# 5.4: Graphs of Linear Inequalities in Two Variables (optional challenge practice)

Learning Objectives

By the end of this section, you will be able to:

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

Note

Before you get started, take this readiness quiz.

1. Solve: (4x+3>23.)
If you missed this problem, review Exercise 2.7.22.
2. Translate from algebra to English: (x<5.)
If you missed this problem, review Exercise 1.3.1.
3. Evaluate (3x−2y) when (x=1, , y=−2.)
If you missed this problem, review Exercise 1.5.28.

## 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.

LINEAR INEQUALITY

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

[A x+B y>C quad A x+B y geq C quad A x+B y

where (A) and (B) are not both zero.

Do you remember that an inequality with one variable had many solutions? The solution to the inequality (x>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 (PageIndex{1}). Similarly, inequalities in two variables have many solutions. Any ordered pair ( (x, y)) that makes the inequality true when we substitute in the values is a solution of the inequality.

Solution OF A LINEAR INEQUALITY

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

Exercise (PageIndex{1})

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

1. ((0,0))
2. ((1,6))
3. ((2,6))
4. ((−5,−15))
5. ((−8,12))
1.
 ((0,0))   Simplify. So, ((0,0)) is not a solution to (y>x+4).
2.
 ((1,6))   Simplify. So, ((1,6)) is a solution to (y>x+4).
3.
 ((2,6))   Simplify. So, ((2,6)) is not a solution to (y>x+4).
4.
 ((−5,−15))   Simplify. So, ((−5,−15)) is not a solution to (y>x+4).
5.
 (−8,12)   Simplify. So, ((−8,12)) is a solution to (y>x+4).

Exercise (PageIndex{2})

Determine whether each ordered pair is a solution to the inequality (y>x−3):

1. ((0,0))
2. ((4,9))
3. ((−2,1))
4. ((−5,−3))
5. ((5,1))
1. yes
2. yes
3. yes
4. yes
5. no

Exercise (PageIndex{3})

Determine whether each ordered pair is a solution to the inequality (y

1. ((0,0))
2. ((8,6))
3. ((−2,−1))
4. ((3,4))
5. ((−1,−4))
1. yes
2. yes
3. no
4. no
5. yes

## 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 (PageIndex{1}) again. The point (x=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 (PageIndex{2}). The solution to (x>3) is the shaded part of the number line to the right of (x=3).

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

BOUNDARY LINE

The line with equation (Ax+By=C) is the boundary line that separates the region where (Ax+By>C) from the region where (Ax+By

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 (PageIndex{1}).

 (Ax+By (Ax+Byleq C) (Ax+By>C) (Ax+Bygeq C) Boundary line is not included in solution. Boundary line is included in solution. Boundary line is dashed. Boundary line is solid.

Now, let’s take a look at what we found in Exercise (PageIndex{1}). We’ll start by graphing the line (y=x+4), and then we’ll plot the five points we tested. See Figure (PageIndex{3}). In Exercise (PageIndex{1}) we found that some of the points were solutions to the inequality (y>x+4) and some were not.

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

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

What about the point ((2,6))? Because (6=2+4), the point is a solution to the equation (y=x+4). So the point ((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+4). The point ((0,10)) clearly looks to be to the left of the boundary line, doesn’t it? Is it a solution to the inequality?

[egin{array}{l}{y>x+4} {10stackrel{?}{>}0+4} {10>4} &{ ext{So, }(0,10) ext{ 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+4). All points on the left are solutions.

Similarly, all points on the right side of the boundary line, the side with ((0,0)) and ((−5,−15)), are not solutions to (y>x+4). See Figure (PageIndex{4}). The graph of the inequality (y>x+4) is shown in Figure (PageIndex{5}) below. The line (y=x+4) divides the plane into two regions. The shaded side shows the solutions to the inequality (y>x+4).

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

The boundary line shown is (y=2x−1). Write the inequality shown by the graph.

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

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

At ((0,0)), which inequality is true:

[egin{array}{ll}{y>2 x-1} & { ext { or }} & {y<2 x-1 ?} {y>2 x-1} && {y<2 x-1} {0>2 cdot 0-1} && {0<2 cdot 0-1} {0>-1 ext { True }} && {0<-1 ext { False }}end{array}]

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

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

The graph shows the inequality (ygeq 2x−1).

We could use any point as a test point, provided it is not on the line. Why did we choose ((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).

Exercise (PageIndex{5})

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

(ygeq −2x+3)

Exercise (PageIndex{6})

Write the inequality shown by the graph with the boundary line (y=frac{1}{2}x−4).

(y leq frac{1}{2}x - 4)

Exercise (PageIndex{7})

The boundary line shown is (2x+3y=6). Write the inequality shown by the graph.

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

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

At ((0,0)), which inequality is true:

[egin{array}{rr}{2 x+3 y>6} && { ext { or } quad 2 x+3 y<6 ?} {2 x+3 y>6} && {2 x+3 y<6} {2(0)+3(0)>6} & & {2(0)+3(0)<6} {0} >6 & { ext { False }} & {0<6}&{ ext { True }}end{array}]

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

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

Exercise (PageIndex{8})

Write the inequality shown by the shaded region in the graph with the boundary line (x−4y=8).

(x-4 y leq 8)

Exercise (PageIndex{9})

Write the inequality shown by the shaded region in the graph with the boundary line (3x−y=6).

(3 x-y leq 6)

## Graph Linear Inequalities

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

Exercise (PageIndex{10}): How to Graph Linear Inequalities

Graph the linear inequality (y geq frac{3}{4} x-2).   Exercise (PageIndex{11})

Graph the linear inequality (y geq frac{5}{2} x-4). Exercise (PageIndex{12})

Graph the linear inequality (y The steps we take to graph a linear inequality are summarized here.

GRAPH A LINEAR INEQUALITY.

1. Identify and graph the boundary line.
• If the inequality is (≤) or (≥), the boundary line is solid.
• If the inequality is (<) or (>), the boundary line is dashed.
2. Test a point that is not on the boundary line. Is it a solution of the inequality?
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.

Exercise (PageIndex{13})

Graph the linear inequality (x−2y<5).

First we graph the boundary line (x−2y=5). The inequality is (<) so we draw a dashed line. Then we test a point. We’ll use ((0,0)) again because it is easy to evaluate and it is not on the boundary line.

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

Exercise (PageIndex{14})

Graph the linear inequality (2x−3yleq 6). Exercise (PageIndex{15})

Graph the linear inequality (2x−y>3). What if the boundary line goes through the origin? Then we won’t be able to use ((0,0)) as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

Exercise (PageIndex{16})

Graph the linear inequality (yleq −4x).

First we graph the boundary line (y=−4x). It is in slope–intercept form, with (m=−4) and (b=0). The inequality is (≤) so we draw a solid line. Now, we need a test point. We can see that the point ((1,0)) is not on the boundary line.

Is ((1,0)) a solution of (y≤−4x)? The point ((1,0)) is not a solution to (y≤−4x), so we shade in the opposite side of the boundary line. See Figure (PageIndex{6}). Exercise (PageIndex{17})

Graph the linear inequality (y>−3x). Exercise (PageIndex{18})

Graph the linear inequality (ygeq −2x). 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?

(egin{array}{ll}{x=a} & { ext { vertical line }} {y=b} & { ext { horizontal line }}end{array})

Exercise (PageIndex{19})

Graph the linear inequality (y>3).

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

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

[y>3 0 ot>3]

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

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

Exercise (PageIndex{20})

Graph the linear inequality (y<5). Exercise (PageIndex{21})

Graph the linear inequality (y leq-1). ## Key Concepts

• To Graph a Linear Inequality
1. Identify and graph the boundary line.
If the inequality is (≤) or (≥), the boundary line is solid.
If the inequality is (<) or (>), the boundary line is dashed.
2. Test a point that is not on the boundary line. Is it a solution of the inequality?
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.

## Glossary

boundary line
The line with equation (A x+B y=C) that separates the region where (A x+B y>C) from the region where (A x+B y
linear inequality
An inequality that can be written in one of the following forms:

[A x+B y>C quad A x+B y geq C quad A x+B ywhere (A) and (B) are not both zero.

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

## 5.4 Solve Applications with Systems of Equations

Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem Solving Strategy for Systems of Linear Equations.

### How To

#### Use a problem solving strategy for systems of linear equations.

1. Step 1. Read the problem. Make sure all the words and ideas are understood.
2. Step 2. Identify what we are looking for.
3. Step 3. Name what we are looking for. Choose variables to represent those quantities.
4. Step 4. Translate into a system of equations.
5. Step 5. Solve the system of equations using good algebra techniques.
6. Step 6. Check the answer in the problem and make sure it makes sense.
7. Step 7. Answer the question with a complete sentence.

### Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models.

• The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
• A married couple together earns $110,000 a year. The wife earns$16,000 less than twice what her husband earns. What does the husband earn?

In that chapter we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?

Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

### Example 5.35

#### How to Translate to a System of Equations

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

#### Solution

Translate to a system of equations:

The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

Translate to a system of equations:

The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.

We’ll do another example where we stop after we write the system of equations.

### Example 5.36

Translate to a system of equations:

A married couple together earns $110,000 a year. The wife earns$16,000 less than twice what her husband earns. What does the husband earn?

#### Solution

Translate to a system of equations:

A couple has a total household income of $84,000. The husband earns$18,000 less than twice what the wife earns. How much does the wife earn?

Translate to a system of equations:

A senior employee makes $5 less than twice what a new employee makes per hour. Together they make$43 per hour. How much does each employee make per hour?

### Solve Direct Translation Applications

We set up, but did not solve, the systems of equations in Example 5.35 and Example 5.36 Now we’ll translate a situation to a system of equations and then solve it.

### Example 5.37

Translate to a system of equations and then solve:

Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

#### Solution

 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the ages of Devon and Cooper. Step 3. Name what we are looking for. Let d = d = Devon’s age. c = c = Cooper’s age Step 4. Translate into a system of equations. Devon is 26 years older than Cooper. The sum of their ages is 50. The system is: Step 5. Solve the system of equations. Solve by substitution. Substitute c + 26 into the second equation. Solve for c. Substitute c = 12 into the first equation and then solve for d. Step 6. Check the answer in the problem. Is Devon’s age 26 more than Cooper’s? Yes, 38 is 26 more than 12. Is the sum of their ages 50? Yes, 38 plus 12 is 50. Step 7. Answer the question. Devon is 38 and Cooper is 12 years old.

Translate to a system of equations and then solve:

Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.

Translate to a system of equations and then solve:

Jake’s dad is 6 more than 3 times Jake’s age. The sum of their ages is 42. Find their ages.

### Example 5.38

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

#### Solution

 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the number of calories burned each minute on the elliptical trainer and each minute of circuit training. Step 3. Name what we are looking for. Let e = e = number of calories burned per minute on the elliptical trainer. c = c = number of calories burned per minute while circuit training Step 4. Translate into a system of equations. 10 minutes on the elliptical and circuit training for 20 minutes, burned 278 calories 20 minutes on the elliptical and 30 minutes of circuit training burned 473 calories The system is: Step 5. Solve the system of equations. Multiply the first equation by −2 to get opposite coefficients of e. Simplify and add the equations. Solve for c. Substitute c = 8.3 into one of the original equations to solve for e. Step 6. Check the answer in the problem. Check the math on your own. Step 7. Answer the question. Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.

Translate to a system of equations and then solve:

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Translate to a system of equations and then solve:

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

### Solve Geometry Applications

When we learned about Math Models, we solved geometry applications using properties of triangles and rectangles. Now we’ll add to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

### Complementary and Supplementary Angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

### Example 5.39

Translate to a system of equations and then solve:

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

#### Solution

 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the measure of each angle. Step 3. Name what we are looking for. Let x = x = the measure of the first angle x = x = . m = m = the measure of the second angle. Step 4. Translate into a system of equations. The angles are complementary. x + y = 90 x + y = 90 The difference of the two angles is 26 degrees. x - y = 26 x - y = 26 The system is < x + y = 90 x − y = 26 < x + y = 90 x − y = 26 Step 5. Solve the system of equations by elimination. < x + y = 90 x − y = 26 _________ 2 x = 116 < x + y = 90 x − y = 26 _________ 2 x = 116 Substitute x = 58 x = 58 into the first equation. x + y = 90 58 + y = 90 y = 32 x + y = 90 58 + y = 90 y = 32 Step 6. Check the answer in the problem. 58 + 32 = 90 ✓ 58 − 32 = 26 ✓ 58 + 32 = 90 ✓ 58 − 32 = 26 ✓ Step 7. Answer the question. The angle measures are 58 degrees and 42 degrees.

Translate to a system of equations and then solve:

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

Translate to a system of equations and then solve:

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

### Example 5.40

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

#### Solution

 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the measure of each angle. Step 3. Name what we are looking for. Let x = x = the measure of the first angle. y = y = the measure of the second angle Step 4. Translate into a system of equations. The angles are supplementary. The larger angle is twelve less than five times the smaller angle The system is: Step 5. Solve the system of equations substitution. Substitute 5x − 12 for y in the first equation. Solve for x. Substitute 32 for in the second equation, then solve for y. Step 6. Check the answer in the problem. 32 + 158 = 180 ✓ 5 · 32 − 12 = 147 ✓ 32 + 158 = 180 ✓ 5 · 32 − 12 = 147 ✓ Step 7. Answer the question. The angle measures are 148 and 32.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

### Example 5.41

Translate to a system of equations and then solve:

Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

#### Solution

 Step 1. Read the problem. Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what we are looking for. Let L = L = the length of the fenced yard. W = W = the width of the fenced yard Step 4. Translate into a system of equations. One length and two widths equal 125. The length will be 5 feet more than four times the width. The system is: Step 5. Solve the system of equations by substitution. Substitute L = 4W + 5 into the first equation, then solve for W. Substitute 20 for W in the second equation, then solve for L. Step 6. Check the answer in the problem. 20 + 28 + 20 = 125 ✓ 85 = 4 · 20 + 5 ✓ 20 + 28 + 20 = 125 ✓ 85 = 4 · 20 + 5 ✓ Step 7. Answer the equation. The length is 85 feet and the width is 20 feet.

Translate to a system of equations and then solve:

Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

Translate to a system of equations and then solve:

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

### Solve Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance travelled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

### Example 5.42

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

#### Solution

A diagram is useful in helping us visualize the situation.

 Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k, and Joni, j, will each drive. Since D = r · t D = r · t we can fill in the Distance column. Translate into a system of equations. To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, 65 j = 78 k . 65 j = 78 k . Also, since Kelly left later, her time will be 1 2 1 2 hour less than Joni’s time. So, k = j − 1 2 . k = j − 1 2 . Now we have the system. Solve the system of equations by substitution. Substitute k = j − 1 2 k = j − 1 2 into the second equation, then solve for j. To find Kelly’s time, substitute j = 3 into the first equation, then solve for k. Check the answer in the problem. Joni 3 hours (65 mph) = 195 miles. Kelly 2 1 2 2 1 2 hours (78 mph) = 195 miles. Yes, they will have traveled the same distance when they meet. Answer the question. Kelly will catch up to Joni in 2 1 2 2 1 2 hours. By then, Joni will have traveled 3 hours.

Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

Translate to a system of equations and then solve: Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes (1/4 hour) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

Figure 5.7 and Figure 5.8 show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c.

In Figure 5.7 the boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c.

In Figure 5.8 the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b − c b − c .

We’ll put some numbers to this situation in Example 5.43.

### Example 5.43

Translate to a system of equations and then solve:

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

#### Solution

This is a uniform motion problem and a picture will help us visualize the situation.

 Identify what we are looking for. We are looking for the speed of the ship in still water and the speed of the current. Name what we are looking for. Let s = s = the rate of the ship in still water. c = c = the rate of the current A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship therefore, the ship’s actual rate is s + c. Going upstream, the current slows the ship therefore, the actual rate is s − c. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles. Translate into a system of equations. Since rate times time is distance, we can write the system of equations. Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination. Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for s. Substitute s = 13.5 into one of the original equations. Check the answer in the problem. The downstream rate would be 13.5 + 1.5 = 15 mph. In 4 hours the ship would travel 15 · 4 = 60 miles. The upstream rate would be 13.5 − 1.5 = 12 mph. In 5 hours the ship would travel 12 · 5 = 60 miles. Answer the question. The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

Translate to a system of equations and then solve: A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in Example 5.44. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.

### Example 5.44

Translate to a system of equations and then solve:

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

#### Solution

This is a uniform motion problem and a picture will help us visualize.

 Identify what we are looking for. We are looking for the speed of the jet in still air and the speed of the wind. Name what we are looking for. Let j = j = the speed of the jet in still air. w = w = the speed of the wind A chart will help us organize the information. The jet makes two trips-one in a tailwind and one in a headwind. In a tailwind, the wind helps the jet and so the rate is j + w. In a headwind, the wind slows the jet and so the rate is j − w. Each trip takes 3 hours. In a tailwind the jet flies 1095 miles. In a headwind the jet flies 987 miles. Translate into a system of equations. Since rate times time is distance, we get the system of equations. Solve the system of equations. Distribute, then solve by elimination. Add, and solve for j. Substitute j = 347 into one of the original equations, then solve for w. Check the answer in the problem. With the tailwind, the actual rate of the jet would be 347 + 18 = 365 mph. In 3 hours the jet would travel 365 · 3 = 1095 miles. Going into the headwind, the jet’s actual rate would be 347 − 18 = 329 mph. In 3 hours the jet would travel 329 · 3 = 987 miles. Answer the question. The rate of the jet is 347 mph and the rate of the wind is 18 mph.

Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Translate to a system of equations and then solve: A commercial jet can fly 1728 miles in 4 hours with a tailwind but only 1536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

### Section 5.4 Exercises

#### Practice Makes Perfect

Translate to a System of Equations

In the following exercises, translate to a system of equations and solve the system.

The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.

The sum of two numbers is twenty-five. One number is five less than the other. Find the numbers.

The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.

The sum of two numbers is negative sixteen. One number is seven times the other. Find the numbers.

Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.

Six times a number plus twice a second number is four. Twice the first number plus four times the second number is eighteen. Find the numbers.

Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.

Twice a number plus three times a second number is negative one. The first number plus four times the second number is two. Find the numbers.

A married couple together earn $75,000. The husband earns$15,000 more than five times what his wife earns. What does the wife earn?

During two years in college, a student earned $9,500. The second year she earned$500 more than twice the amount she earned the first year. How much did she earn the first year?

Daniela invested a total of $50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was$5000 more than twice the amount she put into the CD. How much did she invest in each account?

Jorge invested $28,000 into two accounts. The amount he put in his money market account was$2,000 less than twice what he put into a CD. How much did he invest in each account?

In her last two years in college, Marlene received $42,000 in loans. The first year she received a loan that was$6,000 less than three times the amount of the second year’s loan. What was the amount of her loan for each year?

Jen and David owe $22,000 in loans for their two cars. The amount of the loan for Jen’s car is$2000 less than twice the amount of the loan for David’s car. How much is each car loan?

Solve Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.

Robert is 15 years older than his sister, Helen. The sum of their ages is sixty-three. Find their ages.

The age of Noelle’s dad is six less than three times Noelle’s age. The sum of their ages is seventy-four. Find their ages.

The age of Mark’s dad is 4 less than twice Marks’s age. The sum of their ages is ninety-five. Find their ages.

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball?

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for$68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for$12. What was the cost per pound of the oranges and the bananas?

Solve Geometry Applications In the following exercises, translate to a system of equations and solve.

The difference of two complementary angles is 30 degrees. Find the measures of the angles.

The difference of two complementary angles is 68 degrees. Find the measures of the angles.

The difference of two supplementary angles is 70 degrees. Find the measures of the angles.

The difference of two supplementary angles is 24 degrees. Find the measure of the angles.

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his rectangular front yard. The length, the side along the house, is five feet less than three times the width. Find the length and width of the fencing.

A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a rectangular toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solve Uniform Motion Applications In the following exercises, translate to a system of equations and solve.

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.).

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current. (Round to the nearest hundredth.).

A small jet can fly 1,072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1,435 miles in 5 hours with a tailwind but only 1215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1,170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

At a school concert, 425 tickets were sold. Student tickets cost $5 each and adult tickets cost$8 each. The total receipts for the concert were $2,851. Solve the system The first graders at one school went on a field trip to the zoo. The total number of children and adults who went on the field trip was 115. The number of adults was 1 4 1 4 the number of children. Solve the system #### Writing Exercises Write an application problem similar to Example 5.37 using the ages of two of your friends or family members. Then translate to a system of equations and solve it. Write a uniform motion problem similar to Example 5.42 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it. #### Self Check ⓐ 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? As an Amazon Associate we earn from qualifying purchases. Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax. If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: • Use the information below to generate a citation. We recommend using a citation tool such as this one. • Authors: Lynn Marecek, MaryAnne Anthony-Smith • Publisher/website: OpenStax • Book title: Elementary Algebra • Publication date: Feb 22, 2017 • Location: Houston, Texas • Book URL: https://openstax.org/books/elementary-algebra/pages/1-introduction • Section URL: https://openstax.org/books/elementary-algebra/pages/5-4-solve-applications-with-systems-of-equations © Nov 10, 2020 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. ## 3.4 Graph Linear Inequalities 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. ### Linear Inequality A linear inequality is an inequality that can be written in one of the following forms: Where A and B are not both zero. 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 . ### Solution to a Linear Inequality ### Example 3.34 Determine whether each ordered pair is a solution to the inequality y > x + 4 : y > x + 4 : #### Solution Determine whether each ordered pair is a solution to the inequality y > x − 3 : y > x − 3 : Determine whether each ordered pair is a solution to the inequality y < x + 1 : y < x + 1 : ### 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. ### Boundary Line 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. 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 + 4 y > x + 4 and some were not. Which of the points we plotted are solutions to the inequality y > x + 4 ? y > x + 4 ? 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? 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. The graph of the inequality y > x + 4 y > x + 4 is shown in below. ### Example 3.35 The boundary line shown in this graph is y = 2 x − 1 . y = 2 x − 1 . Write the inequality shown by the graph. #### Solution Since the boundary line is graphed with a solid line, the inequality includes the equal sign. The graph shows the inequality y ≥ 2 x − 1 . y ≥ 2 x − 1 . Write the inequality shown by the graph with the boundary line y = −2 x + 3 . y = −2 x + 3 . Write the inequality shown by the graph with the boundary line y = 1 2 x − 4 . y = 1 2 x − 4 . ### Example 3.36 The boundary line shown in this graph is 2 x + 3 y = 6 . 2 x + 3 y = 6 . Write the inequality shown by the graph. #### Solution (You may want to pick a point on the other side of the boundary line and check that 2 x + 3 y > 6 . 2 x + 3 y > 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 2 x + 3 y < 6 . 2 x + 3 y < 6 . Write the inequality shown by the shaded region in the graph with the boundary line x − 4 y = 8 . x − 4 y = 8 . Write the inequality shown by the shaded region in the graph with the boundary line 3 x − y = 6 . 3 x − y = 6 . ### 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. ### Example 3.37 #### How to Graph a Linear Equation in Two Variables Graph the linear inequality y ≥ 3 4 x − 2 . y ≥ 3 4 x − 2 . #### Solution Graph the linear inequality y ≥ 5 2 x − 4 . y ≥ 5 2 x − 4 . Graph the linear inequality y ≤ 2 3 x − 5 . y ≤ 2 3 x − 5 . The steps we take to graph a linear inequality are summarized here. ### How To #### Graph a linear inequality in two variables. ### Example 3.38 Graph the linear inequality x − 2 y < 5 . x − 2 y < 5 . #### Solution All points in the shaded region, but not those on the boundary line, represent the solutions to x − 2 y < 5 . x − 2 y < 5 . Graph the linear inequality: 2 x − 3 y < 6 . 2 x − 3 y < 6 . Graph the linear inequality: 2 x − y > 3 . 2 x − y > 3 . 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. ### Example 3.39 Graph the linear inequality: y ≤ ​ − 4 x . y ≤ ​ − 4 x . #### Solution Now we need a test point. We can see that the point ( 1 , 0 ) ( 1 , 0 ) is not on the boundary line. All points in the shaded region and on the boundary line represent the solutions to y ≤ ​ − 4 x . y ≤ ​ − 4 x . Graph the linear inequality: y > − 3 x . y > − 3 x . Graph the linear inequality: y ≥ −2 x . y ≥ −2 x . 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. ### Example 3.40 Graph the linear inequality: y > 3 . y > 3 . #### Solution 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 . Graph the linear inequality: y < 5 . y < 5 . Graph the linear inequality: y ≤ −1 . y ≤ −1 . ### 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. ### Example 3.41 Hilaria works two part time jobs in order to earn enough money to meet her obligations of at least$240 a week. Her job in food service pays $10 an hour and her tutoring job on campus pays$15 an hour. How many hours does Hilaria need to work at each job to earn at least $240? #### 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$10 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. 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.

Hugh works two part time jobs. One at a grocery store that pays $10 an hour and the other is babysitting for$13 hour. Between the two jobs, Hugh wants to earn at least $260 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.

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

Veronica works two part time jobs in order to earn enough money to meet her obligations of at least $280 a week. Her job at the day spa pays$10 an hour and her administrative assistant job on campus pays $17.50 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.

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

### Media

Access this online resource for additional instruction and practice with graphing linear inequalities in two variables.

### Section 3.4 Exercises

#### Practice Makes Perfect

Verify Solutions to an Inequality in Two Variables

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

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

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

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

Determine whether each ordered pair is a solution to the inequality y < − 2 x + 5 : y < − 2 x + 5 :

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

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

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

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

Write the inequality shown by the graph with the boundary line y = 3 x − 4 . y = 3 x − 4 .

Write the inequality shown by the graph with the boundary line y = 2 x − 4 . y = 2 x − 4 .

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

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

Write the inequality shown by the shaded region in the graph with the boundary line x + y = 5 . x + y = 5 .

Write the inequality shown by the shaded region in the graph with the boundary line x + y = 3 . x + y = 3 .

Write the inequality shown by the shaded region in the graph with the boundary line 3 x − y = 6 . 3 x − y = 6 .

Write the inequality shown by the shaded region in the graph with the boundary line 2 x − y = 4 . 2 x − y = 4 .

Graph Linear Inequalities in Two Variables

In the following exercises, graph each linear inequality.

Graph the linear inequality: y > 2 3 x − 1 . y > 2 3 x − 1 .

Graph the linear inequality: y < 3 5 x + 2 . y < 3 5 x + 2 .

Graph the linear inequality: y ≤ − 1 2 x + 4 . y ≤ − 1 2 x + 4 .

Graph the linear inequality: y ≥ − 1 3 x − 2 . y ≥ − 1 3 x − 2 .

Graph the linear inequality: x − y ≤ 3 . x − y ≤ 3 .

Graph the linear inequality: x − y ≥ −2 . x − y ≥ −2 .

Graph the linear inequality: 4 x + y > − 4 . 4 x + y > − 4 .

Graph the linear inequality: x + 5 y < − 5 . x + 5 y < − 5 .

Graph the linear inequality: 3 x + 2 y ≥ −6 . 3 x + 2 y ≥ −6 .

Graph the linear inequality: 4 x + 2 y ≥ −8 . 4 x + 2 y ≥ −8 .

Graph the linear inequality: y > 4 x . y > 4 x .

Graph the linear inequality: y ≤ −3 x . y ≤ −3 x .

Graph the linear inequality: y < − 10 . y < − 10 .

Graph the linear inequality: y ≥ 2 . y ≥ 2 .

Graph the linear inequality: x ≤ 5 . x ≤ 5 .

Graph the linear inequality: x ≥ 0 . x ≥ 0 .

Graph the linear inequality: x − y < 4 . x − y < 4 .

Graph the linear inequality: x − y < − 3 . x − y < − 3 .

Graph the linear inequality: y ≥ 3 2 x . y ≥ 3 2 x .

Graph the linear inequality: y ≤ 5 4 x . y ≤ 5 4 x .

Graph the linear inequality: y > − 2 x + 1 . y > − 2 x + 1 .

Graph the linear inequality: y < − 3 x − 4 . y < − 3 x − 4 .

Graph the linear inequality: 2 x + y ≥ −4 . 2 x + y ≥ −4 .

Graph the linear inequality: x + 2 y ≤ −2 . x + 2 y ≤ −2 .

Graph the linear inequality: 2 x − 5 y > 10 . 2 x − 5 y > 10 .

Graph the linear inequality: 4 x − 3 y > 12 . 4 x − 3 y > 12 .

Solve Applications using Linear Inequalities in Two Variables

Harrison works two part time jobs. One at a gas station that pays $11 an hour and the other is IT troubleshooting for$ 16.50 $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. Elena needs to earn at least$450 a week during her summer break to pay for college. She works two jobs. One as a swimming instructor that pays $9 an hour and the other as an intern in a genetics lab for$22.50 per hour. How many hours does Elena need to work at each job to earn at least $450 per week? ⓐ Let x be the number of hours she works teaching swimming and let y be the number of hours she works as an intern. Write an inequality that would model this situation. 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. ⓒ List three solutions to the inequality. What options do the solutions provide Laura? Armando’s workouts consist of kickboxing and swimming. While kickboxing, he burns 10 calories per minute and he burns 7 calories a minute while swimming. He wants to burn 600 calories each day. ⓐ If x is the number of minutes that Armando will kickbox and y is the number minutes he will swim, find the inequality that will help Armando create a workout for today. ⓒ List three solutions to the inequality. What options do the solutions provide Armando? #### Writing Exercises Explain why, in some graphs of linear inequalities, the boundary line is solid but in other graphs it is dashed. #### Self Check ⓐ 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? As an Amazon Associate we earn from qualifying purchases. Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax. If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: • Use the information below to generate a citation. We recommend using a citation tool such as this one. • Authors: Lynn Marecek, Andrea Honeycutt Mathis • Publisher/website: OpenStax • Book title: Intermediate Algebra 2e • Publication date: May 6, 2020 • Location: Houston, Texas • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/3-4-graph-linear-inequalities-in-two-variables © Jan 21, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. ## How to Represent Inequalities We can represent inequalities in one of three different ways: Inequalities as written expressions use only mathematical symbols and no diagrams. They are exactly what we have been working with above (e.g.,$y > 37$). An inequality number line allows us to visualize the set of numbers that represents our inequality. We use a dark line to show all the numbers that match our inequality, and we mark where the inequality begins and/or ends in two different ways. To mark the beginning of an inequality that is "greater than" or "less than," we use an open circle. This shows that the starting number is NOT included. To mark the beginning of an inequality that is "greater than or equal to" or "less than or equal to," we use a closed circle. This shows that the starting number IS included. We can also combine these symbols if our inequality equation requires us to use two different symbols. For instance, if we have$-3 < x ≤ 3$, our number line would look like: And finally, we can have inequalities in graphs for any and all types of graphs on the coordinate plane ( more on the coordinate plane here ). "Greater than" will be above the line of the graph, while "less than" will be below the line of the graph. This is true no matter which direction the line of the graph extends. In terms of markings, inequality graphs follow the same rules as inequalities on number lines. Just as we use an open circle for "greater than" or "less than" inequalities, we use a dashed line for inequality graphs that are "greater than" or "less than." And just how we use a closed circle for "greater than or equal to" or "less than or equal to" inequalities, we use a solid line for our graphs that are greater or less than or equal to.  And now to dive right in to ACT inequality problems! (Awkward flailing optional). ## Basic instructions for the worksheets Each worksheet is randomly generated and thus unique. The answer key is automatically generated and is placed on the second page of the file. You can generate the worksheets either in html or PDF format &mdash both are easy to print. To get the PDF worksheet, simply push the button titled "Create PDF" or "Make PDF worksheet". To get the worksheet in html format, push the button "View in browser" or "Make html worksheet". This has the advantage that you can save the worksheet directly from your browser (choose File &rarr Save) and then edit it in Word or other word processing program. Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options: • PDF format: come back to this page and push the button again. • Html format: simply refresh the worksheet page in your browser window. ## 5.6 Graphing Systems of Linear Inequalities The definition of a system of linear inequalities is very similar to the definition of a system of linear equations. ### System of Linear Inequalities Two or more linear inequalities grouped together form a system of linear inequalities . A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown below. To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) ( x , y ) that make both inequalities true. ### Solutions of a System of Linear Inequalities Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the x-y coordinate system that includes all the points whose ordered pairs make the inequalities true. To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system. ### Example 5.51 Determine whether the ordered pair is a solution to the system. < x + 4 y ≥ 10 3 x − 2 y < 12 < x + 4 y ≥ 10 3 x − 2 y < 12 #### Solution The ordered pair (−2, 4) made both inequalities true. Therefore (−2, 4) is a solution to this system. The ordered pair (3,1) made one inequality true, but the other one false. Therefore (3,1) is not a solution to this system. Determine whether the ordered pair is a solution to the system. < x − 5 y > 10 2 x + 3 y > −2 < x − 5 y > 10 2 x + 3 y > −2 Determine whether the ordered pair is a solution to the system. < y > 4 x − 2 4 x − y < 20 < y > 4 x − 2 4 x − y < 20 ### Solve a System of Linear Inequalities by Graphing The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. ### Example 5.52 #### How to Solve a System of Linear inequalities Solve the system by graphing. #### Solution Solve the system by graphing. < y < 3 x + 2 y > − x − 1 < y < 3 x + 2 y > − x − 1 Solve the system by graphing. < y < − 1 2 x + 3 y < 3 x − 4 < y < − 1 2 x + 3 y < 3 x − 4 ### How To #### Solve a system of linear inequalities by graphing. 1. Step 1. Graph the first inequality. • Graph the boundary line. • Shade in the side of the boundary line where the inequality is true. 2. Step 2. On the same grid, graph the second inequality. • Graph the boundary line. • Shade in the side of that boundary line where the inequality is true. 3. Step 3. The solution is the region where the shading overlaps. 4. Step 4. Check by choosing a test point. ### Example 5.53 Solve the system by graphing. < x − y > 3 y < − 1 5 x + 4 < x − y > 3 y < − 1 5 x + 4 #### Solution  Graph x − y > 3, by graphing x − y = 3 and testing a point. The intercepts are x = 3 and y = −3 and the boundary line will be dashed. Test (0, 0). It makes the inequality false. So, shade the side that does not contain (0, 0) red. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which is the darker-shaded region. Solve the system by graphing. < x + y ≤ 2 y ≥ 2 3 x − 1 < x + y ≤ 2 y ≥ 2 3 x − 1 Solve the system by graphing. < 3 x − 2 y ≤ 6 y > − 1 4 x + 5 < 3 x − 2 y ≤ 6 y > − 1 4 x + 5 ### Example 5.54 Solve the system by graphing. < x − 2 y < 5 y > −4 < x − 2 y < 5 y > −4 #### Solution  Graph y > −4, by graphing y = −4 and recognizing that it is a horizontal line through y = −4. The boundary line will be dashed. Test (0, 0). It makes the inequality true. So, shade (blue) the side that contains (0, 0) blue. The point (0, 0) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which is the darker-shaded region. Solve the system by graphing. < y ≥ 3 x − 2 y < −1 < y ≥ 3 x − 2 y < −1 Solve the system by graphing. < x > −4 x − 2 y ≤ − 4 < x > −4 x − 2 y ≤ − 4 Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in Example 5.55. ### Example 5.55 Solve the system by graphing. < 4 x + 3 y ≥ 12 y < − 4 3 x + 1 < 4 x + 3 y ≥ 12 y < − 4 3 x + 1 #### Solution There is no point in both shaded regions, so the system has no solution. This system has no solution. Solve the system by graphing. < 3 x − 2 y ≤ 12 y ≥ 3 2 x + 1 < 3 x − 2 y ≤ 12 y ≥ 3 2 x + 1 Solve the system by graphing. < x + 3 y > 8 y < − 1 3 x − 2 < x + 3 y > 8 y < − 1 3 x − 2 ### Example 5.56 Solve the system by graphing. < y > 1 2 x − 4 x − 2 y < −4 < y > 1 2 x − 4 x − 2 y < −4 #### Solution No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice, which is also the solution to x − 2 y < −4 x − 2 y < −4 . Solve the system by graphing. < y ≥ 3 x + 1 −3 x + y ≥ −4 < y ≥ 3 x + 1 −3 x + y ≥ −4 Solve the system by graphing. < y ≤ − 1 4 x + 2 x + 4 y ≤ 4 < y ≤ − 1 4 x + 2 x + 4 y ≤ 4 ### Solve Applications of Systems of Inequalities The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system as we did above to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so their graphs will only show Quadrant I. ### Example 5.57 Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her$4 and each large photo costs her $10. She doesn’t want to spend more than$200 on photos to display.

ⓐ Write a system of inequalities to model this situation.

ⓒ Could she display 15 small and 5 large photos?

ⓓ Could she display 3 large and 22 small photos?

#### Solution

Notice that we could also test the possible solutions by substituting the values into each inequality.

A trailer can carry a maximum weight of 160 pounds and a maximum volume of 15 cubic feet. A microwave oven weighs 30 pounds and has 2 cubic feet of volume, while a printer weighs 20 pounds and has 3 cubic feet of space.

1. ⓐ Write a system of inequalities to model this situation.
2. ⓑ Graph the system.
3. ⓒ Could 4 microwaves and 2 printers be carried on this trailer?
4. ⓓ Could 7 microwaves and 3 printers be carried on this trailer?

Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost $2 and the answer sheets cost$1. Mary’s budget for these supplies allows for a maximum cost of $400. ⓐ Write a system of inequalities to model this situation. ⓑ Graph the system. ⓒ Could Mary purchase 100 pencils and 100 answer sheets? ⓓ Could Mary purchase 150 pencils and 150 answer sheets? 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Could Mary purchase 100 pencils and 100 answer sheets? 4. ⓓ Could Mary purchase 150 pencils and 150 answer sheets? ### Example 5.58 Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than$5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs .50. 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Could he eat 3 hamburgers and 1 cookie? 4. ⓓ Could he eat 2 hamburgers and 4 cookies? #### Solution The amount spent on hamburgers at$1.40 each, plus the amount spent on cookies at .50 each must be no more than $5.00. We have our system of inequalities. < 240 h + 160 c ≥ 800 1.40 h + 0.50 c ≤ 5 < 240 h + 160 c ≥ 800 1.40 h + 0.50 c ≤ 5 The solution of the system is the region of the graph that is double shaded and so is shaded darker. ⓒ To determine if 3 hamburgers and 2 cookies would meet Omar’s criteria, we see if the point (3, 1) is in the solution region. It is. He might choose to eat 3 hamburgers and 2 cookies. ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is. He might choose to eat 2 hamburgers and 4 cookies. We could also test the possible solutions by substituting the values into each inequality. Tension needs to eat at least an extra 1,000 calories a day to prepare for running a marathon. He has only$25 to spend on the extra food he needs and will spend it on .75 donuts which have 360 calories each and $2 energy drinks which have 110 calories. 1. ⓐ Write a system of inequalities that models this situation. 2. ⓑ Graph the system. 3. ⓒ Can he buy 8 donuts and 4 energy drinks? 4. ⓓ Can he buy 1 donut and 3 energy drinks? Philip’s doctor tells him he should add at least 1000 more calories per day to his usual diet. Philip wants to buy protein bars that cost$1.80 each and have 140 calories and juice that costs $1.25 per bottle and have 125 calories. He doesn’t want to spend more than$12.

1. ⓐ Write a system of inequalities that models this situation.
2. ⓑ Graph the system.
3. ⓒ Can he buy 3 protein bars and 5 bottles of juice?
4. ⓒ Can he buy 5 protein bars and 3 bottles of juice?

### Media

Access these online resources for additional instruction and practice with graphing systems of linear inequalities.

### Section 5.6 Exercises

#### Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

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

Solve a System of Linear Inequalities by Graphing

In the following exercises, solve each system by graphing.

< 2 x − 5 y < 10 3 x + 4 y ≥ 12 < 2 x − 5 y < 10 3 x + 4 y ≥ 12

Solve Applications of Systems of Inequalities

In the following exercises, translate to a system of inequalities and solve.

Caitlyn sells her drawings at the county fair. She wants to sell at least 60 drawings and has portraits and landscapes. She sells the portraits for $15 and the landscapes for$10. She needs to sell at least $800 worth of drawings in order to earn a profit. 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Will she make a profit if she sells 20 portraits and 35 landscapes? 4. ⓓ Will she make a profit if she sells 50 portraits and 20 landscapes? Jake does not want to spend more than$50 on bags of fertilizer and peat moss for his garden. Fertilizer costs $2 a bag and peat moss costs$5 a bag. Jake’s van can hold at most 20 bags.

1. ⓐ Write a system of inequalities to model this situation.
2. ⓑ Graph the system.
3. ⓒ Can he buy 15 bags of fertilizer and 4 bags of peat moss?
4. ⓓ Can he buy 10 bags of fertilizer and 10 bags of peat moss?

Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than $500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is$3 and for a package the cost is $7. 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Can she mail 60 cards and 26 packages? 4. ⓓ Can she mail 90 cards and 40 packages? Juan is studying for his final exams in Chemistry and Algebra. He knows he only has 24 hours to study, and it will take him at least three times as long to study for Algebra than Chemistry. 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Can he spend 4 hours on Chemistry and 20 hours on Algebra? 4. ⓓ Can he spend 6 hours on Chemistry and 18 hours on Algebra? Jocelyn is pregnant and needs to eat at least 500 more calories a day than usual. When buying groceries one day with a budget of$15 for the extra food, she buys bananas that have 90 calories each and chocolate granola bars that have 150 calories each. The bananas cost .35 each and the granola bars cost $2.50 each. 1. ⓐ Write a system of inequalities to model this situation. 2. ⓑ Graph the system. 3. ⓒ Could she buy 5 bananas and 6 granola bars? 4. ⓓ Could she buy 3 bananas and 4 granola bars? Mark is attempting to build muscle mass and so he needs to eat at least an additional 80 grams of protein a day. A bottle of protein water costs$3.20 and a protein bar costs $1.75. The protein water supplies 27 grams of protein and the bar supplies 16 gram. If he has$ 10 dollars to spend

1. ⓐ Write a system of inequalities to model this situation.
2. ⓑ Graph the system.
3. ⓒ Could he buy 3 bottles of protein water and 1 protein bar?
4. ⓓ Could he buy no bottles of protein water and 5 protein bars?

Jocelyn desires to increase both her protein consumption and caloric intake. She desires to have at least 35 more grams of protein each day and no more than an additional 200 calories daily. An ounce of cheddar cheese has 7 grams of protein and 110 calories. An ounce of parmesan cheese has 11 grams of protein and 22 calories.

1. ⓐ Write a system of inequalities to model this situation.
2. ⓑ Graph the system.
3. ⓒ Could she eat 1 ounce of cheddar cheese and 3 ounces of parmesan cheese?
4. ⓓ Could she eat 2 ounces of cheddar cheese and 1 ounce of parmesan cheese?

Mark is increasing his exercise routine by running and walking at least 4 miles each day. His goal is to burn a minimum of 1,500 calories from this exercise. Walking burns 270 calories/mile and running burns 650 calories.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he meet his goal by walking 3 miles and running 1 mile?
ⓓ Could he meet his goal by walking 2 miles and running 2 mile?

#### Everyday Math

Tickets for an American Baseball League game for 3 adults and 3 children cost less than $75, while tickets for 2 adults and 4 children cost less than$62.

1. ⓐ Write a system of inequalities to model this problem.
2. ⓑ Graph the system.
3. ⓒ Could the tickets cost $20 for adults and$8 for children?
4. ⓓ Could the tickets cost $15 for adults and$5 for children?

Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and$4 per adult. Evening tickets cost $6 per child and$8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than$100 on the evening tickets.

1. ⓐ Write a system of inequalities to model this situation.
2. ⓑ Graph the system.
3. ⓒ Could they take 9 children and 4 adults to both shows?
4. ⓓ Could they take 8 children and 5 adults to both shows?

#### Writing Exercises

Graph the system < x + 2 y ≤ 6 y ≥ − 1 2 x − 4 < x + 2 y ≤ 6 y ≥ − 1 2 x − 4 . What does the solution mean?

#### Self Check

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

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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## Graphing Linear Inequalities Worksheets

This ensemble of printable graphing linear inequalities worksheets drafted for high school students assists in comprehending the use of inequalities when there is a range of possible answers. Identify the shaded region, and the type of boundary line as dashed or solid, complete the inequality, check if the ordered pair is a solution, identify the inequalities, shade the region, graph the inequalities and much more. The inequality is expressed in slope-intercept form in level 1 and has to be converted to slope-intercept form in Level 2. Our free graphing linear inequalities worksheets are a good place to begin practice. The boundary lines in this set of graphing two-variable linear inequalities worksheets are in the slope-intercept form. Observe the inequality and complete the table in Part A. Analyze the properties of the line and write the inequality in Part B. Rearrange the inequality in the slope-intercept form. High school students are expected to write if the line is dashed or solid and if the shaded region is above or below in Part A and complete the inequality in Part B in these pdfs. Determine if an ordered pair is a solution of the graph. Figure out if the pair lies in the shaded region. If it does, then it is a solution and not otherwise. Try a variant by checking if a set of coordinates are solutions. The boundary line of each graph is specified in these printable worksheets. Check if the line is solid or dashed, and the position of the bounded region is above or below and write the inequality in two variables based on the stated properties. Augment skills with this batch of linear inequalities worksheets, where the equation of the boundary line is not indicated. Direct students to determine the boundary line equation and then identify the inequality. High school learners are expected to answer a set of questions based on the graph. Shade the half-plane that contains or doesn't contain the ordered pair as specified in the question and identify the inequality as well. Map each inequality on a coordinate plane. Substitute a list of random coordinates in the inequality given and graph the line as dotted or solid on the x-y plane and shade the region of possible solutions in these graphing linear inequalities worksheet pdfs. The line is not expressed in the slope-intercept form. Rearrange the equation, so that it solves for y and graph the inequality in two variables using a coordinate plane and find the full range of possible solutions.

## Introduction

We have seen that the graphs of linear equations are straight lines. Graphs of other types of equations, called polynomial equations, are curves, like the outline of this suspension bridge. Architects use polynomials to design the shape of a bridge like this and to draw the blueprints for it. Engineers use polynomials to calculate the stress on the bridge’s supports to ensure they are strong enough for the intended load. In this chapter, you will explore operations with and properties of polynomials.

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• Book title: Elementary Algebra 2e
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• Location: Houston, Texas
• Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/elementary-algebra-2e/pages/6-introduction

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## Lesson 22

In a previous lesson, students learned to graphically represent the set of solutions to a linear inequality in two variables. They made a connection between the solutions to a linear inequality and the solutions to a related linear equation.

In this lesson, students deepen their understanding of the solutions to linear inequalities by studying them in context. They write inequalities in two variables to represent constraints, and interpret the points on a boundary line and on either side of it in terms of the situation.

The work here illustrates that the solution region represents the set of values that satisfy the constraint in a situation (MP2). Interpreting the solutions contextually also engages students in an aspect of mathematical modeling (MP4). It enables students to see that, while some values might make an inequality true, they might not be feasible or appropriate in the situation. The activity Rethinking Landscaping is an opportunity to make a generalization based on repeated reasoning (MP8), since students first find numbers that meet a constraint, and then use variables in place of those numbers to write an equation and an inequality.

Because reasoning about the solution region of an inequality is important here, graphing technology should not be used. Students will have opportunities to use graphing technology to solve inequalities in two variables in upcoming lessons.

## Chapter 2

Parallel lines: equations are written in slope-intercept form.

### 2.4 Complex Numbers

( x − 6 ) ( x + 1 ) = 0 x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 x = 6 , x = − 1

### 2.1 Section Exercises

Answers may vary. Yes. It is possible for a point to be on the x-axis or on the y-axis and therefore is considered to NOT be in one of the quadrants.

The y-intercept is the point where the graph crosses the y-axis.

### 2.2 Section Exercises

It means they have the same slope.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

m 1 = − 1 3 , m 2 = 3 Perpendicular . m 1 = − 1 3 , m 2 = 3 Perpendicular .

The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 . The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 .

Yes they are perpendicular.

### 2.3 Section Exercises

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

She traveled for 2 h at 20 mi/h, or 40 miles.

$5,000 at 8% and$15,000 at 12%

W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14

f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21

length = 360 ft width = 160 ft

### 2.4 Section Exercises

Add the real parts together and the imaginary parts together.

### 2.5 Section Exercises

It is a second-degree equation (the highest variable exponent is 2).

We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.

a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a

The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.

### 2.6 Section Exercises

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

### 2.7 Section Exercises

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ ) x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ )

It is never less than zero. No solution.

Where the blue line is above the orange line point of intersection is x = − 3. x = − 3.

Where the blue line is above the orange line always. All real numbers.

Where the blue is below the orange always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .

Where the blue is below the orange ( 1 , 7 ) . ( 1 , 7 ) .

80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400

### Review Exercises

Where the blue is below the orange line point of intersection is x = 3.5. x = 3.5.

### Practice Test

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• Book title: College Algebra
• Publication date: Feb 13, 2015
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