Linear Inequalities

Introduction to Linear Inequalities

Equations vs. Inequalities

In the realm of algebra, we frequently work with mathematical statements that express relationships between quantities. A primary type of these statements is the equation. An equation is a declaration that two mathematical expressions are equal in value. It is fundamentally characterized by the presence of the equals sign, $=$. For example, the statement $2x + 5 = 11$ is a linear equation in one variable, asserting that the expression $2x+5$ has the same value as the expression $11$.

The goal when dealing with equations is typically to find the specific value(s) of the variable(s) that make the equality true. For instance, the equation $2x + 5 = 11$ is only true when $x = 3$. Linear equations in one variable usually have exactly one solution. Quadratic equations can have up to two solutions, cubic equations up to three, and so on, but the number of solutions is generally finite for polynomial equations.

However, many situations, both in mathematics and the real world, involve comparing quantities that are not necessarily equal but have a clear relative order. Consider scenarios like requiring a speed to be "greater than" a certain limit, a budget to be "less than or equal to" a specific amount, or a height to be "at least" a minimum value. These comparisons are modeled using inequalities.

An inequality is a mathematical statement that indicates that two expressions are not equal, showing the relative order or size of the expressions using specific symbols. Instead of equality, inequalities describe relationships like one expression being strictly greater than, strictly less than, greater than or equal to, or less than or equal to another.

Inequality Symbols

The fundamental symbols used to denote the relationship between two expressions in an inequality are based on their order on the real number line:

• $a < b$: This symbol means that $a$ is less than $b$. On a number line, the point representing $a$ lies strictly to the left of the point representing $b$.
• $a > b$: This symbol means that $a$ is greater than $b$. On a number line, the point representing $a$ lies strictly to the right of the point representing $b$.
• $a \le b$: This symbol means that $a$ is less than or equal to $b$. It is a compound inequality, meaning either $a < b$ is true or $a = b$ is true. On a number line, $a$ is to the left of or coincides with $b$.
• $a \ge b$: This symbol means that $a$ is greater than or equal to $b$. It means either $a > b$ is true or $a = b$ is true. On a number line, $a$ is to the right of or coincides with $b$.
• $a \neq b$: This symbol means that $a$ is not equal to $b$. This is also a type of inequality, indicating that $a$ and $b$ do not have the same value. However, in the context of linear inequalities, we are typically more interested in the order relationships ($<, >, \le, \ge$).

Linear Inequalities

Similar to how we define linear equations, a linear inequality is an inequality that involves algebraic expressions that are linear. A linear expression is characterized by the fact that any variables within it appear only with an exponent of $1$. Furthermore, there are no terms involving products of variables (like $xy$), variables under a radical sign (like $\sqrt{x}$), or variables in the denominator of a fraction (like $\frac{1}{x}$).

Linear inequalities are classified based on the number of variables they contain. The most common types are:

• Linear inequality in one variable: This type of inequality involves only a single variable. It can generally be simplified into one of the following forms, where $a$ and $b$ are real number coefficients and $a$ is non-zero ($a \neq 0$):

  • $ax + b < 0$
  • $ax + b > 0$
  • $ax + b \le 0$
  • $ax + b \ge 0$

  Examples:

  • $2x + 3 \le 7$
  • $5y - 1 > 9$
  • $4 \ge 1 - p$

• Linear inequality in two variables: This type of inequality involves two distinct variables, typically denoted as $x$ and $y$. It can be written in one of the following forms, where $a, b,$ and $c$ are real number coefficients, and at least one of $a$ or $b$ is non-zero:

  • $ax + by + c < 0$
  • $ax + by + c > 0$
  • $ax + by + c \le 0$
  • $ax + by + c \ge 0$

  Equivalently, these can be written as $ax + by < c$, $ax + by > c$, $ax + by \le c$, or $ax + by \ge c$.

  Examples:

  • $x + y > 5$
  • $2x - 3y \le 6$
  • $y < 4x - 1$ (which can be rearranged to $4x - y - 1 > 0$ or $4x - y > 1$)
• Linear inequality in multiple variables: While less commonly introduced initially, linear inequalities can extend to more than two variables (e.g., $2x + 3y - z \le 10$).

In this section, our primary focus will be on understanding and solving linear inequalities involving one or two variables.

Solutions of a Linear Inequality

A solution to a linear inequality is the value (for inequalities in one variable) or the set of values (like an ordered pair $(x, y)$ for inequalities in two variables) of the variable(s) that makes the inequality statement true.

A significant difference between linear equations and linear inequalities lies in their solution sets. While a linear equation in one variable typically has a single solution (or occasionally no solution or infinitely many, in trivial cases like $x=x$ or $x=x+1$), linear inequalities generally have an infinite number of solutions.

Solutions of a Linear Inequality in One Variable:

Consider the simple inequality $x > 3$. Any real number whose value is strictly greater than $3$ is a solution. This includes numbers like $3.1, 4, 5.5, 10, 1000$, $\sqrt{15} \approx 3.87$, $\pi \approx 3.14159$, etc. There are infinitely many such numbers. The set of all solutions is the collection of all real numbers greater than 3, which is represented by the interval $(3, \infty)$.

Solutions of a Linear Inequality in Two Variables:

Consider the inequality $x + y \le 5$. A solution to this inequality is an ordered pair $(x, y)$ of real numbers such that when the values are substituted into the inequality, the statement becomes true. Let's test some pairs:

• For $(x, y) = (1, 2)$: $1 + 2 = 3$. Since $3 \le 5$, the pair $(1, 2)$ is a solution.
• For $(x, y) = (0, 0)$: $0 + 0 = 0$. Since $0 \le 5$, the pair $(0, 0)$ is a solution.
• For $(x, y) = (-5, 10)$: $-5 + 10 = 5$. Since $5 \le 5$, the pair $(-5, 10)$ is a solution.
• For $(x, y) = (6, -1)$: $6 + (-1) = 5$. Since $5 \le 5$, the pair $(6, -1)$ is a solution.
• For $(x, y) = (3, 3)$: $3 + 3 = 6$. Since $6$ is not less than or equal to $5$ ($6 \not\le 5$), the pair $(3, 3)$ is not a solution.
• For $(x, y) = (5, 1)$: $5 + 1 = 6$. Since $6 \not\le 5$, the pair $(5, 1)$ is not a solution.

There are infinitely many ordered pairs $(x, y)$ that satisfy this inequality. The set of all solutions for a linear inequality in two variables represents a region in the coordinate plane.

Solving a linear inequality means identifying and describing this infinite set of solutions. For inequalities in one variable, the solution set is an interval on the number line. For inequalities in two variables, the solution set is a region in the Cartesian coordinate plane, typically represented by shading the appropriate area.

Linear Inequalities in One Variable and Solving Methods

Form of Linear Inequalities in One Variable

A linear inequality in one variable involves a single variable (let's use $x$ as the variable) within expressions that are linear. This means the variable $x$ appears with an exponent of $1$, and there are no terms like $x^2$, $\sqrt{x}$, $1/x$, or products of variables if there were more than one.

Any linear inequality in one variable can be simplified and written in one of the following standard forms, where $a$ and $b$ are real number coefficients, and importantly, $a$ is non-zero ($a \neq 0$). The non-zero condition for $a$ ensures that the inequality is truly "linear" and involves the variable $x$. If $a=0$, the inequality reduces to a statement about constants, like $b < 0$, which is either always true or always false.

• $ax + b < 0$
• $ax + b > 0$
• $ax + b \le 0$
• $ax + b \ge 0$

Other appearances of linear inequalities in one variable can be transformed into one of these standard forms by applying basic algebraic operations (addition, subtraction, multiplication, division) to both sides of the inequality, similar to how we manipulate equations. The key is to gather all terms involving the variable on one side and all constant terms on the other.

For instance, consider the inequality $2x + 5 \ge x - 1$. To convert this into a standard form, we can move terms around:

Subtract $x$ from both sides of the inequality:

Now, subtract $5$ from both sides:

This is a simplified form. To write it in the standard form $ax + b \ge 0$, we can move the constant $-6$ to the left side by adding $6$ to both sides:

This matches the form $ax + b \ge 0$ with $a=1$ and $b=6$. Thus, understanding how to solve the standard forms allows us to solve any linear inequality in one variable.

Solving Linear Inequalities in One Variable

Solving a linear inequality in one variable means determining the complete set of real numbers for which the inequality holds true. The process for solving linear inequalities closely mirrors the steps for solving linear equations, involving isolating the variable. However, there is one critical difference that must be remembered: the rule regarding multiplying or dividing by a negative number.

Fundamental Principles for Manipulating Inequalities:

These principles are based on maintaining the truth of the inequality while transforming its appearance:

1. Adding or Subtracting a Number (or Expression): The direction of an inequality sign remains unchanged if the same real number or algebraic expression is added to or subtracted from both sides of the inequality.

   If $a < b$, then for any real number $k$, $a + k < b + k$ and $a - k < b - k$.

   Similarly, if $a > b$, then $a + k > b + k$ and $a - k > b - k$. These principles also apply to $\le$ and $\ge$ inequalities.

   Example: We know $7 < 10$. Add 5 to both sides: $7+5 < 10+5 \implies 12 < 15$, which is true. Subtract 2 from both sides: $7-2 < 10-2 \implies 5 < 8$, which is also true.

2. Multiplying or Dividing by a Positive Number: If both sides of an inequality are multiplied or divided by the same positive real number, the direction of the inequality sign remains unchanged.

   If $a < b$ and $k$ is a positive real number ($k > 0$), then $ak < bk$ and $\frac{a}{k} < \frac{b}{k}$.

   Similarly, if $a > b$ and $k > 0$, then $ak > bk$ and $\frac{a}{k} > \frac{b}{k}$. These principles apply to $\le$ and $\ge$ inequalities as well.

   Example: We know $4 < 6$. Multiply by 3: $4 \times 3 < 6 \times 3 \implies 12 < 18$, true. Divide by 2: $\frac{4}{2} < \frac{6}{2} \implies 2 < 3$, true.

3. Multiplying or Dividing by a Negative Number: If both sides of an inequality are multiplied or divided by the same negative real number, the direction of the inequality sign must be reversed.

   If $a < b$ and $k$ is a negative real number ($k < 0$), then $ak > bk$ and $\frac{a}{k} > \frac{b}{k}$.

   Similarly, if $a > b$ and $k < 0$, then $ak < bk$ and $\frac{a}{k} < \frac{b}{k}$. If the original inequality was $\le$, it becomes $\ge$; if it was $\ge$, it becomes $\le$.

   Example: We know $2 < 5$. Multiply by $-1$: $2 \times (-1) = -2$ and $5 \times (-1) = -5$. Comparing $-2$ and $-5$, we see that $-2$ is greater than $-5$. So, $2 < 5$ becomes $-2 > -5$. The inequality sign is reversed. Example: We know $-6 \le -2$. Divide by $-2$: $\frac{-6}{-2} = 3$ and $\frac{-2}{-2} = 1$. Comparing 3 and 1, we see that $3$ is greater than $1$. So, $-6 \le -2$ becomes $3 \ge 1$. The inequality sign is reversed from $\le$ to $\ge$.

The standard procedure for solving a linear inequality in one variable is to apply these principles to isolate the variable term first, and then the variable itself, on one side of the inequality sign. The goal is to simplify the inequality into one of the basic forms like $x < c$, $x > c$, $x \le c$, or $x \ge c$, where $c$ is a specific real number.

Examples of Solving Linear Inequalities

Answer:

Answer:

Answer:

Solving linear inequalities in one variable follows standard algebraic procedures for isolating the variable, with the essential rule of reversing the inequality sign when multiplying or dividing by a negative number. The solution set is always an infinite interval (or ray) on the real number line, excluding the boundary point for strict inequalities ($<, >$) and including it for non-strict inequalities ($\le, \ge$).

Graphical Solution of Linear Inequalities (One Variable)

Representing Solutions on a Number Line

We have learned that the solution set for a linear inequality in one variable (such as $x < 5$ or $x \ge -3$) is an interval or a ray of real numbers. While interval notation like $(-\infty, 5)$ or $[-3, \infty)$ is a concise way to represent these solution sets, a graphical representation on a number line offers a visual understanding of which real numbers satisfy the inequality.

The number line is essentially a one-dimensional coordinate system where every real number corresponds to a unique point. Representing the solution set of a one-variable inequality on a number line involves indicating the range of numbers that make the inequality true.

Steps for Graphical Representation

To effectively represent the solution set of a linear inequality in one variable (like $x < c$, $x > c$, $x \le c$, or $x \ge c$, where $c$ is a specific real number) on a number line, follow these steps:

1. Draw a Number Line and Mark the Endpoint: Draw a horizontal line representing the set of real numbers. Mark the specific value $c$ (the number on the right side of the solved inequality, e.g., 5 in $x < 5$, or -3 in $x \ge -3$) on this number line. It's helpful to also mark 0 and perhaps a few other values to provide context.

2. Determine How to Mark the Endpoint ($c$): The nature of the inequality symbol tells us whether the endpoint $c$ itself is included in the solution set or not. We use different graphical symbols at point $c$ to indicate this:

   • If the inequality is strict ($<$ or $>$), the number $c$ is not included in the solution set. Represent this by drawing an open circle at the point corresponding to $c$ on the number line. Alternatively, some notations use a parenthesis $(\ )$ at $c$ facing the direction of the solution.

   • If the inequality is non-strict ($\le$ or $\ge$), the number $c$ is included in the solution set. Represent this by drawing a closed circle (a filled-in circle) at the point corresponding to $c$. Alternatively, a square bracket $[\ ]$ at $c$ facing the direction of the solution can be used.

3. Shade the Appropriate Region: The direction of the inequality symbol indicates which side of $c$ on the number line contains the solutions. Shade the part of the number line that represents the solution set:

   • If the inequality is $x > c$ or $x \ge c$, the solutions are all numbers greater than or greater than or equal to $c$. These numbers lie to the right of $c$ on the number line. So, shade the number line to the right of the marked endpoint.
   • If the inequality is $x < c$ or $x \le c$, the solutions are all numbers less than or less than or equal to $c$. These numbers lie to the left of $c$ on the number line. So, shade the number line to the left of the marked endpoint.

4. Add an Arrow: Since the solution set extends infinitely in one direction, draw an arrow at the end of the shaded region to indicate that it continues indefinitely.

Examples of Graphical Representation

Answer:

The solution to the inequality is $x < 5$.

The endpoint value is $c = 5$.

The inequality symbol is $<$ (less than), which is a strict inequality. Therefore, the endpoint $5$ is not included in the solution set. We will use an open circle at 5.

The inequality $x < 5$ means all numbers strictly less than 5. These numbers are located to the left of 5 on the number line.

Draw a number line, mark 5, place an open circle at 5, and shade the line to the left of 5, adding an arrow pointing left.

*(Visual Description: The image shows a horizontal number line. The number 5 is marked on the line. An open circle is placed directly above 5. The portion of the number line to the left of 5 is shaded, and a left-pointing arrow is at the end of the shaded region, indicating it continues towards negative infinity.)*

Answer:

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

The endpoint value is $c = -3$.

The inequality symbol is $\ge$ (greater than or equal to), which is a non-strict inequality. Therefore, the endpoint $-3$ is included in the solution set. We will use a closed circle at -3.

The inequality $x \ge -3$ means all numbers greater than or equal to -3. These numbers are located to the right of -3 on the number line (including -3 itself).

Draw a number line, mark -3, place a closed circle at -3, and shade the line to the right of -3, adding an arrow pointing right.

*(Visual Description: The image shows a horizontal number line. The number -3 is marked on the line. A closed, filled-in circle is placed directly above -3. The portion of the number line to the right of -3 is shaded, and a right-pointing arrow is at the end of the shaded region, indicating it continues towards positive infinity.)*

Using these visual conventions (open vs. closed circles, shading direction), the graphical representation on a number line clearly illustrates the infinite set of real numbers that satisfy a linear inequality in one variable. This provides a complementary way to understand the solution set alongside interval notation.

Graphical Solution of Linear Inequalities in Two Variables

Representing Solutions as Regions

In the Cartesian coordinate plane, a linear equation involving two variables, typically written as $ax + by = c$ (where $a, b,$ and $c$ are real numbers, and $a$ and $b$ are not both zero), geometrically represents a straight line. This line serves as a boundary that divides the entire plane into two distinct regions, often called half-planes.

A linear inequality in two variables, such as $ax + by < c$, $ax + by > c$, $ax + by \le c$, or $ax + by \ge c$, does not represent a single line. Instead, its solution set comprises all the ordered pairs $(x, y)$ that satisfy the inequality. When plotted on the coordinate plane, these solution points fill an entire half-plane. The solution to a linear inequality in two variables is therefore a continuous region in the plane.

Procedure for Graphing a Linear Inequality in Two Variables

To graphically represent the solution set of a linear inequality in two variables on the coordinate plane, follow these steps:

1. Graph the Boundary Line: The first step is to identify the boundary of the solution region. This boundary is the line associated with the linear inequality. To find the boundary line, temporarily replace the inequality symbol ($<, >, \le, \ge$) with an equals sign $(=)$ and graph the resulting linear equation $ax + by = c$. You can graph this line by finding any two points that satisfy the equation (e.g., the intercepts by setting $x=0$ and $y=0$ successively) and drawing the line passing through them.

2. Determine the Type of Boundary Line: The original inequality symbol tells us whether the points lying directly on the boundary line are included in the solution set or not:

   • If the original inequality is strict ($<$ or $>$), the points on the boundary line are *not* solutions. The line itself is just a visual boundary. Represent this by drawing the boundary line as a dashed or dotted line.
   • If the original inequality is non-strict ($\le$ or $\ge$), the points on the boundary line *are* solutions. Represent this by drawing the boundary line as a solid line.

3. Choose a Test Point: Select any point in the coordinate plane that does *not* lie on the boundary line. The easiest point to use, if the boundary line does not pass through it, is the origin $(0, 0)$. If the boundary line passes through $(0, 0)$, choose any other convenient point (e.g., $(1, 0)$ or $(0, 1)$).

4. Test the Inequality: Substitute the coordinates $(x, y)$ of the chosen test point into the original linear inequality. Evaluate both sides to see if the inequality holds true for that point.

5. Determine and Shade the Solution Region: Based on the result of the test in Step 4, decide which half-plane represents the solution set and shade it:

   • If the test point satisfies the inequality (i.e., makes it a true statement), then the half-plane containing the test point is the solution region. Shade this half-plane.
   • If the test point does not satisfy the inequality (i.e., makes it a false statement), then the half-plane on the side *opposite* to the test point is the solution region. Shade that half-plane.

The shaded region, along with the boundary line (if it's solid), represents all the points $(x, y)$ in the coordinate plane that satisfy the given linear inequality.

Examples of Graphing Linear Inequalities in Two Variables

Answer:

The given inequality is $x + y \ge 5$.

Step 1: Graph the boundary line. The corresponding equation is $x + y = 5$.

To find points on this line:

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

Plot the points $(0, 5)$ and $(5, 0)$ and draw a line through them.

Step 2: Determine Line Type. The inequality symbol is $\ge$ (greater than or equal to), which is non-strict. So, the boundary line $x + y = 5$ is included in the solution set. Draw the line as a solid line.

Step 3: Choose a Test Point. Let's use the origin $(0, 0)$, as it is not on the line $x + y = 5$ (since $0+0=0 \neq 5$).

Step 4: Test the Inequality. Substitute $(0, 0)$ into the original inequality:

$x + y \ge 5$

... (i)

Substitute $x=0, y=0$ into (i):

$$ 0 + 0 \ge 5 $$ $$ 0 \ge 5 $$

This statement is false.

Step 5: Determine and Shade the Solution Region. Since the test point $(0, 0)$ does not satisfy the inequality, the solution region is the half-plane that does *not* contain the origin. The origin is below and to the left of the line $x+y=5$. Therefore, we shade the region above and to the right of the line.

Answer: The solution is the solid line $x+y=5$ and the shaded region above and to its right.

Answer:

The given inequality is $2x - y < 4$.

Step 1: Graph the boundary line. The corresponding equation is $2x - y = 4$.

To find points on this line:

• If $x=0$, then $2(0) - y = 4 \implies -y = 4 \implies y = -4$. Point is $(0, -4)$.
• If $y=0$, then $2x - 0 = 4 \implies 2x = 4 \implies x = 2$. Point is $(2, 0)$.

Plot the points $(0, -4)$ and $(2, 0)$ and draw a line through them.

Step 2: Determine Line Type. The inequality symbol is $<$ (less than), which is strict. So, the boundary line $2x - y = 4$ is not included in the solution set. Draw the line as a dashed line.

Step 3: Choose a Test Point. Let's use the origin $(0, 0)$, as it is not on the line $2x - y = 4$ (since $2(0)-0=0 \neq 4$).

Step 4: Test the Inequality. Substitute $(0, 0)$ into the original inequality:

$2x - y < 4$

... (i)

Substitute $x=0, y=0$ into (i):

$$ 2(0) - 0 < 4 $$ $$ 0 < 4 $$

This statement is true.

Step 5: Determine and Shade the Solution Region. Since the test point $(0, 0)$ satisfies the inequality, the solution region is the half-plane that contains the origin. The origin is above and to the left of the dashed line $2x-y=4$. Therefore, we shade the region above and to the left of the line.

Answer: The solution is the shaded region above and to the left of the dashed line $2x-y=4$.

Graphing linear inequalities in two variables provides a visual representation of their infinite solution sets as regions in the coordinate plane, clearly indicating which points satisfy the given condition. The boundary line's type (solid or dashed) distinguishes between non-strict and strict inequalities.

Solution of a System of Linear Inequalities (One and Two Variables)

Solving Multiple Inequalities Simultaneously

In previous sections, we learned how to solve and graph a single linear inequality in one or two variables. Often, real-world problems involve multiple constraints or conditions that must be satisfied simultaneously. When we have a collection of two or more linear inequalities involving the same variable(s), this is called a system of linear inequalities.

A solution to a system of linear inequalities is a value (if in one variable) or a set of values (an ordered pair $(x, y)$ if in two variables) that satisfies *every* inequality in the system at the same time. Geometrically, finding the solution to a system of inequalities involves finding the intersection of the solution sets of all the individual inequalities in the system. This means finding the region or interval where the solutions of all inequalities overlap.

System of Linear Inequalities in One Variable

A system of linear inequalities in one variable consists of two or more linear inequalities that all involve the same single variable (e.g., $x$). A simple example would be:

• $x + 3 < 7$
• $2x - 1 \ge 5$

To find the solution set for such a system, we use the methods for solving single variable inequalities and then find the values that satisfy all conditions simultaneously.

Procedure to solve a system of linear inequalities in one variable:

1. Solve Each Inequality Individually: Solve each linear inequality in the system independently using the methods discussed earlier (isolating the variable). This will give you a solution set for each inequality, typically in the form of an interval or ray on the number line (e.g., $x < c$ or $x \ge d$).
2. Represent Individual Solutions Graphically: Draw a single number line. On this number line, graphically represent the solution set of *each* inequality from the system. You can use different lines above the number line or different shading methods if needed to keep track of the individual solutions.
3. Find the Intersection: The solution set of the entire system is the set of points that are common to the solution sets of all the individual inequalities. On the graphical representation from Step 2, this is the portion of the number line where the shadings (or lines) for *all* the inequalities overlap. Write this intersection in interval notation or as a single inequality.

Example

Answer:

We need to solve the system: $$ x + 3 < 7 \quad \text{... (1)} $$ $$ 2x - 1 \ge 5 \quad \text{... (2)} $$

Step 1: Solve each inequality individually.

For inequality (1):

$$ x + 3 < 7 $$

Subtract 3 from both sides:

$$ x < 7 - 3 $$ $$ x < 4 $$

The solution set for $x+3<7$ is $(-\infty, 4)$.

For inequality (2):

$$ 2x - 1 \ge 5 $$

Add 1 to both sides:

$$ 2x \ge 5 + 1 $$ $$ 2x \ge 6 $$

Divide both sides by 2 (positive number, no sign change):

$$ \frac{2x}{2} \ge \frac{6}{2} $$ $$ x \ge 3 $$

The solution set for $2x-1 \ge 5$ is $[3, \infty)$.

Step 2: Represent the solutions on a single number line.

The solution $x < 4$ is represented by an open circle at 4 and shading to the left.

The solution $x \ge 3$ is represented by a closed circle at 3 and shading to the right.

Step 3: Find the intersection of the solution sets.

The solution for the system is the set of values of $x$ that are in *both* $(-\infty, 4)$ and $[3, \infty)$. Looking at the number line, the overlap occurs for values of $x$ that are greater than or equal to 3 and strictly less than 4.

The intersection is $[3, \infty) \cap (-\infty, 4) = [3, 4)$.

This can be written as an inequality $3 \le x < 4$.

Answer: The solution to the system is $3 \le x < 4$, or in interval notation, $[3, 4)$.

System of Linear Inequalities in Two Variables

A system of linear inequalities in two variables consists of two or more linear inequalities involving the same two variables (typically $x$ and $y$). A solution to this system is an ordered pair $(x, y)$ that satisfies all inequalities simultaneously. The solution set of such a system is a region in the coordinate plane where the solution regions of all individual inequalities overlap.

Procedure to solve a system of linear inequalities in two variables graphically:

1. Graph Each Inequality: Graph each linear inequality in the system on the *same* Cartesian coordinate plane. For each inequality, follow the steps from the previous section:
   • Graph the boundary line (solid for $\le, \ge$; dashed for $<, >$).
   • Choose a test point (usually the origin if not on the line).
   • Test the point in the inequality to determine which half-plane is the solution.
   • Shade the appropriate half-plane. Use different shading patterns or colors for each inequality to make the overlapping region clear.
2. Identify the Solution Region: The solution set of the system is the region in the coordinate plane where the shaded areas of *all* the inequalities in the system overlap. This overlapping region represents the set of points $(x, y)$ that satisfy every inequality simultaneously. This region is often called the feasible region.
3. Indicate the Solution: Clearly mark or re-shade the overlapping region as the final solution. If the boundary lines are solid, they are part of the solution. If they are dashed, they are not. If a point is the intersection of a solid and a dashed line, that intersection point is not part of the solution.

Example

Answer:

We need to graph and find the intersection of the solution regions for the two inequalities:

$$ x + y \ge 5 \quad \text{... (1)} $$ $$ x - y \le 1 \quad \text{... (2)} $$

Step 1: Graph each inequality.

For inequality (1): $x + y \ge 5$.

• Boundary line: $x + y = 5$. Points: $(0, 5)$ and $(5, 0)$.
• Line type: $\ge$ means solid line.
• Test point: $(0, 0)$. $0 + 0 \ge 5 \implies 0 \ge 5$ (False).
• Shade: The half-plane opposite to $(0, 0)$ (above the line).

For inequality (2): $x - y \le 1$.

• Boundary line: $x - y = 1$. Points: $(0, -1)$ and $(1, 0)$.
• Line type: $\le$ means solid line.
• Test point: $(0, 0)$. $0 - 0 \le 1 \implies 0 \le 1$ (True).
• Shade: The half-plane containing $(0, 0)$ (above the line, which can also be seen by rewriting as $y \ge x-1$).

Plot both lines and shade the respective regions on the same coordinate plane.

Step 2 & 3: Identify the solution region.

The solution to the system is the region where the shaded area for $x + y \ge 5$ and the shaded area for $x - y \le 1$ overlap. This overlapping region is the feasible region for the system.

The boundary lines intersect at a vertex of the feasible region. Let's find the intersection point by solving the system of equations:

$$ x + y = 5 $$ $$ x - y = 1 $$

Adding the two equations: $(x+y) + (x-y) = 5 + 1 \implies 2x = 6 \implies x = 3$.

Substitute $x=3$ into the first equation: $3 + y = 5 \implies y = 2$.

The intersection point is $(3, 2)$. Since both boundary lines are solid, this vertex is part of the solution region.

The feasible region is the area above the line $x+y=5$ and also above the line $x-y=1$. It is an unbounded region starting from the vertex $(3,2)$ and extending upwards between the two lines.

Answer: The solution is the unbounded region in the coordinate plane defined by the overlap of the shaded areas for $x+y \ge 5$ and $x-y \le 1$, including the solid boundary lines $x+y=5$ and $x-y=1$.

Solving systems of linear inequalities, especially graphically in two variables, is a fundamental concept used in fields like economics, operations research, and business for modeling resource allocation and optimization problems, commonly known as linear programming. The overlapping region represents all possible combinations of the variables that satisfy all the given constraints.

Numerical Inequalities (Applied Context)

Inequalities in Real-World Numbers

Before delving into algebraic inequalities involving variables, it's helpful to understand the most basic type: numerical inequalities. These involve a direct comparison between two specific numerical values using the inequality symbols ($<, >, \le, \ge, \neq$). They represent simple statements about the relative size or order of numbers and arise naturally from everyday observations and measurements.

Numerical inequalities are statements that are either definitively true or definitively false.

Consider these examples:

• Comparing Temperatures: If the temperature in Delhi today is $35^\circ\text{C}$ and the temperature yesterday was $32^\circ\text{C}$, we can state this relationship as $35^\circ\text{C} > 32^\circ\text{C}$, or simply $35 > 32$. This is a true numerical inequality.
• Comparing Costs vs. Budget: If you want to buy a notebook that costs $\textsf{₹} 40$ and your budget for notebooks is $\textsf{₹} 50$, the cost is less than or equal to your budget. This is expressed as $\textsf{₹} 40 \le \textsf{₹} 50$, or $40 \le 50$. This is a true numerical inequality.
• Comparing Counts: In a class of 50 students, if 48 students are present on a particular day, the number present is not equal to the total number enrolled: $48 \neq 50$. This is a true numerical inequality.
• Comparing Weights: If Parcel A weighs $1.7 \text{ kg}$ and Parcel B weighs $1.9 \text{ kg}$, we can state that Parcel A's weight is less than Parcel B's weight: $1.7 < 1.9$. This is a true numerical inequality.

These simple comparisons form the foundation for understanding more complex inequalities involving variables.

Applications of Numerical Inequalities

Although seemingly simple, numerical inequalities play a crucial role in expressing conditions, limits, and comparisons in various applied contexts. They are the outcome when specific values are known and need to be compared according to some rule.

Some common applications include:

1. Direct Comparison of Quantities: This is the most straightforward application, where we compare the magnitudes of two known values.

   Example: A shop offers a discount if the quantity purchased of a certain item is more than 10 units. If a customer buys 8 units, did they qualify for the discount? We compare the purchased quantity (8) with the required quantity (10). The condition is "quantity purchased > 10". Checking for the customer: $8 > 10$. This is false. So, they do not qualify for the discount.

2. Checking if a Condition is Met: Numerical inequalities are used to verify if a specific value or the result of a calculation meets a predefined criterion, limit, or requirement.

   Example: The minimum marks required to pass an exam is 33. If a student scores 29 marks, did they pass? We check if their score (29) is greater than or equal to the passing marks (33). The condition is "score $\ge$ 33". Checking for the student: $29 \ge 33$. This is false. The student did not pass.

3. Stating Facts or Observations Quantitatively: Inequalities provide a precise way to record observations where exact equality is not the relevant characteristic.

   Example: The height of Mount Everest is approximately $8848.86$ metres. The height of K2 is approximately $8611$ metres. We can state the factual relationship using an inequality: $8848.86 > 8611$.

4. Evaluating Solutions in Algebraic Inequalities: When we work with inequalities involving variables, substituting specific values for the variables reduces the algebraic inequality to a numerical inequality. Checking the truth value of this resulting numerical inequality tells us whether the substituted values constitute a solution to the original inequality.

   Example: Is the pair $(x, y) = (2, 3)$ a solution to the linear inequality $2x + y \le 7$? Substitute $x=2$ and $y=3$ into the inequality: $2(2) + 3 \le 7 \implies 4 + 3 \le 7 \implies 7 \le 7$. This is a true numerical inequality. Therefore, the pair $(2, 3)$ is a solution to the inequality $2x + y \le 7$.

Numerical inequalities, while basic, are fundamental building blocks in quantitative reasoning and serve as the simplest form of expressing comparison and constraint, forming the basis for understanding and applying algebraic inequalities.