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# 6.7: Linear Inequalities in Two Variables

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph linear inequalities in one variable on the coordinate plane.
• Graph linear inequalities in two variables.
• Solve real-world problems using linear inequalities

## Introduction

A linear inequality in two variables takes the form

y>mx+b\begin{align*} y > mx+b\end{align*} or y<mx+b\begin{align*} y < mx+b\end{align*}

Linear inequalities are closely related to graphs of straight lines. A straight line has the equation y=mx+b\begin{align*} y = mx+b\end{align*}. When we graph a line in the coordinate plane, we can see that it divides the plane in two halves.

The solution to a linear inequality includes all the points in one of the plane halves. We can tell which half of the plane the solution is by looking at the inequality sign.

> The solution is the half plane above the line.

\begin{align*}\geq\end{align*} The solution is the half plane above the line and also all the points on the line.

< The solution is the half plane below the line.

\begin{align*}\leq\end{align*} The solution is the half plane below the line and also all the points on the line.

(Above the line means for a given x\begin{align*}x-\end{align*}coordinate, all points with y\begin{align*}y-\end{align*}values greater than the y\begin{align*}y-\end{align*}value are on the line)

For a strict inequality, we draw a dashed line to show that the points on the line are not part of the solution.

For an inequality that includes the equal sign, we draw a solid line to show that the points on the line are part of the solution.

Here is what you should expect linear inequality graphs to look like.

The solution of y>mx+b\begin{align*} y>mx+b\end{align*} is the half plane above the line. The dashed line shows that the points on the line are not part of the solution.

The solution of ymx+b\begin{align*} y \geq mx+b\end{align*} is the half plane above the line and all the points on the line.

The solution of y<mx+b\begin{align*} y is the half plane below the line.

The solution of ymx+b\begin{align*} y\leq mx+b\end{align*} is the half plane below the line and all the points on the line.

## Graph Linear Inequalities in One Variable in the Coordinate Plane

In the last few sections, we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type x=a\begin{align*}x=a\end{align*} we get a vertical line and when we graph an equation of the type y=b\begin{align*}y=b\end{align*} we get a horizontal line.

Example 1

Graph the inequality x>4\begin{align*}x>4\end{align*} on the coordinate plane.

Solution

First, let’s remember what the solution to x>4\begin{align*}x>4\end{align*} looks like on the number line.

The solution to this inequality is the set of all real numbers x\begin{align*}x\end{align*} that are bigger than four but not including four. The solution is represented by a line.

In two dimensions we are also concerned with values of y\begin{align*} y\end{align*}, and the solution to x>4\begin{align*} x>4\end{align*} consists of all coordinate points for which the value of x\begin{align*}x\end{align*} is bigger than four. The solution is represented by the half plane to the right of x=4\begin{align*}x=4\end{align*}.

The line x=4\begin{align*}x=4\end{align*} is dashed because the equal sign is not included in the inequality and therefore points on the line are not included in the solution.

Example 2

Graph the inequality y6\begin{align*} y \leq 6\end{align*} on the coordinate plane.

Solution

The solution is all coordinate points for which the value of y\begin{align*}y\end{align*} is less than or equal than 6. This solution is represented by the half plane below the line y=6\begin{align*}y=6\end{align*}.

The line y=6\begin{align*}y=6\end{align*} is solid because the equal sign is included in the inequality sign and the points on the line are included in the solution.

Example 3

Graph the inequality 6<5\begin{align*} \mid 6 \mid <5\end{align*}

Solution

The absolute value inequality 6<5\begin{align*} \mid 6 \mid <5\end{align*} can be re-written as 5<y<5\begin{align*}-5 < y <5\end{align*}. This is a compound inequality which means

y>5\begin{align*}y>-5\end{align*} and y<5\begin{align*}y<5\end{align*}

In other words, the solution is all the coordinate points for which the value of y\begin{align*}y\end{align*} is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines y=5\begin{align*}y=-5\end{align*} and y=5\begin{align*}y=5\end{align*}.

Both horizontal lines are dashed because points on the line are not included in the solution.

Example 4

Graph the inequality x2\begin{align*}\mid x \mid \geq 2\end{align*}.

Solution

The absolute value inequality x2\begin{align*} \mid x \mid \geq 2\end{align*} can be re-written as a compound inequality:

x2\begin{align*} x \leq -2 \end{align*} or x2\begin{align*} x \geq 2 \end{align*}

In other words, the solution is all the coordinate points for which the value of x\begin{align*}x\end{align*} is smaller than or equal to -2 and greater than or equal to 2. The solution is represented by the plane to the left of the vertical line x=2\begin{align*}x= -2\end{align*} and the plane to the right of line x=2\begin{align*}x=2\end{align*}.

Both vertical lines are solid because points on the line are included in the solution.

## Graph Linear Inequalities in Two Variables

The general procedure for graphing inequalities in two variables is as follows.

Step 1: Re-write the inequality in slope-intercept form y=mx+b\begin{align*}y=mx+b\end{align*}. Writing the inequality in this form lets you know the direction of the inequality

Step 2 Graph the line of equation y=mx+b\begin{align*}y=mx+b\end{align*} using your favorite method. (For example, plotting two points, using slope and y\begin{align*}y-\end{align*}intercept, using y\begin{align*}y-\end{align*}intercept and another point, etc.). Draw a dashed line if the equal sign in not included and a solid line if the equal sign is included.

Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than.

Example 5

Graph the inequality y2x3\begin{align*} y \geq 2x-3\end{align*}.

Solution

Step 1

The inequality is already written in slope-intercept form y2x3\begin{align*} y \geq 2x-3\end{align*}.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
1\begin{align*}-1\end{align*} 2(1)3=5\begin{align*}2(-1) - 3 = -5\end{align*}
0 2(0)3=3\begin{align*}2(0) - 3 = -3\end{align*}
1 2(1)3=1\begin{align*}2(1) - 3 = -1\end{align*}

Step 2

Graph the equation y=2x3\begin{align*}y=2x-3\end{align*} by making a table of values.

Step 3

Graph the inequality. We shade the plane above the line because y\begin{align*}y\end{align*} is greater than. The value 2x3\begin{align*}2x-3\end{align*} defines the line. The line is solid because the equal sign is included.

Example 6

Graph the inequality 5x2y>4\begin{align*} 5x - 2y > 4\end{align*}.

Solution

Step 1

Rewrite the inequality in slope-intercept form.

2yy>5x+4>52x2\begin{align*} -2y & > -5x + 4\\ y & > \frac{5} {2}x - 2\end{align*}

Notice that the inequality sign changed direction due to division of negative sign.

Step 2

Graph the equation y>52x2\begin{align*} y>\frac{5} {2}x - 2\end{align*} by making a table of values.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
2\begin{align*}-2\end{align*} 52(2)2=7\begin{align*}\frac{5} {2}(-2) - 2 = -7\end{align*}
0 52(0)2=2\begin{align*}\frac{5} {2}(0) - 2 = -2\end{align*}
2 52(2)2=3\begin{align*}\frac{5} {2}(2) - 2 = 3\end{align*}

Step 3

Graph the inequality. We shade the plane below the line because the inequality in slope-intercept form is less than. The line is dashed because the equal sign in not included.

Example 7

Graph the inequality y+4x3+5\begin{align*} y+4 \leq -\frac{x} {3}+5\end{align*}.

Solution

Step 1

Rewrite the inequality in slope-intercept form yx3+1\begin{align*} y \leq -\frac{x} {3}+1\end{align*}

Step 2

Graph the equation y=x3+1\begin{align*} y= -\frac{x} {3}+1\end{align*} by making a table of values.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
3\begin{align*}-3\end{align*} (3)3+1=2\begin{align*}-\frac{(-3)} {3}+1=2\end{align*}
0 03(0)+1=1\begin{align*}-\frac{0} {3}(0)+1=1\end{align*}
3 33+1=0\begin{align*}-\frac{3} {3}+1=0\end{align*}

Step 3

Graph the inequality. We shade the plane below the line. The line is solid because the equal sign in included.

## Solve Real-World Problems Using Linear Inequalities

In this section, we see how linear inequalities can be used to solve real-world applications.

Example 8

A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9 per pound and another type costs$7 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blend costs 8.50 per pound or less. Solution Let’s apply our problem solving plan to solve this problem. Step 1: Let x=\begin{align*}x =\end{align*} weight of9 per pound coffee beans in pounds

Let y=\begin{align*}y =\end{align*} weight of 7 per pound coffee beans in pounds Step 2 The cost of a pound of coffee blend is given by 9x+7y\begin{align*}9x + 7y\end{align*}. We are looking for the mixtures that cost8.50 or less.

We write the inequality 9x+7y8.50\begin{align*}9x + 7y \leq 8.50\end{align*}.

Step 3

To find the solution set, graph the inequality 9x+7y8.50\begin{align*}9x + 7y \leq 8.50\end{align*}.

Rewrite in slope-intercept y1.29x+1.21\begin{align*}y \leq -1.29x + 1.21\end{align*}.

Graph y=1.29x+1.21\begin{align*}y=-1.29x+1.21\end{align*} by making a table of values.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
0 1.21
1 -0.08
2 -1.37

Step 4

Graph the inequality. The line will be solid. We shade below the line.

Notice that we show only the first quadrant of the coordinate plane because the weight values should be positive.

The blue-shaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equal to $8.50. Example 9 Julian has a job as an appliance salesman. He earns a commission of$60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell in order to make$1000 or more in commission?

Solution Let’s apply our problem solving plan to solve this problem.

Step 1

Let x=\begin{align*}x =\end{align*} number of washing machines Julian sells

Let y=\begin{align*}y =\end{align*} number of refrigerators Julian sells

Step 2

The total commission is given by the expression 60x+130y\begin{align*}60x+130y\end{align*}.

We are looking for total commission of 1000 or more. We write the inequality. 60x+130y1000\begin{align*}60x + 130y \geq 1000\end{align*}. Step 3 To find the solution set, graph the inequality 60x+130y1000\begin{align*}60x + 130y \geq 1000\end{align*}. Rewrite it in slope-intercept y.46x+7.7\begin{align*}y \geq -.46x + 7.7\end{align*}. Graph y=.46x+7.7\begin{align*}y = -.46x + 7.7\end{align*} by making a table of values. x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} 0 7.7 2 6.78 4 5.86 Step 4 Graph the inequality. The line will be solid. We shade above the line. Notice that we show only the first quadrant of the coordinate plane because dollar amounts should be positive. Also, only the points with integer coordinates are possible solutions. ## Lesson Summary • The general procedure for graphing inequalities in two variables is as follows: Step 1 Rewrite the inequality in slope-intercept form y=mx+b\begin{align*}y=mx+b\end{align*}. Step 2 Graph the line of equation y=mx+b\begin{align*}y=mx+b\end{align*} by building a table of values. Draw a dashed line if the equal sign in not included and a solid line if the it is included. Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than. ## Review Questions Graph the following inequalities on the coordinate plane. 1. x<20\begin{align*}x<20\end{align*} 2. y5\begin{align*}y \geq -5\end{align*} 3. x>10\begin{align*} \mid x \mid >10 \end{align*} 4. y7\begin{align*} \mid y \mid \leq 7 \end{align*} 5. y4x+3\begin{align*} y \leq 4x+3\end{align*} 6. y>x26\begin{align*}y > -\frac{x}{2}-6\end{align*} 7. 3x4y12\begin{align*}3x-4y \geq 12\end{align*} 8. x+7y<5\begin{align*} x+7y <5 \end{align*} 9. 6x+5y>1\begin{align*}6x + 5y>1\end{align*} 10. y+54x+10\begin{align*}y + 5 \leq -4x + 10\end{align*} 11. x12y5\begin{align*}x-\frac{1}{2}y \geq 5\end{align*} 12. 30x+5y<100\begin{align*}30x + 5y< 100\end{align*} 13. An ounce of gold costs670 and an ounce of silver costs $13. Find all possible weights of silver and gold that makes an alloy that costs less than$600 per ounce.
14. A phone company charges 50 cents cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and night time minutes would you have to use to pay more than \$20 over a 24 hour period?

## Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.

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