Elementary Algebra 2e · free preview

4.7 Graphs of Linear Inequalities

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

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

2. Verify Solutions to an Inequality in Two Variables

Verify Solutions to an Inequality in Two Variables

We have learned how to solve inequalities in one variable. Now, we will look at inequalities in two variables. 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 would make 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}{cccccccccc} Ax+By>C & & & Ax+By\ge{}C & & & Ax+By<C & & & Ax+By\le{}C \\ \end{array}

where A and BA\,\mathrm{and}\,B are not both zero.

Do you remember that an inequality with one variable had many solutions? 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 4.30.

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

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

3. Example 4.69

Example 4.69

Example 4.69

Determine whether each ordered pair is a solution to the inequality y>x+4y>x+4:

ⓐ (0,0)(0,0) ⓑ (1,6)(1,6) ⓒ (2,6)(2,6) ⓓ (−5,−15)(-5,-15) ⓔ (−8,12)(-8,12)

Solution

  1. ⓐ (0,0)(0,0) Simplify. So, (0,0)(0,0) is not a solution to y>x+4y>x+4.
  1. ⓑ (1,6)(1,6) Simplify. So, (1,6)(1,6) is a solution to y>x+4y>x+4.
  1. ⓒ (2,6)(2,6) Simplify. So, (2,6)(2,6) is not a solution to y>x+4y>x+4.
  1. ⓓ (−5,−15)(-5,-15) Simplify. So, (−5,−15)(-5,-15) is not a solution to y>x+4y>x+4.
  1. ⓔ (−8,12)(-8,12) Simplify. So, (−8,12)(-8,12) is a solution to y>x+4y>x+4.

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 Figure 4.30 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 4.31.

The solution to x>3x>3 is the shaded part of the number line to the right of x=3x=3.

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+4y<x+4. On the other side of the line are the points with y>x+4y>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<CAx+By<C.

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

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in Table 4.48

Ax+By<CAx+By<CAx+By≤CAx+By\le{}C
Ax+By>CAx+By>CAx+By≥CAx+By\ge{}C
Boundary line is not included in solution.Boundary line is included in solution.
Boundary line is dashed.Boundary line is solid.

Table

4.48

Now, let’s take a look at what we found in Example 4.69. We’ll start by graphing the line y=x+4y=x+4, and then we’ll plot the five points we tested. See Figure 4.32.

In Example 4.69 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+4y>x+4? The points (1,6)(1,6) and (−8,12)(-8,12) are solutions to the inequality y>x+4y>x+4. Notice that they are both on the same side of the boundary line y=x+4y=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+4y=x+4, and they are not solutions to the inequality y>x+4y>x+4. For those two points, y<x+4y<x+4.

What about the point (2,6)(2,6)? Because 6=2+46=2+4, the point is a solution to the equation y=x+4y=x+4. So the point (2,6)(2,6) is on the boundary line.

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

y>x+410>?0+410>4So, (0,10) isasolutionto y>x+4.\begin{array}{lccc} y>x+4 & & & \\ 10\overset{?}{>}0+4 & & & \\ 10>4 & & & \mathrm{So},\,(0,10)\,\mathrm{is a solution to}\,y>x+4. \\ \end{array}

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

Similarly, all points on the right side of the boundary line, the side with (0,0)(0,0) and (−5,−15)(-5,-15), are not solutions to y>x+4y>x+4. See Figure 4.33.

The graph of the inequality y>x+4y>x+4 is shown in Figure 4.34 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+4y>x+4.

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

5. Example 4.70

Example 4.70

Example 4.70

The boundary line shown is y=2x−1y=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−1y<2x-1.

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:

y>2x−1ory<2x−1?y>2x−1y<2x−10>?2⋅0−10<?2⋅0−10>−1 True0<−1 False\begin{array}{ccccc} y>2x-1 & & \mathrm{or} & & y<2x-1? \\ 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−1y>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−1y\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−1y<2x-1.

6. Example 4.71

Example 4.71

Example 4.71

The boundary line shown is 2x+3y=62x+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<62x+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>6or2x+3y<6?2x+3y>62x+3y<62(0)+3(0)>?62(0)+3(0)<?60>6 False0<6 True\begin{array}{rllcccccrll} 2x+3y & > & 6 & & & \mathrm{or} & & & 2x+3y & < & 6? \\ 2x+3y & > & 6 & & & & & & 2x+3y & < & 6 \\ 2(0)+3(0) & \overset{?}{>} & 6 & & & & & & 2(0)+3(0) & \overset{?}{<} & 6 \\ 0 & > & 6\,\mathrm{False} & & & & & & 0 & < & 6\,\mathrm{True} \\ \end{array}

So the side with (0,0)(0,0) is the side where 2x+3y<62x+3y<6.

(You may want to pick a point on the other side of the boundary line and check that 2x+3y>62x+3y>6.)

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

The graph shows the solution to the inequality 2x+3y<62x+3y<6.

7. Graph Linear Inequalities

Graph Linear Inequalities

Now, we’re ready to put all this together to graph linear inequalities.

8. Example 4.72

Example 4.72

Example 4.72

How to Graph Linear Inequalities

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

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

Graph a linear inequality.

  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 >, 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 4.73

Example 4.73

Example 4.73

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

Solution

First we graph the boundary line x−2y=5x-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<5x-2y<5?

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

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 4.74

Example 4.74

Example 4.74

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

Solution

First we graph the boundary line y=−4xy=-4x. It is in slope–intercept form, with m=−4 and b=0m=-4\,\mathrm{and}\,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≤−4xy\le-4x?

The point (1,0)(1,0) is not a solution to y≤−4xy\le-4x, so we shade in the opposite side of the boundary line. See Figure 4.35.

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. Do you remember?

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

11. Example 4.75

Example 4.75

Example 4.75

Graph the linear inequality y>3y>3.

Solution

First we graph the boundary line y=3y=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}

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

So we shade the side that does not include (0, 0).

12. Section 4.7 Exercises

Section 4.7 Exercises

Section 4.7 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 the following exercises, graph each linear inequality.

Everyday Math

Writing Exercises

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

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

13. Practice 1

  1. Determine whether each ordered pair is a solution to the inequality y>x−3y>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+1y<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+3y=-2x+3.

14. Practice 2

  1. Write the inequality shown by the graph with the boundary line y=12x−4y=\frac{1}{2}x-4.
  1. Write the inequality shown by the shaded region in the graph with the boundary line x−4y=8x-4y=8.
  1. Write the inequality shown by the shaded region in the graph with the boundary line 3x−y=63x-y=6.

15. Practice 3

  1. Graph the linear inequality y>52x−4y>\frac{5}{2}x-4.
  1. Graph the linear inequality y<23x−5y<\frac{2}{3}x-5.
  1. Graph the linear inequality 2x−3y≤62x-3y\le{}6.

16. Practice 4

  1. Graph the linear inequality 2x−y>32x-y>3.
  1. Graph the linear inequality y>−3xy>-3x.
  1. Graph the linear inequality y≥−2xy\ge-2x.

17. Practice 5

  1. Graph the linear inequality y<5y<5.
  1. Graph the linear inequality y≤−1y\le-1.
  1. Determine whether each ordered pair is a solution to the inequality y>x−3y>x-3:
  1. ⓐ (0,0)(0,0)
  1. ⓑ (2,1)(2,1)
  1. ⓒ (−1,−5)(-1,-5)
  1. ⓓ (−6,−3)(-6,-3)
  1. ⓔ (1,0)(1,0)

18. Practice 6

  1. Determine whether each ordered pair is a solution to the inequality y<x+2y<x+2:
  1. ⓐ (0,3)(0,3)
  1. ⓑ (−3,−2)(-3,-2)
  1. ⓒ (−2,0)(-2,0)
  1. ⓓ (0,0)(0,0)
  1. ⓔ (−1,4)(-1,4)
  1. Determine whether each ordered pair is a solution to the inequality y<x+5y<x+5:
  1. ⓐ (−3,0)(-3,0)
  1. ⓑ (1,6)(1,6)
  1. ⓒ (−6,−2)(-6,-2)
  1. ⓓ (0,1)(0,1)
  1. ⓔ (5,−4)(5,-4)
  1. Determine whether each ordered pair is a solution to the inequality x+y>2x+y>2:
  1. ⓐ (1,1)(1,1)
  1. ⓑ (4,−3)(4,-3)
  1. ⓒ (0,0)(0,0)
  1. ⓓ (−8,12)(-8,12)
  1. ⓔ (3,0)(3,0)

19. Practice 7

  1. Write the inequality shown by the graph with the boundary line y=2x−4.y=2x-4.
  1. Write the inequality shown by the graph with the boundary line y=−13x−2.y=-\frac{1}{3}x-2.
  1. Write the inequality shown by the shaded region in the graph with the boundary line x+y=3.x+y=3.

20. Practice 8

  1. Write the inequality shown by the shaded region in the graph with the boundary line x+2y=−2.x+2y=-2.
  1. Write the inequality shown by the shaded region in the graph with the boundary line 2x−y=4.2x-y=4.
  1. Write the inequality shown by the shaded region in the graph with the boundary line 4x−3y=12.4x-3y=12.

21. Practice 9

  1. Graph the linear inequality y<35x+2y<\frac{3}{5}x+2.
  1. Graph the linear inequality y≥−13x−2y\ge-\frac{1}{3}x-2.
  1. Graph the linear inequality x−y≥−2x-y\ge-2.

22. Practice 10

  1. Graph the linear inequality 4x+y>−44x+y>-4.
  1. Graph the linear inequality x+5y<−5x+5y<-5.
  1. Graph the linear inequality 4x+2y≥−84x+2y\ge-8.

23. Practice 11

  1. Graph the linear inequality y>xy>x.
  1. Graph the linear inequality y≤−xy\le-x.
  1. Graph the linear inequality y≤−3xy\le-3x.

24. Practice 12

  1. Graph the linear inequality y<−1y<-1.
  1. Graph the linear inequality y≥2y\ge{}2.
  1. Graph the linear inequality x>−2x>-2.

25. Practice 13

  1. Graph the linear inequality x≤4x\le{}4.
  1. Graph the linear inequality x−y<−3x-y<-3.
  1. Graph the linear inequality y≤54xy\le\frac{5}{4}x.

26. Practice 14

  1. Graph the linear inequality y<−3x−4y<-3x-4.
  1. Graph the linear inequality x≥0x\ge{}0.
  1. Shopping. Tula has 20tospendattheusedbooksale.Hardcoverbookscost20 to spend at the used book sale. Hardcover books cost 2 each and paperback books cost 0.50each.If∗x∗isthenumberofhardcoverbooksTulacanbuyand∗y∗isthenumberofpaperbackbooksshecanbuy,theinequality0.50 each. If *x* is the number of hardcover books Tula can buy and *y* is the number of paperback books she can buy, the inequality 2x+\frac{1}{2}y\le{}20$ models the situation.
  1. ⓐ Graph the inequality.
  1. ⓑ List three solutions to the inequality 2x+12y≤202x+\frac{1}{2}y\le{}20 where both x and y are whole numbers.

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