3$, $y = 2(3) - 4 = 6 - 4 = 2$. Point: $(3, 2)$.
• If $x = 5$, $y = 2(5) - 4 = 10 - 4 = 6$. Point: $(5, 6)$.

The x-axis is the line $y = 0$. Points on this line have a y-coordinate of 0, e.g., (3, 0), (5, 0), (2, 0), etc.

---

When these lines are plotted on a graph, the quadrilateral is formed by the intersection points of these four lines: $x=3$, $x=5$, $y=2x-4$, and $y=0$.

The vertices of the quadrilateral are the points of intersection:

• Intersection of $x=3$ and $y=0$: $(3, 0)$. Let's call this vertex A.
• Intersection of $x=5$ and $y=0$: $(5, 0)$. Let's call this vertex B.
• Intersection of $x=5$ and $y=2x-4$: Substitute $x=5$ into $y=2x-4 \implies y = 2(5) - 4 = 6$. Point: $(5, 6)$. Let's call this vertex C.
• Intersection of $x=3$ and $y=2x-4$: Substitute $x=3$ into $y=2x-4 \implies y = 2(3) - 4 = 2$. Point: $(3, 2)$. Let's call this vertex D.

The vertices of the quadrilateral are (3, 0), (5, 0), (5, 6), and (3, 2).

---

This quadrilateral ABCD, with vertices A(3,0), B(5,0), C(5,6), and D(3,2), is a trapezoid. The parallel sides are the vertical segments AD and BC, as they lie on the vertical lines $x=3$ and $x=5$. The height of the trapezoid is the perpendicular distance between these parallel lines, which is the distance between $x=3$ and $x=5$.

Length of parallel side AD (on $x=3$) = $|y_D - y_A| = |2 - 0| = 2$ units.

Length of parallel side BC (on $x=5$) = $|y_C - y_B| = |6 - 0| = 6$ units.

Height of the trapezoid = Distance between $x=3$ and $x=5$ = $|5 - 3| = 2$ units.

The area of a trapezoid is given by the formula: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.

The area of the quadrilateral formed by the lines and the x-axis is 8 square units.

Given:

The cost of 4 pens and 4 pencil boxes is $\textsf{₹} 100$.

Three times the cost of a pen is $\textsf{₹} 15$ more than the cost of a pencil box.

---

To Find:

The pair of linear equations representing the given situation.

The cost of a pen and the cost of a pencil box.

---

Solution:

Let the cost of a pen be $\textsf{₹} x$.

Let the cost of a pencil box be $\textsf{₹} y$.

According to the first condition:

The cost of 4 pens is $4x$.

The cost of 4 pencil boxes is $4y$.

Total cost is $\textsf{₹} 100$.

Dividing the equation by 4, we get:

According to the second condition:

Three times the cost of a pen is $3x$.

This is $\textsf{₹} 15$ more than the cost of a pencil box ($y$).

Rearranging the terms, we get:

The pair of linear equations is $x + y = 25$ and $3x - y = 15$.

Now, we solve this system of equations to find the values of $x$ and $y$.

$x + y = 25$ (i)

$3x - y = 15$ (ii)

Adding equation (i) and equation (ii):

Substitute the value of $x = 10$ into equation (i):

Thus, the cost of a pen is $\textsf{₹} 10$ and the cost of a pencil box is $\textsf{₹} 15$.

Given:

The three linear equations:

$3x - y = 3$ (1)

$2x - 3y = 2$ (2)

$x + 2y = 8$ (3)

---

To Find:

The vertices of the triangle formed by the given lines using algebraic methods.

---

Solution:

To find the vertices of the triangle, we need to find the intersection points of each pair of lines by solving the corresponding system of linear equations.

1. Intersection of lines (1) and (2):

Equation (1): $3x - y = 3 \implies y = 3x - 3$

Substitute this expression for $y$ into equation (2):

Substitute $x = 1$ back into $y = 3x - 3$:

The first vertex is (1, 0).

---

2. Intersection of lines (1) and (3):

Equation (1): $3x - y = 3 \implies y = 3x - 3$

Substitute this expression for $y$ into equation (3):

Substitute $x = 2$ back into $y = 3x - 3$:

The second vertex is (2, 3).

---

3. Intersection of lines (2) and (3):

Equation (3): $x + 2y = 8 \implies x = 8 - 2y$

Substitute this expression for $x$ into equation (2):

Substitute $y = 2$ back into $x = 8 - 2y$:

The third vertex is (4, 2).

---

The vertices of the triangle formed by the given lines are (1, 0), (2, 3), and (4, 2).

Given:

Total distance to travel = 14 km.

Case 1: 2 km by rickshaw, remaining (12 km) by bus. Total time = half an hour (30 minutes).

Case 2: 4 km by rickshaw, remaining (10 km) by bus. Total time = half an hour + 9 minutes = 39 minutes.

---

To Find:

The speed of the rickshaw and the speed of the bus in km/hour.

---

Solution:

Let the speed of the rickshaw be $v_r$ km/minute.

Let the speed of the bus be $v_b$ km/minute.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

According to the first condition:

Time taken by rickshaw for 2 km = $\frac{2}{v_r}$ minutes.

Time taken by bus for 12 km = $\frac{12}{v_b}$ minutes.

Total time = 30 minutes.

According to the second condition:

Time taken by rickshaw for 4 km = $\frac{4}{v_r}$ minutes.

Time taken by bus for 10 km = $\frac{10}{v_b}$ minutes.

Total time = 39 minutes.

Let $u = \frac{1}{v_r}$ and $v = \frac{1}{v_b}$. Substituting these values into equations (i) and (ii), we get:

From equation (iii), we can write $2u = 30 - 12v$, or $u = 15 - 6v$.

Substitute the expression for $u$ into equation (iv):

Substitute the value of $v$ into the expression for $u$:

We have $u = \frac{1}{v_r} = 6$ and $v = \frac{1}{v_b} = \frac{3}{2}$.

From $u = 6$, we get $v_r = \frac{1}{6}$ km/minute.

From $v = \frac{3}{2}$, we get $v_b = \frac{1}{\frac{3}{2}} = \frac{2}{3}$ km/minute.

To convert speed from km/minute to km/hour, we multiply by 60 (since 1 hour = 60 minutes).

Speed of rickshaw in km/hour = $v_r \times 60 = \frac{1}{6} \times 60 = 10$ km/hour.

Speed of bus in km/hour = $v_b \times 60 = \frac{2}{3} \times 60 = 2 \times 20 = 40$ km/hour.

The speed of the rickshaw is 10 km/hour and the speed of the bus is 40 km/hour.

Given:

Speed of the person in still water = 5 km/h.

Distance travelled upstream = 40 km.

Distance travelled downstream = 40 km.

Time taken upstream is thrice the time taken downstream.

---

To Find:

The speed of the stream.

---

Solution:

Let the speed of the stream be $x$ km/h.

Speed of the person in still water = 5 km/h.

Speed upstream = (Speed in still water) - (Speed of stream)

Speed downstream = (Speed in still water) + (Speed of stream)

We know that $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

Time taken to go 40 km upstream ($T_{upstream}$) = $\frac{40}{5 - x}$ hours.

Time taken to go 40 km downstream ($T_{downstream}$) = $\frac{40}{5 + x}$ hours.

According to the given condition, $T_{upstream} = 3 \times T_{downstream}$.

Multiply both sides by $(5 - x)(5 + x)$ (assuming $x \neq 5$ and $x \neq -5$, which is true for speed):

Divide both sides by 40:

Distribute 3 on the right side:

Bring terms with $x$ to one side and constant terms to the other:

The speed of the stream must be less than the speed of the person in still water for upstream motion to be possible, i.e., $5 - x > 0 \implies x < 5$. Our calculated value $x = 2.5$ satisfies this condition.

The speed of the stream is 2.5 km/h.

Given:

Case 1: Time taken to travel 30 km upstream and 28 km downstream is 7 hours.

Case 2: Time taken to travel 21 km upstream and 21 km downstream (return) is 5 hours.

---

To Find:

The speed of the boat in still water and the speed of the stream.

---

Solution:

Let the speed of the boat in still water be $x$ km/h.

Let the speed of the stream be $y$ km/h.

The speed of the boat upstream is $(x - y)$ km/h.

The speed of the boat downstream is $(x + y)$ km/h.

We know that $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

According to the first condition:

Time taken to travel 30 km upstream = $\frac{30}{x - y}$ hours.

Time taken to travel 28 km downstream = $\frac{28}{x + y}$ hours.

Total time = 7 hours.

According to the second condition:

Time taken to travel 21 km upstream = $\frac{21}{x - y}$ hours.

Time taken to travel 21 km downstream = $\frac{21}{x + y}$ hours.

Total time = 5 hours.

These are non-linear equations. Let $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$. Substituting these into equations (i) and (ii), we get a system of linear equations:

We can solve this system for $u$ and $v$. Multiply equation (iii) by 21 and equation (iv) by 30:

Subtract equation (v) from equation (vi):

Substitute the value of $v = \frac{1}{14}$ into equation (iv):

Now substitute back $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$:

Now we have a simple system of linear equations for $x$ and $y$. Add equation (vii) and equation (viii):

Substitute the value of $x = 10$ into equation (viii):

The speed of the boat in still water is 10 km/h, and the speed of the stream is 4 km/h.

The speed of the boat in still water is 10 km/h and the speed of the stream is 4 km/h.

Given:

A two-digit number satisfies two conditions based on its digits:

Condition 1: The number is 8 times the sum of its digits, minus 5.

Condition 2: The number is 16 times the difference of its digits, plus 3.

---

To Find:

The two-digit number.

---

Solution:

Let the tens digit of the two-digit number be $t$ and the units digit be $u$.

The two-digit number can be represented as $10t + u$.

According to the first condition:

Simplify the equation:

According to the second condition, the number is 16 times the difference of the digits, plus 3.

The "difference of the digits" can mean $(t - u)$ or $(u - t)$. We consider both cases.

Case 1: Assume the difference of the digits is $(t - u)$, where $t \ge u$.

Simplify the equation:

Now we solve the system of linear equations formed by (1) and (2a):

$2t - 7u = -5$ (1)

$-6t + 17u = 3$ (2a)

Multiply equation (1) by 3:

Add equation (3) and equation (2a):

Substitute the value of $u = 3$ into equation (1):

The digits are $t = 8$ and $u = 3$. Since $8 \ge 3$, this is consistent with our assumption for Case 1.

The number is $10t + u = 10(8) + 3 = 83$.

Let's check if this number satisfies the original conditions:

Condition 1: Sum of digits = $8+3=11$. $8(11) - 5 = 88 - 5 = 83$. Correct.

Condition 2: Difference of digits = $8-3=5$. $16(5) + 3 = 80 + 3 = 83$. Correct.

So, 83 is a valid solution.

---

Case 2: Assume the difference of the digits is $(u - t)$, where $u > t$.

Simplify the equation:

Now we solve the system of linear equations formed by (1) and (2b):

$2t - 7u = -5$ (1)

$26t - 15u = 3$ (2b)

Multiply equation (1) by 13:

Subtract equation (4) from equation (2b):

Since $u$ must be an integer digit between 0 and 9, this case does not yield a valid two-digit number.

---

The only valid solution is from Case 1.

The two-digit number is 83.

Given:

Cost of a reserved full first class ticket from A to B = $\textsf{₹} 2530$.

Cost of one reserved full first class ticket and one reserved first class half ticket from A to B = $\textsf{₹} 3810$.

A half ticket costs half the full fare.

Reservation charges are the same for a full ticket and a half ticket.

---

To Find:

The full first class fare from station A to B.

The reservation charges for a ticket.

---

Solution:

Let the full first class fare from station A to B be $\textsf{₹} F$.

Let the reservation charges for a ticket be $\textsf{₹} R$.

The cost of a reserved full first class ticket is the sum of the full fare and the reservation charges.

According to the first condition:

The cost of a reserved first class half ticket is the sum of half the full fare and the reservation charges.

According to the second condition, the cost of one reserved full ticket and one reserved half ticket is $\textsf{₹} 3810$.

Combine like terms:

Now we have a system of two linear equations:

$F + R = 2530$ (1)

$3F + 4R = 7620$ (2)

From equation (1), we can express $R$ in terms of $F$:

Substitute this expression for $R$ into equation (2):

Distribute 4:

Combine like terms:

Now substitute the value of $F$ back into the expression for $R$:

The full first class fare from station A to B is $\textsf{₹} 2500$ and the reservation charges are $\textsf{₹} 30$.

Given:

Case 1: Saree sold at 8% profit and sweater at 10% discount. Total selling price = $\textsf{₹} 1008$.

Case 2: Saree sold at 10% profit and sweater at 8% discount. Total selling price = $\textsf{₹} 1028$.

---

To Find:

The cost price of the saree and the list price of the sweater.

---

Solution:

Let the cost price of the saree be $\textsf{₹} s$.

Let the list price of the sweater be $\textsf{₹} w$.

When an item is sold at a profit of $P\%$ on the cost price (CP), the selling price (SP) is given by $\text{SP} = \text{CP} + \frac{P}{100} \times \text{CP} = \text{CP} \left(1 + \frac{P}{100}\right)$.

When an item is sold at a discount of $D\%$ on the list price (LP), the selling price (SP) is given by $\text{SP} = \text{LP} - \frac{D}{100} \times \text{LP} = \text{LP} \left(1 - \frac{D}{100}\right)$.

According to the first condition:

Selling price of saree = $s \left(1 + \frac{8}{100}\right) = s(1 + 0.08) = 1.08s$.

Selling price of sweater = $w \left(1 - \frac{10}{100}\right) = w(1 - 0.10) = 0.90w$.

Total selling price in Case 1:

Multiply by 100 to remove decimals:

Divide by 18:

According to the second condition:

Selling price of saree = $s \left(1 + \frac{10}{100}\right) = s(1 + 0.10) = 1.10s$.

Selling price of sweater = $w \left(1 - \frac{8}{100}\right) = w(1 - 0.08) = 0.92w$.

Total selling price in Case 2:

Multiply by 100 to remove decimals:

Divide by 2:

Now we have a system of two linear equations:

6s + 5w = 5600 (1)

55s + 46w = 51400 (2)

We can solve this system using the elimination method. Multiply equation (1) by 46 and equation (2) by 5 to make the coefficients of $w$ equal:

Subtract equation (4) from equation (3):

Substitute the value of $s = 600$ into equation (1):

The cost price of the saree is $\textsf{₹} 600$ and the list price of the sweater is $\textsf{₹} 400$.

Given:

Annual interest rate for Scheme A = 8%.

Annual interest rate for Scheme B = 9%.

Case 1: Investment in A + Investment in B gives annual interest = $\textsf{₹} 1860$.

Case 2: Interchanged investments give annual interest = $\textsf{₹} 1860 + \textsf{₹} 20 = \textsf{₹} 1880$.

---

To Find:

The amount of money invested in each scheme.

---

Solution:

Let the amount invested in Scheme A be $\textsf{₹} x$.

Let the amount invested in Scheme B be $\textsf{₹} y$.

The annual interest from Scheme A is 8% of $x$, which is $0.08x$.

The annual interest from Scheme B is 9% of $y$, which is $0.09y$.

According to the first condition:

Multiply by 100 to remove decimals:

According to the second condition, if the investments were interchanged ($y$ in A and $x$ in B), the total interest would be $\textsf{₹} 1880$.

Interest from Scheme A (with amount $y$) = 8% of $y$ = $0.08y$.

Interest from Scheme B (with amount $x$) = 9% of $x$ = $0.09x$.

Multiply by 100 to remove decimals:

Now we have a system of two linear equations:

8$x$ + 9$y$ = 186000 (1)

9$x$ + 8$y$ = 188000 (2)

To solve by elimination, multiply equation (1) by 9 and equation (2) by 8:

Subtract equation (4) from equation (3):

Substitute the value of $y = 10000$ into equation (1):

The amount invested in Scheme A was $\textsf{₹} 12000$, and the amount invested in Scheme B was $\textsf{₹} 10000$.

Given:

Bananas are divided into two lots, A and B.

Case 1: Lot A sold at $\textsf{₹} 2$ for 3 bananas, Lot B sold at $\textsf{₹} 1$ per banana. Total collection = $\textsf{₹} 400$.

Case 2: Lot A sold at $\textsf{₹} 1$ per banana, Lot B sold at $\textsf{₹} 4$ for 5 bananas. Total collection = $\textsf{₹} 460$.

---

To Find:

The total number of bananas Vijay had.

---

Solution:

Let the number of bananas in Lot A be $x$.

Let the number of bananas in Lot B be $y$.

In Case 1:

Selling price per banana in Lot A = $\textsf{₹} \frac{2}{3}$.

Selling price per banana in Lot B = $\textsf{₹} 1$.

Total collection = (Price of bananas in Lot A) + (Price of bananas in Lot B)

Multiply by 3 to clear the fraction:

In Case 2:

Selling price per banana in Lot A = $\textsf{₹} 1$.

Selling price per banana in Lot B = $\textsf{₹} \frac{4}{5}$.

Total collection = (Price of bananas in Lot A) + (Price of bananas in Lot B)

Multiply by 5 to clear the fraction:

Now we solve the system of linear equations:

2x + 3y = 1200 (1)

5x + 4y = 2300 (2)

To eliminate $x$, multiply equation (1) by 5 and equation (2) by 2:

Subtract equation (4) from equation (3):

Substitute the value of $y = 200$ into equation (1):

The number of bananas in Lot A is 300, and the number of bananas in Lot B is 200.

The total number of bananas Vijay had is the sum of the bananas in Lot A and Lot B.

Vijay had a total of 500 bananas.