Intermediate Algebra 2e · free preview

3.4 Graph Linear Inequalities in Two Variables

Learning objectives: Verify solutions to an inequality in two variables.; Recognize the relation between the solutions of an inequality and its graph.; Graph linear inequalities in two variables; Solve applications using linear inequalities in two variables

1. Learning objectives

Learning objectives

Learning objectives

  • Verify solutions to an inequality in two variables.
  • Recognize the relation between the solutions of an inequality and its graph.
  • Graph linear inequalities in two variables
  • Solve applications using linear inequalities in two variables

2. Verify Solutions to an Inequality in Two Variables

Verify Solutions to an Inequality in Two Variables

Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables. In particular we will look at linear inequalities in two variables which are very similar to linear equations in two variables.

Linear inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business made a profit.

A linear inequality is an inequality that can be written in one of the following forms:

Ax+By>CAx+By≥CAx+By<CAx+By≤C\begin{array}{lcclcclccl} Ax+By>C & & & Ax+By\ge{}C & & & Ax+By<C & & & Ax+By\le{}C \\ \end{array}

Where A and B are not both zero.

Recall that an inequality with one variable had many solutions. For example, the solution to the inequality x>3x>3 is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3. See Figure 3.10.

Similarly, linear inequalities in two variables have many solutions. Any ordered pair (x,y)(x,y) that makes an inequality true when we substitute in the values is a solution to a linear inequality.

An ordered pair (x,y)(x,y) is a solution to a linear inequality if the inequality is true when we substitute the values of x and y.

3. Example 3.34

Example 3.34

Example 3.34

Determine whether each ordered pair is a solution to the inequality y>x+4:y>x+4: ⓐ (0,0)(0,0) ⓑ (1,6)(1,6) ⓒ (2,6)(2,6) ⓓ (−5,−15)(-5,-15) ⓔ (−8,12)(-8,12)

Solution

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4. Recognize the Relation Between the Solutions of an Inequality and its Graph

Recognize the Relation Between the Solutions of an Inequality and its Graph

Now, we will look at how the solutions of an inequality relate to its graph.

Let’s think about the number line in shown previously again. The point x=3x=3 separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See Figure 3.11.

Similarly, the line y=x+4y=x+4 separates the plane into two regions. On one side of the line are points with y<x+4.y<x+4. On the other side of the line are the points with y>x+4.y>x+4. We call the line y=x+4y=x+4 a boundary line.

The line with equation Ax+By=CAx+By=C is the boundary line that separates the region where Ax+By>CAx+By>C from the region where Ax+By<C.Ax+By<C.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not a is included in the solution:

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to show whether or not it the line is included in the solution.

Ax+By<CAx+By≤CAx+By>CAx+By≥CBoundarylineis Ax+By=CBoundarylineis Ax+By=CBoundarylineisnotincludedinsolution.Boundarylineisincludedinsolution.Boundarylineisdashed.Boundarylineissolid.\begin{array}{lccccl} Ax+By<C & & & & & Ax+By\le{}C \\ Ax+By>C & & & & & Ax+By\ge{}C \\ \mathrm{Boundary line is}\,Ax+By=C & & & & & \mathrm{Boundary line is}\,Ax+By=C \\ \mathrm{Boundary line is not included in solution.} & & & & & \mathrm{Boundary line is included in solution.} \\ \mathrm{Boundary line is dashed.} & & & & & \mathrm{Boundary line is solid.} \\ \end{array}

Now, let’s take a look at what we found in Example 3.34. We’ll start by graphing the line y=x+4,y=x+4, and then we’ll plot the five points we tested, as shown in the graph. See Figure 3.12.

In Example 3.34 we found that some of the points were solutions to the inequality y>x+4y>x+4 and some were not.

Which of the points we plotted are solutions to the inequality y>x+4?y>x+4?

The points (1,6)(1,6) and (−8,12)(-8,12) are solutions to the inequality y>x+4.y>x+4. Notice that they are both on the same side of the boundary line y=x+4.y=x+4.

The two points (0,0)(0,0) and (−5,−15)(-5,-15) are on the other side of the boundary line y=x+4,y=x+4, and they are not solutions to the inequality y>x+4.y>x+4. For those two points, y<x+4.y<x+4.

What about the point (2,6)?(2,6)? Because 6=2+4,6=2+4, the point is a solution to the equation y=x+4,y=x+4, but not a solution to the inequality y>x+4.y>x+4. So the point (2,6)(2,6) is on the boundary line.

Let’s take another point above the boundary line and test whether or not it is a solution to the inequality y>x+4.y>x+4. The point (0,10)(0,10) clearly looks to above the boundary line, doesn’t it? Is it a solution to the inequality?

y>x+410>?0+410>4\begin{array}{rll} y & > & x+4 \\ 10 & \overset{?}{>} & 0+4 \\ 10 & > & 4 \\ \end{array}

So, (0,10)(0,10) is a solution to y>x+4.y>x+4.

Any point you choose above the boundary line is a solution to the inequality y>x+4.y>x+4. All points above the boundary line are solutions.

Similarly, all points below the boundary line, the side with (0,0)(0,0) and (−5,−15),(-5,-15), are not solutions to y>x+4,y>x+4, as shown in Figure 3.13.

The graph of the inequality y>x+4y>x+4 is shown in below.

The line y=x+4y=x+4 divides the plane into two regions. The shaded side shows the solutions to the inequality y>x+4.y>x+4.

The points on the boundary line, those where y=x+4,y=x+4, are not solutions to the inequality y>x+4,y>x+4, so the line itself is not part of the solution. We show that by making the line dashed, not solid.

5. Example 3.35

Example 3.35

Example 3.35

The boundary line shown in this graph is y=2x−1.y=2x-1. Write the inequality shown by the graph.

Solution

The line y=2x−1y=2x-1 is the boundary line. On one side of the line are the points with y>2x−1y>2x-1 and on the other side of the line are the points with y<2x−1.y<2x-1.

Let’s test the point (0,0)(0,0) and see which inequality describes its position relative to the boundary line.

At (0,0),(0,0), which inequality is true: y>2x−1y>2x-1 or y<2x−1?y<2x-1?

y>2x−1y<2x−10>?2⋅0−10<?2⋅0−10>−1 True0<−1 False\begin{array}{lccccl} y>2x-1 & & & & & y<2x-1 \\ 0\overset{?}{>}2\cdot{}0-1 & & & & & 0\overset{?}{<}2\cdot{}0-1 \\ 0>-1\,\mathrm{True} & & & & & 0<-1\,\mathrm{False} \\ \end{array}

Since, y>2x−1y>2x-1 is true, the side of the line with (0,0),(0,0), is the solution. The shaded region shows the solution of the inequality y>2x−1.y>2x-1.

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality y≥2x−1.y\ge{}2x-1.

We could use any point as a test point, provided it is not on the line. Why did we choose (0,0)?(0,0)? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that y<2x−1.y<2x-1.

6. Example 3.36

Example 3.36

Example 3.36

The boundary line shown in this graph is 2x+3y=6.2x+3y=6. Write the inequality shown by the graph.

Solution

The line 2x+3y=62x+3y=6 is the boundary line. On one side of the line are the points with 2x+3y>62x+3y>6 and on the other side of the line are the points with 2x+3y<6.2x+3y<6.

Let’s test the point (0,0)(0,0) and see which inequality describes its side of the boundary line.

At (0,0),(0,0), which inequality is true: 2x+3y>62x+3y>6 or 2x+3y<6?2x+3y<6?

2x+3y>62(0)+3(0)>?60>6 False2x+3y<62(0)+3(0)<?60<6 True\begin{array}{lcccclcccccccccccc} \begin{array}{rll} 2x+3y & > & 6 \\ 2(0)+3(0) & \overset{?}{>} & 6 \\ 0 & > & 6\,\mathrm{False} \\ \end{array} & & & & & \begin{array}{rll} 2x+3y & < & 6 \\ 2(0)+3(0) & \overset{?}{<} & 6 \\ 0 & < & 6\,\mathrm{True} \\ \end{array} \\ \end{array}

So the side with (0,0)(0,0) is the side where 2x+3y<6.2x+3y<6.

(You may want to pick a point on the other side of the boundary line and check that 2x+3y>6.2x+3y>6.)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The shaded region shows the solution to the inequality 2x+3y<6.2x+3y<6.

7. Graph Linear Inequalities in Two Variables

Graph Linear Inequalities in Two Variables

Now that we know what the graph of a linear inequality looks like and how it relates to a boundary equation we can use this knowledge to graph a given linear inequality.

8. Example 3.37

Example 3.37

Example 3.37

How to Graph a Linear Equation in Two Variables

Graph the linear inequality y≥34x−2.y\ge\frac{3}{4}x-2.

The steps we take to graph a linear inequality are summarized here.

Graph a linear inequality in two variables.

  1. Step 1. Identify and graph the boundary line. If the inequality is ≤or≥,\le\mathrm{or}\ge, the boundary line is solid. If the inequality is <or>,<\mathrm{or}>, the boundary line is dashed.
  2. Step 2. Test a point that is not on the boundary line. Is it a solution of the inequality?
  3. Step 3. Shade in one side of the boundary line. If the test point is a solution, shade in the side that includes the point. If the test point is not a solution, shade in the opposite side.

9. Example 3.38

Example 3.38

Example 3.38

Graph the linear inequality x−2y<5.x-2y<5.

Solution

First, we graph the boundary line x−2y=5.x-2y=5. The inequality is << so we draw a dashed line.

Then, we test a point. We’ll use (0,0)(0,0) again because it is easy to evaluate and it is not on the boundary line.

Is (0,0)(0,0) a solution of x−2y<5?x-2y<5?

The point (0,0)(0,0) is a solution of x−2y<5,x-2y<5, so we shade in that side of the boundary line.

All points in the shaded region, but not those on the boundary line, represent the solutions to x−2y<5.x-2y<5.

What if the boundary line goes through the origin? Then, we won’t be able to use (0,0)(0,0) as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

10. Example 3.39

Example 3.39

Example 3.39

Graph the linear inequality: y≤−4x.y\le-4x.

Solution

First, we graph the boundary line y=−4x.y=-4x. It is in slope–intercept form, with m=−4m=-4 and b=0.b=0. The inequality is ≤\le so we draw a solid line.

Now we need a test point. We can see that the point (1,0)(1,0) is not on the boundary line.

Is (1,0)(1,0) a solution of y≤−4x?y\le-4x?

The point (1,0)(1,0) is not a solution to y≤−4x,y\le-4x, so we shade in the opposite side of the boundary line.

All points in the shaded region and on the boundary line represent the solutions to y≤−4x.y\le-4x.

Some linear inequalities have only one variable. They may have an x but no y, or a y but no x. In these cases, the boundary line will be either a vertical or a horizontal line.

Recall that:

x=averticalliney=bhorizontalline\begin{array}{lccccl} x=a & & & & & \mathrm{vertical line} \\ y=b & & & & & \mathrm{horizontal line} \\ \end{array}

11. Example 3.40

Example 3.40

Example 3.40

Graph the linear inequality: y>3.y>3.

Solution

First, we graph the boundary line y=3.y=3. It is a horizontal line. The inequality is >> so we draw a dashed line.

We test the point (0,0).(0,0).

y>30>3\begin{array}{l} y>3 \\ 0>3 \\ \end{array}

So, (0,0)(0,0) is not a solution to y>3.y>3.

So we shade the side that does not include (0,0)(0,0) as shown in this graph.

All points in the shaded region, but not those on the boundary line, represent the solutions to y>3.y>3.

12. Solve Applications using Linear Inequalities in Two Variables

Solve Applications using Linear Inequalities in Two Variables

Many fields use linear inequalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used.

13. Example 3.41

Example 3.41

Example 3.41

Hilaria works two part time jobs in order to earn enough money to meet her obligations of at least 240aweek.Herjobinfoodservicepays240 a week. Her job in food service pays 10 an hour and her tutoring job on campus pays 15anhour.HowmanyhoursdoesHilarianeedtoworkateachjobtoearnatleast15 an hour. How many hours does Hilaria need to work at each job to earn at least 240?

ⓐ Let xx be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x,y)(x,y) that would be solutions to the inequality. Then, explain what that means for Hilaria.

Solution

ⓐ We let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring.

She earns 10perhouratthejobinfoodserviceand10 per hour at the job in food service and 15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will gives the amount earned at that job.

ⓑ To graph the inequality, we put it in slope–intercept form.

10x+15y≥24015y≥−10x+240y≥−23x+16\begin{array}{rll} 10x+15y & \ge & 240 \\ 15y & \ge & -10x+240 \\ y & \ge & -\frac{2}{3}x+16 \\ \end{array}

ⓒ From the graph, we see that the ordered pairs (15,10),(0,16),(24,0)(15,10),(0,16),(24,0) represent three of infinitely many solutions. Check the values in the inequality.

For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her fast-food job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service.

14. Section 3.4 Exercises

Section 3.4 Exercises

Section 3.4 Exercises

Verify Solutions to an Inequality in Two Variables

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Recognize the Relation Between the Solutions of an Inequality and its Graph

In the following exercises, write the inequality shown by the shaded region.

Graph Linear Inequalities in Two Variables

In the following exercises, graph each linear inequality.

Solve Applications using Linear Inequalities in Two Variables

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

15. Practice 1

  1. Determine whether each ordered pair is a solution to the inequality y>x−3:y>x-3:

ⓐ (0,0)(0,0) ⓑ (4,9)(4,9) ⓒ (−2,1)(-2,1) ⓓ (−5,−3)(-5,-3) ⓔ (5,1)(5,1)

  1. Determine whether each ordered pair is a solution to the inequality y<x+1:y<x+1:

ⓐ (0,0)(0,0) ⓑ (8,6)(8,6) ⓒ (−2,−1)(-2,-1) ⓓ (3,4)(3,4) ⓔ (−1,−4)(-1,-4)

  1. Write the inequality shown by the graph with the boundary line y=−2x+3.y=-2x+3.

16. Practice 2

  1. Write the inequality shown by the graph with the boundary line y=12x−4.y=\frac{1}{2}x-4.
  1. Write the inequality shown by the shaded region in the graph with the boundary line x−4y=8.x-4y=8.
  1. Write the inequality shown by the shaded region in the graph with the boundary line 3x−y=6.3x-y=6.

17. Practice 3

  1. Graph the linear inequality y≥52x−4.y\ge\frac{5}{2}x-4.
  1. Graph the linear inequality y<23x−5.y<\frac{2}{3}x-5.
  1. Graph the linear inequality: 2x−3y<6.2x-3y<6.

18. Practice 4

  1. Graph the linear inequality: 2x−y>3.2x-y>3.
  1. Graph the linear inequality: y>−3x.y>-3x.
  1. Graph the linear inequality: y≥−2x.y\ge-2x.

19. Practice 7

  1. Determine whether each ordered pair is a solution to the inequality 3x−4y>4:3x-4y>4:

ⓐ (5,1)(5,1)ⓑ (−2,6)(-2,6)ⓒ (3,2)(3,2)ⓓ (10,−5)(10,-5)ⓔ (0,0)(0,0)

  1. Write the inequality shown by the graph with the boundary line y=3x−4.y=3x-4.
  1. Write the inequality shown by the graph with the boundary line y=12x+1.y=\frac{1}{2}x+1.

20. Practice 8

  1. Write the inequality shown by the shaded region in the graph with the boundary line x+y=5.x+y=5.
  1. Write the inequality shown by the shaded region in the graph with the boundary line 3x−y=6.3x-y=6.
  1. Graph the linear inequality: y>23x−1.y>\frac{2}{3}x-1.

21. Practice 9

  1. Graph the linear inequality: y≤−12x+4.y\le-\frac{1}{2}x+4.
  1. Graph the linear inequality: x−y≤3.x-y\le{}3.
  1. Graph the linear inequality: 4x+y>−4.4x+y>-4.

22. Practice 10

  1. Graph the linear inequality: 3x+2y≥−6.3x+2y\ge-6.
  1. Graph the linear inequality: y>4x.y>4x.
  1. Graph the linear inequality: y<−10.y<-10.

23. Practice 11

  1. Graph the linear inequality: x≤5.x\le{}5.
  1. Graph the linear inequality: x−y<4.x-y<4.
  1. Graph the linear inequality: y≥32x.y\ge\frac{3}{2}x.

24. Practice 12

  1. Graph the linear inequality: y>−2x+1.y>-2x+1.
  1. Graph the linear inequality: 2x+y≥−4.2x+y\ge-4.
  1. Graph the linear inequality: 2x−5y>10.2x-5y>10.

25. Practice 5

  1. Graph the linear inequality: y<5.y<5.
  1. Graph the linear inequality: y≤−1.y\le-1.
  1. Hugh works two part time jobs. One at a grocery store that pays 10anhourandtheotherisbabysittingfor10 an hour and the other is babysitting for 13 hour. Between the two jobs, Hugh wants to earn at least 260aweek.HowmanyhoursdoesHughneedtoworkateachjobtoearnatleast260 a week. How many hours does Hugh need to work at each job to earn at least 260?

ⓐ Let x be the number of hours he works at the grocery store and let y be the number of hours he works babysitting. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x, y) that would be solutions to the inequality. Then, explain what that means for Hugh.

26. Practice 6

  1. Veronica works two part time jobs in order to earn enough money to meet her obligations of at least 280aweek.Herjobatthedayspapays280 a week. Her job at the day spa pays 10 an hour and her administrative assistant job on campus pays 17.50anhour.HowmanyhoursdoesVeronicaneedtoworkateachjobtoearnatleast17.50 an hour. How many hours does Veronica need to work at each job to earn at least 280?

ⓐ Let x be the number of hours she works at the day spa and let y be the number of hours she works as administrative assistant. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x, y) that would be solutions to the inequality. Then, explain what that means for Veronica

  1. Determine whether each ordered pair is a solution to the inequality y>x−1:y>x-1:

ⓐ (0,1)(0,1)ⓑ (−4,−1)(-4,-1)ⓒ (4,2)(4,2)ⓓ (3,0)(3,0)ⓔ (−2,−3)(-2,-3)

  1. Determine whether each ordered pair is a solution to the inequality y<3x+2:y<3x+2:

ⓐ (0,3)(0,3)ⓑ (−3,−2)(-3,-2)ⓒ (−2,0)(-2,0)ⓓ (0,0)(0,0)ⓔ (−1,4)(-1,4)

27. Practice 13

  1. Harrison works two part time jobs. One at a gas station that pays 11anhourandtheotherisITtroubleshootingfor$16.50anhour.Betweenthetwojobs,Harrisonwantstoearnatleast11 an hour and the other is IT troubleshooting for \$16.50 an hour. Between the two jobs, Harrison wants to earn at least 330 a week. How many hours does Harrison need to work at each job to earn at least $330?

ⓐ Let x be the number of hours he works at the gas station and let y be the number of (hours he works troubleshooting. Write an inequality that would model this situation.

ⓑ Graph the inequality.

ⓒ Find three ordered pairs (x,y)(x,y) that would be solutions to the inequality. Then, explain what that means for Harrison.

  1. The doctor tells Laura she needs to exercise enough to burn 500 calories each day. She prefers to either run or bike and burns 15 calories per minute while running and 10 calories a minute while biking.

ⓐ If x is the number of minutes that Laura runs and y is the number minutes she bikes, find the inequality that models the situation.

ⓑ Graph the inequality.

ⓒ List three solutions to the inequality. What options do the solutions provide Laura?

  1. Lester thinks that the solution of any inequality with a >> sign is the region above the line and the solution of any inequality with a << sign is the region below the line. Is Lester correct? Explain why or why not.

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