# 9.8: Graphs of Linear Inequalities

**At Grade**Created by: CK-12

## Introduction

*A Traveling Dilemma*

The students finished up their trip to the Omni by presenting real – world problems to the class. Some has to do with conservation and some had to do with travel or with animals. Each student wrote a problem and had a solution for it. Here is the problem that Kenya wrote about the rainforest.

You are going on a trip in a jeep out to the rainforest. You need to have enough fuel to get there and back. It is 115 miles one way. You can buy two different kinds of fuel for your truck, gasoline or ethanol. Gasoline gets better gas mileage at 21 mpg but ethanol is cleaner burning and gives you 17 mpg. Use a linear inequality to figure at how many gallons of each you would need.

**To write an inequality, you will first need to know about them. This lesson will teach you everything that you need to know. By the end of it, you should be able to solve Kenya’s problem.**

*What You Will Learn*

In this lesson, you will learn how to complete the following skills.

- Identify the meaning of inequality symbols relating expressions involving two variables.
- Recognize a solution of a linear inequality in two variables as an ordered pair of values which make the inequality true.
- Graph linear inequalities by rewriting as an equation in slope – intercept form graphing the equation as a dashed line for \begin{align*} < \end{align*}
< or \begin{align*} > \end{align*}> , a solid line for \begin{align*}\le\end{align*}≤ or \begin{align*}\ge\end{align*}≥ , and shadowing the half-plane that tests true for the inequality. - Model and solve real - world problems using graphs of linear inequalities.

*Teaching Time*

I. **Identify the Meaning of Inequality Symbols Relating Expressions Involving Two Variables**

Our task in equations has been to find any and all of the solutions—the solution set—that make the equation true. As we consider inequalities, the task remains the same. We have to find all the values that make the statement true.

Remember the inequality symbols that you have used before.

less than \begin{align*} < \end{align*}

less than or equal to \begin{align*} \le \end{align*}

You may have seen a simple inequality like “a number is less than five” or \begin{align*} x < 5 \end{align*}

**Just as equations can be in one or two variables, so can inequalities. However, because there are five inequality signs, we must be aware of their meanings. Additionally, there are oftentimes many ways to say the same thing—“is less than” could be said “is not as much as”—although they use the same inequality symbol.**

Example

Translate the following expressions to inequalities:

- The sum of two numbers is more than 10. \begin{align*} x+y>10 \end{align*}
x+y>10 - The difference between two numbers is at least 32. \begin{align*} x-y \ge 32 \end{align*}
x−y≥32 - Four less than a number is less than one-third another number. \begin{align*} x-4< \frac{1}{3}y \end{align*}
x−4<13y - Negative 5 times the sum of two numbers is not 18. \begin{align*} -5(x+y) \ne 18 \end{align*}
−5(x+y)≠18

**You can see that anytime that we are talking about two numbers or two unknown numbers that we can use two variables in an equation. This allows us to express an inequality in two variables.**

II. **Recognize a Solution of a Linear Inequality in Two Variables as an Ordered Pair of Values which Make the Inequality True**

Just as with equations, a solution to an inequality will be the value(s) that make the inequality true. When you saw the inequality, \begin{align*} x<5\end{align*}

There are infinitely many values that make the inequality true.

When we had equations in two variables like \begin{align*} x+y=7\end{align*}

**For the same reason, we will show solutions to inequalities in two variables as ordered pairs. We can find solutions by guessing and checking or by using mathematical reasoning.**

Example

Which solution makes this inequality a true statement?

\begin{align*}3x+2>y\end{align*}

\begin{align*}& (3, 5) && 3 \cdot 3 +2>5? && \text{yes}\\
& (-6, 0) && 3 \cdot -6+2>0? && \text{no}\\
& (-1, -1) && 3 \cdot -1+2>-1? && \text{no}\\
& (10, 31) && 3 \cdot 10+2>31? && \text{yes}\end{align*}

**In this example, there are two possible solutions for this inequality. Remember that oftentimes, you will have more than one solution for an inequality.**

III. **Graph Linear Inequalities by Rewriting as an Equation in Slope – Intercept Form graphing the equation as a dashed line for \begin{align*}\underline{<}\end{align*} or \begin{align*}\underline{>}\end{align*}, a solid line for \begin{align*}\underline{\le}\end{align*} or \begin{align*}\underline{\ge}\end{align*}, and shadowing the half-plane that tests true for the inequality**

Listing solutions to single variable inequalities is useful but, because there are infinitely many solutions, it is impossible to show the entire solution set with a list. For that reason, we use number lines. So when \begin{align*}x<2\end{align*}, we show it like this.

With the less than \begin{align*}(<)\end{align*}, greater than \begin{align*}(>)\end{align*}, and not equal to \begin{align*}(\ne)\end{align*} symbols, we use an open circle because, although the solution set is infinitely close to the endpoint, the endpoint itself is not actually a solution. With the less than or equal to \begin{align*}(\le)\end{align*} and greater than or equal to \begin{align*}(\ge)\end{align*} symbols, the endpoint is a solution so we use a closed circle.

**Just as we graphed linear equations, we can also graph linear inequalities. We will graph the linear inequalities using slope-intercept form. As the circle on a number line marks the end of the solution set of a single variable inequality, so the line on the coordinate plane will mark the boundary of the solution set of a linear inequality.**

**The solution set will be on one side of the line or the other. We will take a test point to figure out which side makes the inequality true and then shade that half of the coordinate plane to indicate the solution set.**

**With the less than \begin{align*}(<)\end{align*}, greater than \begin{align*}(>)\end{align*}, and not equal to \begin{align*}(\ne)\end{align*} symbols, we will use a dotted line instead of an open circle because, although the solution set is infinitely close to the line, the points on the line itself are not actually solutions. With the less than or equal to \begin{align*}(\le)\end{align*} and greater than or equal to \begin{align*}(\ge)\end{align*} symbols, the points on the line are solutions so we use a solid line.**

Let’s look at an example.

Example

Graph the solution set of the inequality \begin{align*}y > x+3\end{align*}.

**Graph using \begin{align*}m=1, b=3\end{align*}. Use a dotted line because the symbol is \begin{align*} > \end{align*}.**

Now, the solution set is on one side of the line or the other. In order to determine which side, we will just try a point that is not on the line itself. Try, for example, (1, 1). Does the point make the inequality \begin{align*}y>x+3\end{align*} true? \begin{align*}1>1+3?\end{align*} No! The solution set must be on the other side of the line. As you can see in the graph, we shaded the opposite side.

**We can graph any inequality in this way. First, graph the equation of the line. Then check if it is a solid or dashed line. Then shade above or below the line based on the inequality symbol.**

*Write these steps down in your notebook.*

IV. **Model and Solve Real – World Problems Using Graphs of Linear Inequalities**

There are many different real life situations that require inequalities or can be described by an inequality. Let’s look at an example.

Example

You’re going to a party! You’re supposed to bring sodas and chips but you only have $20 to spend. Sodas cost $1.50 per bottle and chips cost $2.50 per bag. How many of each can you buy?

**This situation can be modeled with a linear inequality.**

**Let \begin{align*}s\end{align*} equal the number of sodas you buy and \begin{align*}c\end{align*} the number of chips that you buy.**

**The inequality is \begin{align*} 1.5s + 2.5c \le 20 \end{align*} because the cost of the sodas plus the cost of the chips must not be more than $20.**

**Graph the inequality and shade the correct region. Find 5 combinations of sodas and chips that you could buy by looking at the ordered pairs within the solution on the graph.**

**You can buy any combination that is one of the ordered pairs in the shaded region. Possible answers are (11, 1) (9, 2) (7, 3) (6, 4) etc.**

Now let’s look at how we can apply what we have learned to the problem from the introduction.

## Real-Life Example Completed

*A Traveling Dilemma*

**Here is the problem from the introduction. Reread it and then use an inequality and a graph to solve the problem.**

The students finished up their trip to the Omni by presenting real – world problems to the class. Some has to do with conservation and some had to do with travel or with animals. Each student wrote a problem and had a solution for it. Here is the problem that Kenya wrote about the rainforest.

You are going on a trip in a jeep out to the rainforest. You need to have enough fuel to get there and back. It is 115 miles one way. You can buy two different kinds of fuel for your truck, gasoline or ethanol. Gasoline gets better gas mileage at 21 mpg but ethanol is cleaner burning and gives you 17 mpg. Use a linear inequality to figure at how many gallons of each you would need.

*Remember, there are two parts to your problem.*

*Solution to Real – Life Example*

**To write the solution, we first need to write an inequality. We use \begin{align*}e\end{align*} for ethanol and \begin{align*}g\end{align*} for gasoling and we write the number of miles per gallon that each gets. We want to be able to travel there and back. Since it is 115 miles per way, that is a sum of 230 miles. Here is the inequality.**

\begin{align*}21g + 17e \ge 230 \end{align*}

**Using the ordered pair \begin{align*}(e, g)\end{align*}, you could have (13, 2), (12, 3), (6, 8). The first value in each pair represents the ethanol usuage, the second value represents the gasoline usuage.**

## Time to Practice

Directions: translate each statement into an inequality.

- The difference between two numbers is greater than 8.
- Half of one number is at least 3 times another number.
- A quarter the sum of two numbers is less than 15.
- Seven times one number plus 3 less than another is not more than -16.
- Six times a number is greater than negative thirty.
- Five times a number and six is less than or equal to 39.
- Twelve divided by a number is less than seven.
- Six times a number and two less than another number is less than or equal to -12.

Directions: Which ordered pairs make the inequalities true?

\begin{align*}2x-5y \ge 20\end{align*}

- (10, 5)
- (10, 4)
- (-10, -10)
- (0, 0)

\begin{align*}y>-4x-2\end{align*}

- (1, 1)
- (0, -6)
- (2, -9)
- (-1, 0)

Directions: Graph the following inequalities. Be sure to choose a dotted or a solid line and shade the correct region based on a test point.

- \begin{align*}y < - \frac{2}{3} x-2 \end{align*}
- \begin{align*}6x - 3y \le -9 \end{align*}
- \begin{align*}3x + y > 3x -4 \end{align*}
- You’re going to make a fruit salad with mangos and bananas. Bananas cost $.60 per pound and mangos cost $.90 per pound. If you can only spend $7.50, how many pounds of each can you buy? Write an inequality for the situation and solve it.
- Julio is going to Europe and wants to have at least $500 worth of cash. He has Euros and dollars. Each Euro is worth about one and a half dollars. What are 3 combinations he could have to maintain his goal? Show your inequality and solve it.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |