problem solving involving system of linear inequalities in two variables

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4.5: Solving Systems of Linear Inequalities (Two Variables)

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Learning Objectives

• Check solutions to systems of linear inequalities with two variables.
• Solve systems of linear inequalities.

Solutions to Systems of Linear Inequalities

A system of linear inequalities consists of a set of two or more linear inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets defines the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately, we have

This is “Solving Systems of Linear Inequalities (Two Variables)”, section 4.5 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here .

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4.5 Solving Systems of Linear Inequalities (Two Variables)

Learning objectives.

• Check solutions to systems of linear inequalities with two variables.
• Solve systems of linear inequalities.

Solutions to Systems of Linear Inequalities

A system of linear inequalities A set of two or more linear inequalities that define the conditions to be considered simultaneously. consists of a set of two or more linear inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets defines the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately, we have

When graphed on the same set of axes, the intersection can be determined.

The intersection is shaded darker and the final graph of the solution set is presented as follows:

The graph suggests that (3, 2) is a solution because it is in the intersection. To verify this, show that it solves both of the original inequalities:

Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed boundary are not. Consider the point (−1, 0) on the solid boundary defined by y = 2 x + 2 and verify that it solves the original system:

Notice that this point satisfies both inequalities and thus is included in the solution set. Now consider the point (2, 0) on the dashed boundary defined by y = x − 2 and verify that it does not solve the original system:

This point does not satisfy both inequalities and thus is not included in the solution set.

Solving Systems of Linear Inequalities

Solutions to a system of linear inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems, graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, defines the region of common ordered pair solutions.

Example 1: Graph the solution set: { − 2 x + y > − 4 3 x − 6 y ≥ 6 .

Solution: To facilitate the graphing process, we first solve for y .

For the first inequality, we use a dashed boundary defined by y = 2 x − 4 and shade all points above the line. For the second inequality, we use a solid boundary defined by y = 1 2 x − 1 and shade all points below. The intersection is darkened.

Now we present the solution with only the intersection shaded.

Example 2: Graph the solution set: { − 2 x + 3 y > 6 4 x − 6 y > 12 .

Solution: Begin by solving both inequalities for y .

Use a dashed line for each boundary. For the first inequality, shade all points above the boundary. For the second inequality, shade all points below the boundary.

As you can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

Answer: No solution, ∅

Example 3: Graph the solution set: { y ≥ − 4 y < x + 3 y ≤ − 3 x + 3 .

Solution: The intersection of all the shaded regions forms the triangular region as pictured darkened below:

After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured.

The graphic suggests that (−1, 1) is a common point. As a check, substitute that point into the inequalities and verify that it solves all three conditions.

Key Takeaway

• To solve systems of linear inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.

Topic Exercises

Part A: Solving Systems of Linear Inequalities

Determine whether the given point is a solution to the given system of linear equations.

1. (3, 2); { y ≤ x + 3 y ≥ − x + 3

2. (−3, −2); { y < − 3 x + 4 y ≥ 2 x − 1

3. (5, 0); { y > − x + 5 y ≤ 3 4 x − 2

4. (0, 1); { y < 2 3 x + 1 y ≥ 5 2 x − 2

5. ( − 1 ,   8 3 ) ; { − 4 x + 3 y ≥ − 12 2 x + 3 y < 6

6. (−1, −2); { − x + y < 0 x + y < 0 x + y < − 2

Part B: Solving Systems of Linear Inequalities

Graph the solution set.

7. { y ≤ x + 3 y ≥ − x + 3

8. { y < − 3 x + 4 y ≥ 2 x − 1

9. { y > x y < − 1

10. { y < 2 3 x + 1 y ≥ 5 2 x − 2

11. { y > − x + 5 y ≤ 3 4 x − 2

12. { y > 3 5 x + 3 y < 3 5 x − 3

13. { x + 4 y < 12 − 3 x + 12 y ≥ − 12

14. { − x + y ≤ 6 2 x + y ≥ 1

15. { − 2 x + 3 y > 3 4 x − 3 y < 15

16. { − 4 x + 3 y ≥ − 12 2 x + 3 y < 6

17. { 5 x + y ≤ 4 − 4 x + 3 y < − 6

18. { 3 x + 5 y < 15 − x + 2 y ≤ 0

19. { x ≥ 0 5 x + y > 5

20. { x ≥ − 2 y ≥ 1

21. { x − 3 < 0 y + 2 ≥ 0

22. { 5 y ≥ 2 x + 5 − 2 x < − 5 y − 5

23. { x − y ≥ 0 − x + y < 1

24. { − x + y ≥ 0 y − x < 1

25. { x > − 2 x ≤ 2

26. { y > − 1 y < 2

27. { − x + 2 y > 8 3 x − 6 y ≥ 18

28. { − 3 x + 4 y ≤ 4 6 x − 8 y > − 8

29. { 2 x + y < 3 − x ≤ 1 2 y

30. { 2 x + 6 y ≤ 6 − 1 3 x − y ≤ 3

31. { y < 3 y > x x > − 4

32. { y < 1 y ≥ x − 1 y < − 3 x + 3

33. { − 4 x + 3 y > − 12 y ≥ 2 2 x + 3 y > 6

34. { − x + y < 0 x + y ≤ 0 x + y > − 2

35. { x + y < 2 x < 3 − x + y ≤ 2

36. { y + 4 ≥ 0 1 2 x + 1 3 y ≤ 1 − 1 2 x + 1 3 y ≤ 1

37. Construct a system of linear inequalities that describes all points in the first quadrant.

38. Construct a system of linear inequalities that describes all points in the second quadrant.

39. Construct a system of linear inequalities that describes all points in the third quadrant.

40. Construct a system of linear inequalities that describes all points in the fourth quadrant.

27: No solution, ∅

37: { x > 0 y > 0

39: { x < 0 y < 0

3.4 Graph Linear Inequalities in Two Variables

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 in two variables
• Solve applications using linear inequalities in two variables

Be Prepared 3.10

Before you get started, take this readiness quiz.

Graph x > 2 x > 2 on a number line. If you missed this problem, review Example 2.48 .

Be Prepared 3.11

Solve: 4 x + 3 > 23 4 x + 3 > 23 . If you missed this problem, review Example 2.52 .

Be Prepared 3.12

Translate: 8 > x > 3 8 > x > 3 . If you missed this problem, review Example 2.56 .

Verify Solutions to an Inequality in Two Variables

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

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

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.

Recall that an inequality with one variable had many solutions. For example, the solution to the inequality x > 3 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 3.10 .

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

Solution to a Linear Inequality

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

Example 3.34

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

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

Try It 3.67

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

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

Try It 3.68

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

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

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

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

Let’s think about the number line in shown previously again. The point x = 3 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 3.11 .

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

Boundary Line

The line with equation A x + B y = C A x + B y = C is the boundary line that separates the region where A x + B y > C A x + B y > C from the region where A x + B y < C . A x + B y < C .

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

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

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 ?

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

Since, y > 2 x − 1 y > 2 x − 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 > 2 x − 1 . y > 2 x − 1 .

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 .

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 < 2 x − 1 . y < 2 x − 1 .

Try It 3.69

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

Try It 3.70

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.

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

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

(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 .

Try It 3.71

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

Try It 3.72

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 .

Try It 3.73

Graph the linear inequality y ≥ 5 2 x − 4 . y ≥ 5 2 x − 4 .

Try It 3.74

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.

Graph a linear inequality in two variables.

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

Example 3.38

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

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

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

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

Try It 3.75

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

Try It 3.76

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 .

First, we graph the boundary line y = −4 x . y = −4 x . It is in slope–intercept form, with m = −4 m = −4 and b = 0 . 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 ) ( 1 , 0 ) is not on the boundary line.

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

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

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

Try It 3.77

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

Try It 3.78

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.

Recall that:

Example 3.40

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

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

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

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

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

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

Try It 3.79

Graph the linear inequality: y < 5 . y < 5 .

Try It 3.80

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?

ⓐ Let x 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. Write an inequality that would model this situation.

ⓑ Graph the inequality.

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

ⓐ 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.

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

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

Try It 3.81

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.

Try It 3.82

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

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

• Graphing Linear Inequalities in Two Variables

Section 3.4 Exercises

Practice makes perfect.

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 :

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

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

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

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

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

ⓐ ( −3 , 0 ) ( −3 , 0 ) ⓑ ( 1 , 6 ) ( 1 , 6 ) ⓒ ( −6 , −2 ) ( −6 , −2 ) ⓓ ( 0 , 1 ) ( 0 , 1 ) ⓔ ( 5 , −4 ) ( 5 , −4 )

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

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

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

ⓐ ( 1 , 1 ) ( 1 , 1 ) ⓑ ( 4 , −3 ) ( 4 , −3 ) ⓒ ( 0 , 0 ) ( 0 , 0 ) ⓓ ( −8 , 12 ) ( −8 , 12 ) ⓔ ( 3 , 0 ) ( 3 , 0 )

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 2 x − y = 4 . 2 x − y = 4 .

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 .

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.

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

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.

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

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

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

Explain why, in some graphs of linear inequalities, the boundary line is solid but in other graphs it is dashed.

ⓐ 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?

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Course: Algebra 1   >   Unit 7

• Writing two-variable inequalities word problem
• Solving two-variable inequalities word problem
• Interpreting two-variable inequalities word problem

Two-variable inequalities word problems

• Modeling with systems of inequalities
• Writing systems of inequalities word problem
• Solving systems of inequalities word problem
• Graphs of systems of inequalities word problem
• Systems of inequalities word problems
• Graphs of two-variable inequalities word problem
• Inequalities (systems & graphs): FAQ
• Creativity break: What can we do to expand our creative skills?

Grade 8 Mathematics Module: “Solving Systems of Linear Inequalities in Two Variables”

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this module, you will be acquainted with the application of systems of linear inequalities in two variables in solving problems related to real-life. The scope of this module enables you to use it in many different learning situations. The lesson is arranged following the standard sequence of the course. But the order in which you read them can be changed corresponding with the textbook you are now using.

This module is divided into two lessons:

• Lesson 1- Graphing Systems of Linear Inequalities in Two Variables
• Lesson 2– Solving Problems Involving Systems of Linear Inequalities in Two Variables

After going through this module, you are expected to:

1. define systems of linear inequalities in two variables;

2. graph systems of linear inequalities in two variables; and

3. solve problems involving systems of linear inequalities in two variables.

Grade 8 Mathematics Quarter 2 Self-Learning Module: “Solving Systems of Linear Inequalities in Two Variables”

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Systems of Linear Inequalities, Word Problems - Examples - Expii

Systems of linear inequalities, word problems - examples, explanations (3), word problems with systems of linear inequalities.

Solving word problems with systems of linear inequalities involves 2 steps:

Translate the word problem into mathematical expressions. We're finding a system of linear inequalities .

Solve! One strategy you can use to find solutions is to graph each inequality . Then, use the overlapping areas of the graphs to identify the solutions to the word problem.

Let's look at an example:

Image source: by Hannah Bonville

STEP 1: Translate the word problem into inequalities.

We're going to use x for the number of pizzas and y for the number of pounds of cookies.

Which inequality represents the number of pizzas Pilar needs?

Related Lessons

Systems of Linear Inequalities: Word Problems

When you are presented with a word problem that deals with inequalities, it may seem daunting at first to solve the problem. The good news is that there are two simple steps to follow in order to solve systems of linear inequalities!

Image source: by Anusha Rahman

Let's walk through an example problem together.

Jenn and John want to go on a road trip together. Jenn drives an average of 50mph, and John drives an average of 45mph. They want to drive less than 20 hours, but at least 500 miles. Write a system of linear inequalities to figure out how many miles John and Jenn must drive to make that happen.

Step 1: Translate the word problem into inequalities.

Let's start with establishing our variables. x=Jenn y=John

We know that Jenn and John want to drive at least 500 miles. At least generally means greater than or equal to . So if we incorporate the speed at which the two of them drive, we can say that:

50x+45y≥500

We also know that, combined, John and Jenn want to drive less than 20 hours. So, we can say that: x+y=20

Therefore, our system of linear equations is:

50x+45y≥500 x+y=20

Step 2: Solve the inequalities.

Let's solve these inequalities by graphing. Here is a graph of the two inequalities.

Made using Desmos

50x+45y≥500 is represented by the red shading. x+y=20 is represented by the blue shading. The overlap, or the solution, is represented by the purple shading.

(Video) Solving Inequalities: Word Problems

by Anusha Rahman

This video by Anusha Rahman walks you through an example word problem of how to solve a system of inequalities.

Example: Jackson is having a party for his friends on Saturday. Jackson expects at least 7 people to show up on Saturday. He is trying to figure out the total costs of all of the pizzas. He calls his local pizza place, Meredith’s pizza, who says that the cost of his pizzas will be less than 5 more than 2 times the number of people he needs pizza for.

Step 1: Establish variables x=the number of people y=the cost of all pizzas

Step 2: Set up the inequalities x≥7 y<5+2x

Step 3: Graph the inequalities to find the solution

Graph of: x≥7

Graph of y=5+2x

This is the graph of the line, y=5+2x. In order to graph the inequality, y<5+2x, we have to figure out what area on the graph to shade. To do that, we test out some points and see whether the inequality remains true!

Let's test the point (10,2). 2<2(10)+5 2<20+5 2<25 This statement is true ! So we will probably shade in this area, the bottom area of the graph. But, let's just check one more point to make sure.

Let's test the point (−10,2). 2<2(−10)+5 2<−20+5 2<−15 This statement is false .

So, we can see in the above graph that the area under the dotted line is shaded. This is the graph of y<5+2x.

Now, let's put both inequalities on the same graph to see the area of intersection!

The area shaded in purple is the solution to the system of inequalities! Because Jackson probably won't have negative friends or pay negative dollars, we can assume 0">x>0 and 0">y>0.

Jackson won’t know for certain what the price of his pizzas will be. Why? Because Jackson doesn’t know how many of his friends will be coming to his party. Especially when solving real-world problems, a system of inequalities is really helpful when we’re looking at something in the future, and we don’t have exact numbers, but we have a general idea, like Jackson does.

2.7 Solving Inequalities with Two Variables

Learning objectives.

• Identify and check solutions to inequalities with two variables.
• Graph solution sets of linear inequalities with two variables.

Solutions to Inequalities with Two Variables

We know that a linear equation with two variables has infinitely many ordered pair solutions that form a line when graphed. A linear inequality with two variables An inequality relating linear expressions with two variables. The solution set is a region defining half of the plane. , on the other hand, has a solution set consisting of a region that defines half of the plane.

For the inequality, the line defines the boundary of the region that is shaded. This indicates that any ordered pair in the shaded region, including the boundary line, will satisfy the inequality. To see that this is the case, choose a few test points A point not on the boundary of the linear inequality used as a means to determine in which half-plane the solutions lie. and substitute them into the inequality.

Also, we can see that ordered pairs outside the shaded region do not solve the linear inequality.

The graph of the solution set to a linear inequality is always a region. However, the boundary may not always be included in that set. In the previous example, the line was part of the solution set because of the “or equal to” part of the inclusive inequality ≤ . If given a strict inequality < , we would then use a dashed line to indicate that those points are not included in the solution set.

Consider the point (0, 3) on the boundary; this ordered pair satisfies the linear equation. It is the “or equal to” part of the inclusive inequality that makes the ordered pair part of the solution set.

So far we have seen examples of inequalities that were “less than.” Now consider the following graphs with the same boundary:

Given the graphs above, what might we expect if we use the origin (0, 0) as a test point?

Determine whether or not ( 2 , 1 2 ) is a solution to 5 x − 2 y < 10 .

Substitute the x - and y -values into the equation and see if a true statement is obtained.

5 x − 2 y < 10 5 ( 2 ) − 2 ( 1 2 ) < 10 10 − 1 < 10 9 < 10           ✓

Answer: ( 2 , 1 2 ) is a solution.

These ideas and techniques extend to nonlinear inequalities with two variables. For example, all of the solutions to y > x 2 are shaded in the graph below.

The boundary of the region is a parabola, shown as a dashed curve on the graph, and is not part of the solution set. However, from the graph we expect the ordered pair (−1,4) to be a solution. Furthermore, we expect that ordered pairs that are not in the shaded region, such as (−3, 2), will not satisfy the inequality.

Following are graphs of solutions sets of inequalities with inclusive parabolic boundaries.

You are encouraged to test points in and out of each solution set that is graphed above.

Try this! Is ( − 3 , − 2 ) a solution to 2 x − 3 y < 0 ?

Graphing Solutions to Inequalities with Two Variables

Solutions to linear inequalities are a shaded half-plane, bounded by a solid line or a dashed line. This boundary is either included in the solution or not, depending on the given inequality. If we are given a strict inequality, we use a dashed line to indicate that the boundary is not included. If we are given an inclusive inequality, we use a solid line to indicate that it is included. The steps for graphing the solution set for an inequality with two variables are shown in the following example.

Graph the solution set y > − 3 x + 1 .

Step 1: Graph the boundary. Because of the strict inequality, we will graph the boundary y = − 3 x + 1 using a dashed line. We can see that the slope is m = − 3 = − 3 1 = r i s e r u n and the y -intercept is (0, 1).

Step 2: Test a point that is not on the boundary. A common test point is the origin, (0, 0). The test point helps us determine which half of the plane to shade.

• Step 3: Shade the region containing the solutions. Since the test point (0, 0) was not a solution, it does not lie in the region containing all the ordered pair solutions. Therefore, shade the half of the plane that does not contain this test point. In this case, shade above the boundary line.

Consider the problem of shading above or below the boundary line when the inequality is in slope-intercept form. If y > m x + b , then shade above the line. If y < m x + b , then shade below the line. Shade with caution; sometimes the boundary is given in standard form, in which case these rules do not apply.

Graph the solution set 2 x − 5 y ≥ − 10 .

Here the boundary is defined by the line 2 x − 5 y = − 10 . Since the inequality is inclusive, we graph the boundary using a solid line. In this case, graph the boundary line using intercepts.

Next, test a point; this helps decide which region to shade.

Since the test point is in the solution set, shade the half of the plane that contains it.

In this example, notice that the solution set consists of all the ordered pairs below the boundary line. This may seem counterintuitive because the original inequality involved “greater than” ≥ . This illustrates that it is a best practice to actually test a point. Solve for y and you see that the shading is correct.

2 x − 5 y ≥ − 10 2 x − 5 y − 2 x ≥ − 10 − 2 x − 5 y ≥ − 2 x − 10 − 5 y − 5 ≤ − 2 x − 10 − 5                   R e v e r s e   t h e   i n e q u a l i t y . y ≤ 2 5 x + 2

In slope-intercept form, you can see that the region below the boundary line should be shaded. An alternate approach is to first express the boundary in slope-intercept form, graph it, and then shade the appropriate region.

Graph the solution set y < 2 .

First, graph the boundary line y = 2 with a dashed line because of the strict inequality. Next, test a point.

In this case, shade the region that contains the test point.

Try this! Graph the solution set 2 x − 3 y < 0 .

The steps are the same for nonlinear inequalities with two variables. Graph the boundary first and then test a point to determine which region contains the solutions.

Graph the solution set y < ( x + 2 ) 2 − 1 .

The boundary is a basic parabola shifted 2 units to the left and 1 unit down. Begin by drawing a dashed parabolic boundary because of the strict inequality.

Next, test a point.

In this case, shade the region that contains the test point ( 0 , 0 ) .

Graph the solution set y ≥ x 2 + 3 .

The boundary is a basic parabola shifted 3 units up. It is graphed using a solid curve because of the inclusive inequality.

In this case, shade the region that does not contain the test point ( 0 , 0 ) .

Try this! Graph the solution set y < | x − 1 | − 3 .

Key Takeaways

• Linear inequalities with two variables have infinitely many ordered pair solutions, which can be graphed by shading in the appropriate half of a rectangular coordinate plane.
• To graph the solution set of an inequality with two variables, first graph the boundary with a dashed or solid line depending on the inequality. If given a strict inequality, use a dashed line for the boundary. If given an inclusive inequality, use a solid line. Next, choose a test point not on the boundary. If the test point solves the inequality, then shade the region that contains it; otherwise, shade the opposite side.
• Check your answer by testing points in and out of the shading region to verify that they solve the inequality or not.

Topic Exercises

Part a: solutions to inequalities with two variables.

Is the ordered pair a solution to the given inequality?

5 x − y > − 2 ;    ( − 3 , − 4 )

4 x − y < − 8 ;    ( − 3 , − 10 )

• 6 x − 15 y ≥ − 1 ;    ( 1 2 , − 1 3 )
• x − 2 y ≥ 2 ;    ( 2 3 , − 5 6 )
• 3 4 x − 2 3 y < 3 2 ;    ( 1 , − 1 )
• 2 5 x + 4 3 y > 1 2 ;    ( − 2 , 1 )

y ≤ x 2 − 1 ;    ( − 1 , 1 )

y ≥ x 2 + 3 ;    ( − 2 , 0 )

y ≥ ( x − 5 ) 2 + 1 ;    ( 3 , 4 )

y ≤ 2 ( x + 1 ) 2 − 3 ;    ( − 1 , − 2 )

y > 3 − | x | ;    ( − 4 , − 3 )

y < | x | − 8 ;    ( 5 , − 7 )

y > | 2 x − 1 | − 3 ;    ( − 1 , 3 )

y < | 3 x − 2 | + 2 ;    ( − 2 , 10 )

Part B: Graphing Solutions to Inequalities with Two Variables.

Graph the solution set.

y < 2 x − 1

y > − 4 x + 1

y ≥ − 2 3 x + 3

y ≤ 4 3 x − 3

2 x + 3 y ≤ 18

5 x + 2 y ≤ 8

6 x − 5 y > 30

8 x − 6 y < 24

3 x − 4 y < 0

x − 3 y > 0

• 1 6 x + 1 10 y ≤ 1 2
• 3 8 x + 1 2 y ≥ 3 4
• 1 12 x − 1 6 y < 2 3
• 1 3 x − 1 9 y > 4 3

5 x ≤ − 4 y − 12

− 4 x ≤ 12 − 3 y

4 y + 2 < 3 x

8 x < 9 − 6 y

5 ≥ 3 x − 15 y

• 2 x ≥ 6 − 9 y

Write an inequality that describes all points in the upper half-plane above the x -axis.

Write an inequality that describes all points in the lower half-plane below the x -axis.

Write an inequality that describes all points in the half-plane left of the y -axis.

Write an inequality that describes all points in the half-plane right of the y -axis.

Write an inequality that describes all ordered pairs whose y -coordinate is at least k units.

Write an inequality that describes all ordered pairs whose x -coordinate is at most k units.

y ≤ x 2 + 3

y > x 2 − 2

y > ( x + 1 ) 2

y > ( x − 2 ) 2

y ≤ ( x − 1 ) 2 + 2

y ≤ ( x + 3 ) 2 − 1

y < − x 2 + 1

y > − ( x − 2 ) 2 + 1

y ≥ | x | − 2

y < | x | + 1

y < | x − 3 |

y ≤ | x + 2 |

y > − | x + 1 |

y ≤ − | x − 2 |

y ≥ | x + 3 | − 2

y ≥ | x − 2 | − 1

y < − | x + 4 | + 2

y > − | x − 4 | − 1

y > x 3 − 1

y ≤ x 3 + 2

y > x − 1

A rectangular pen is to be constructed with at most 200 feet of fencing. Write a linear inequality in terms of the length l and the width w . Sketch the graph of all possible solutions to this problem.

A company sells one product for $8 and another for $12. How many of each product must be sold so that revenues are at least $2,400? Let x represent the number of products sold at $8 and let y represent the number of products sold at $12. Write a linear inequality in terms of x and y and sketch the graph of all possible solutions.

l + w ≤ 100 ;

• Math Article

Linear Inequalities In Two Variables

Linear inequalities in two variables represent the inequalities between two algebraic expressions where two distinct variables are included. In linear inequalities in two variables, we use greater than (>), less than (<), greater than or equal (≥) and less than or equal (≤) symbols, instead of using equal to a symbol (=).

What is Linear Inequalities?

Any two real numbers or two algebraic expressions associated with the symbol ‘<’, ‘>’, ‘≤’ or ‘≥’ form a linear inequality. For example, 9<11, 18>17 are examples of numerical inequalities and x+7>y, y<10-x, x ≥ y > 11 are examples of algebraic inequalities.

The symbols ‘<‘ and ‘>’ represent the strict inequalities and the symbols ‘≤’ and ‘≥’ represent slack inequalities. To represent linear inequalities in one variable in a number line is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss the graph of a linear inequality in two variables.

What are Linear Inequalities in Two Variables?

Linear inequalities in two variables represent the unequal relation between two algebraic expressions that includes two distinct variables. Hence, the symbols used between the expression in two variables will be ‘<’, ‘>’, ‘≤’ or ‘≥’, but we cannot use equal to ‘=’ symbol here.

The examples of linear inequalities in two variables are:

• 3x < 2y + 5
• 8y – 9x > 10

Note:   4x 2 + 2x + 5 < 0 is not an example of linear inequality in one variable, because the exponent of x is 2 in the first term. It is a quadratic inequality.

How to Solve Linear Inequalities in Two Variables?

The solution for linear inequalities in two variables is an ordered pair that is true for the inequality statement. Let us say if Ax + By > C is a linear inequality where x and y are two variables, then an ordered pair (x, y) satisfying the statement will be the required solution.

The method of solving linear inequalities in two variables is the same as solving linear equations .

For example, if 2x + 3y > 4 is a linear inequality, then we can check the solution, by putting the values of x and y here.

Let x = 1 and y = 2

Taking LHS, we have;

2 (1) + 3 (2) = 2 + 6 = 8

Since, 8 > 4, therefore, the ordered pair (1, 2) satisfy the inequality 2x + 3y > 4. Hence, (1, 2) is the solution.

We can also put different values of x and y to find different solutions here.

Graphical Solution of Linear Inequalities in Two Variables

The statements involving symbols like ‘<’(less than), ‘>’ (greater than), ‘≤’’(less than or equal to), ‘≥’ (greater than or equal to) and two distinct variables are called linear inequalities in two variables. Let us see here, how to find the solution of such expressions, graphically.

Below are the two examples of linear inequalities shown in the figure. The graph of y > x – 2 and y ≤ 2x + 2 are:

Real-life Examples:

Following example validates the difference between equation and inequality:

Statement 1: The distance between your house and school is exactly 4.5 kilometres,

The mathematical expression of the above statement is,

x = 4.5 km, where ‘x’ is the distance between house and the school.

Statement 2: The distance between your house and the school is at least 4.5 kilometers.

Here, the distance can be 4.5 km or more than that. Therefore the mathematical expression for the above statement is,

x ≥ 4.5 km, where ‘x’ is a variable that is equal to the distance between the house and the school.

Important Facts

• We can add, subtract, multiply and divide by the same number to solve the inequalities
• While multiplying and dividing by negative number, the inequality sign get reversed
• In graphical solution, the ordered pair outside the shaded portion does not solve the inequality
• Numerical inequalities: If only numbers are involved in the expression, then it is a numerical inequality. Example:  10 > 8, 5 < 7
• Literal inequalities: x < 2, y > 5, z < 10 are the examples for literal inequalities.
• Double inequalities: 5 < 7 < 9 read as 7 less than 9 and greater than 5 is an example of double inequality.
• Strict inequality: Mathematical expressions involve only ‘<‘ or ‘>’  are called strict inequalities. Example: 2x + 3 < 6, 2x + 3y > 6
• Slack inequality: Mathematical expressions involve only ‘≤′ or ‘≥’ are called slack inequalities. Example: 2x + 3 ≤ 6, 2x + 3y ≥ 6

Related Articles

• Linear Inequalities
• Solving Linear Inequalities
• Represent Linear Inequalities In One Variable On Number Line
• Linear Equations In Two Variables
• Cross Multiplication- Pair Of Linear Equations In Two Variables

Solved Examples on Linear Inequalities in Two Variables

1.) Classify the following expressions into:

• Linear inequality in one variable.
• Linear inequality in two variables.
• Slack inequality.

5x < 6, 8x + 3y ≤ 5, 2x – 5 < 9 , 2x ≤ 9 , 2x + 3y < 10.

2.) Solve y < 2 graphically.

Solution: Graph of y = 2. So we can show it graphically as given below:

Let us select a point, (0, 0) in the lower half-plane I and putting y = 0 in the given inequality, we see that: 1 × 0 < 2 or 0 < 2 which is true. Thus, the solution region is the shaded region below the line y = 2.

Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.

Linear Inequalities in Two Variables Word Problem

In an experiment, a solution of hydrochloric acid is to be kept between 25° and 30° Celsius. What is the range of temperature in degree Fahrenheit if the conversion formula is given by C = 5/9 (F – 32), where C and F represent the temperature in degree Celsius and degree Fahrenheit, respectively.

Solution: As per the question it is given:

25<C<30

Now if we put C = 5/9 (F – 32), we get;

25 < 5/9 (F – 32) < 30

9/5 x 25 < F – 32 < 30 x 9/5

45 < F -32< 54

77 < F < 86

Thus, the required range of temperature is between 77° F and 86° F.

Frequently Asked Questions – FAQs

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Module 3: Quadratic Equations and Functions

Linear inequalities and systems of linear inequalities in two variables, learning objectives.

• Identify and follow steps for graphing a linear inequality in two variables
• Identify whether an ordered pair is in the solution set of a linear inequality
• Graph a system of linear inequalities and define the solutions region
• Verify whether a point is a solution to a system of inequalities
• Identify when a system of inequalities has no solution
• Solve systems of linear inequalities by graphing the solution region
• Graph solutions to a system that contains a compound inequality
• Write and graph a system that models the quantity that must be sold to achieve a given amount of sales
• Write a system of inequalities that represents the profit region for a business
• Interpret the solutions to a system of cost/ revenue inequalities

Inequalities in two variables

Did you know that you use linear inequalities when you shop online? When you use the option to view items within a specific price range, you are asking the search engine to use a linear inequality based on price. Essentially, you are saying “show me all the items for sale between $50 and $100,” which can be written as [latex]{50}\le {x} \le {100}[/latex], where x is price. In this section, you will apply what you know about graphing linear equations to graphing linear inequalities.

So how do you get from the algebraic form of an inequality, like [latex]y>3x+1[/latex], to a graph of that inequality? Plotting inequalities is fairly straightforward if you follow a couple steps.

Graphing Inequalities

To graph an inequality:

• Graph the related boundary line. Replace the <, >, ≤ or ≥ sign in the inequality with = to find the equation of the boundary line.
• Identify at least one ordered pair on either side of the boundary line and substitute those [latex](x,y)[/latex] values into the inequality. Shade the region that contains the ordered pairs that make the inequality a true statement.  
• If points on the boundary line are solutions, then use a solid line for drawing the boundary line. This will happen for ≤ or ≥ inequalities.
• If points on the boundary line aren’t solutions, then use a dotted line for the boundary line. This will happen for < or > inequalities.

Let’s graph the inequality [latex]x+4y\leq4[/latex].

To graph the boundary line, find at least two values that lie on the line [latex]x+4y=4[/latex]. You can use the x – and y -intercepts for this equation by substituting 0 in for x first and finding the value of y ; then substitute 0 in for y and find x .

Plot the points [latex](0,1)[/latex] and [latex](4,0)[/latex], and draw a line through these two points for the boundary line. The line is solid because ≤ means “less than or equal to,” so all ordered pairs along the line are included in the solution set.

The next step is to find the region that contains the solutions. Is it above or below the boundary line? To identify the region where the inequality holds true, you can test a couple of ordered pairs, one on each side of the boundary line.

If you substitute [latex](−1,3)[/latex] into [latex]x+4y\leq4[/latex]:

[latex]\begin{array}{r}−1+4\left(3\right)\leq4\\−1+12\leq4\\11\leq4\end{array}[/latex]

This is a false statement, since 11 is not less than or equal to 4.

On the other hand, if you substitute [latex](2,0)[/latex] into [latex]x+4y\leq4[/latex]:

[latex]\begin{array}{r}2+4\left(0\right)\leq4\\2+0\leq4\\2\leq4\end{array}[/latex]

This is true! The region that includes [latex](2,0)[/latex] should be shaded, as this is the region of solutions.

And there you have it—the graph of the set of solutions for [latex]x+4y\leq4[/latex].

Graphing Linear Inequalities in Two Variables

Graph the inequality [latex]2y>4x–6[/latex].

Solve for y .

[latex] \displaystyle \begin{array}{r}2y>4x-6\\\\\frac{2y}{2}>\frac{4x}{2}-\frac{6}{2}\\\\y>2x-3\\\end{array}[/latex]

Create a table of values to find two points on the line [latex] \displaystyle y=2x-3[/latex], or graph it based on the slope-intercept method, the b value of the y -intercept is [latex]-3[/latex] and the slope is 2.

Plot the points, and graph the line. The line is dotted because the sign in the inequality is >, not ≥ and therefore points on the line are not solutions to the inequality.

[latex] \displaystyle y=2x-3[/latex]

Find an ordered pair on either side of the boundary line. Insert the x – and y -values into the inequality [latex]2y>4x–6[/latex] and see which ordered pair results in a true statement. Since [latex](−3,1)[/latex] results in a true statement, the region that includes [latex](−3,1)[/latex] should be shaded.

[latex]\begin{array}{l}2y>4x–6\\\\\text{Test }1:\left(−3,1\right)\\2\left(1\right)>4\left(−3\right)–6\\\,\,\,\,\,\,\,2>–12–6\\\,\,\,\,\,\,\,2>−18\\\text{TRUE}\\\\\text{Test }2:\left(4,1\right)\\2(1)>4\left(4\right)– 6\\\,\,\,\,\,\,2>16–6\\\,\,\,\,\,\,2>10\\\text{FALSE}\end{array}[/latex]

The graph of the inequality [latex]2y>4x–6[/latex] is:

A quick note about the problem above—notice that you can use the points [latex](0,−3)[/latex] and [latex](2,1)[/latex] to graph the boundary line, but that these points are not included in the region of solutions, since the region does not include the boundary line!

Solution Sets of Inequalities

The graph below shows the region of values that makes the inequality [latex]3x+2y\leq6[/latex] true (shaded red), the boundary line [latex]3x+2y=6[/latex], as well as a handful of ordered pairs. The boundary line is solid because points on the boundary line [latex]3x+2y=6[/latex] will make the inequality [latex]3x+2y\leq6[/latex] true.

You can substitute the x- and y- values in each of the [latex](x,y)[/latex] ordered pairs into the inequality to find solutions. Sometimes making a table of values makes sense for more complicated inequalities.

If substituting [latex](x,y)[/latex] into the inequality yields a true statement, then the ordered pair is a solution to the inequality, and the point will be plotted within the shaded region or the point will be part of a solid boundary line. A false statement means that the ordered pair is not a solution, and the point will graph outside the shaded region, or the point will be part of a dotted boundary line.

Use the graph to determine which ordered pairs plotted below are solutions of the inequality [latex]x–y<3[/latex].

Solutions will be located in the shaded region. Since this is a “less than” problem, ordered pairs on the boundary line are not included in the solution set.

These values are located in the shaded region, so are solutions. (When substituted into the inequality [latex]x–y<3[/latex], they produce true statements.)

[latex](−1,1)[/latex]

[latex](−2,−2)[/latex]

These values are not located in the shaded region, so are not solutions. (When substituted into the inequality [latex]x-y<3[/latex], they produce false statements.)

[latex](1,−2)[/latex]

[latex](3,−2)[/latex]

[latex](4,0)[/latex]

[latex](−1,1)\,\,\,(−2,−2)[/latex]

The following video show an example of determining whether an ordered pair is a solution to an inequality.

Is [latex](2,−3)[/latex] a solution of the inequality [latex]y<−3x+1[/latex]?

If [latex](2,−3)[/latex] is a solution, then it will yield a true statement when substituted into the inequality [latex]y<−3x+1[/latex].

[latex]y<−3x+1[/latex]

Substitute [latex]x=2[/latex] and [latex]y=−3[/latex] into inequality.

[latex]−3<−3\left(2\right)+1[/latex]

[latex]\begin{array}{l}−3<−6+1\\−3<−5\end{array}[/latex]

This statement is not true, so the ordered pair [latex](2,−3)[/latex] is not a solution.

[latex](2,−3)[/latex] is not a solution.

The following video shows another example of determining whether an ordered pair is a solution to an inequality.

Graph a system of two inequalities

Consider the graph of the inequality [latex]y<2x+5[/latex].

The dashed line is [latex]y=2x+5[/latex]. Every ordered pair in the shaded area below the line is a solution to [latex]y<2x+5[/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the x and y coordinates of Points A and B into the inequality—you’ll see that they work. So, the shaded area shows all of the solutions for this inequality.

The boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don’t satisfy the inequality. If the inequality had been [latex]y\leq2x+5[/latex], then the boundary line would have been solid.

Let’s graph another inequality: [latex]y>−x[/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.

To create a system of inequalities, you need to graph two or more inequalities together. Let’s use [latex]y<2x+5[/latex] and [latex]y>−x[/latex] since we have already graphed each of them.

The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the system of inequalities . Any point within this purple region will be true for both [latex]y>−x[/latex] and [latex]y<2x+5[/latex].

In the following video examples, we show how to graph a system of linear inequalities, and define the solution region.

In the next section, we will see that points can be solutions to systems of equations and inequalities.  We will verify algebraically whether a point is a solution to a linear equation or inequality.

Determine whether an ordered pair is a solution to a system of linear inequalities

On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.

In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y>−x[/latex] and point A is a solution for the inequality [latex]y<2x+5[/latex], neither point is a solution for the system . The following example shows how to test a point to see whether it is a solution to a system of inequalities.

Is the point (2, 1) a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex]?

[latex]\begin{array}{r}x+y>1\\2+1>1\\3>1\\\text{TRUE}\end{array}[/latex]

(2, 1) is a solution for [latex]x+y>1[/latex].

[latex]\begin{array}{r}2x+y<8\\2\left(2\right)+1<8\\4+1<8\\5<8\\\text{TRUE}\end{array}[/latex]

(2, 1) is a solution for [latex]2x+y<8.[/latex]

Since (2, 1) is a solution of each inequality, it is also a solution of the system.

The point (2, 1) is a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex].

Here is a graph of the system in the example above. Notice that (2, 1) lies in the purple area, which is the overlapping area for the two inequalities.

Is the point (2, 1) a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex]?

Check the point with each of the inequalities. Substitute 2 for x and 1 for y . Is the point a solution of both inequalities?

[latex]\begin{array}{r}3x+y<4\\3\left(2\right)+1<4\\6+1<4\\7<4\\\text{FALSE}\end{array}[/latex]

(2, 1) is not a solution for [latex]3x+y<4[/latex].

Since (2, 1) is not a solution of one of the inequalities, it is not a solution of the system.

The point (2, 1) is not a solution of the system [latex]x+y>1[/latex]  and [latex]3x+y<4[/latex].

Here is a graph of this system. Notice that (2, 1) is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).

In the following video we show another example of determining whether a point is in the solution of a system of linear inequalities.

As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities.  The general steps are outlined below:

• Graph each inequality as a line and determine whether it will be solid or dashed
• Determine which side of each boundary line represents solutions to the inequality by testing a point on each side
• Shade the region that represents solutions for both inequalities

Systems with no solutions

In the next example, we will show the solution to a system of two inequalities whose boundary lines are parallel to each other.  When the graphs of a system of two linear equations are parallel to each other, we found that there was no solution to the system.  We will get a similar result for the following system of linear inequalities.

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]

The boundary lines for this system are parallel to each other, note how they have the same slopes.

[latex]\begin{array}{c}y=2x+1\\y=2x-3\end{array}[/latex]

Plotting the boundary lines will give the graph below. Note that the inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line, while the inequality [latex]y\ge2x+1[/latex] will require a solid line.

Now we need to add the regions that represent the inequalities.  For the inequality [latex]y\ge2x+1[/latex] we can test a point on either side of the line to see which region to shade. Let’s test [latex]\left(0,0\right)[/latex] to make it easy.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\ge2x+1[/latex]

[latex]\begin{array}{c}y\ge2x+1\\0\ge2\left(0\right)+1\\0\ge{1}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality  [latex]y\ge2x+1[/latex]. The graph will now look like this:

Now let’s shade the region that shows the solutions to the inequality [latex]y\lt2x-3[/latex].  Again, we can pick [latex]\left(0,0\right)[/latex] to test because it makes easy algebra.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\lt2x-3[/latex]

[latex]\begin{array}{c}y\lt2x-3\\0\lt2\left(0,\right)x-3\\0\lt{-3}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\lt2x-3[/latex]. The graph will now look like this:

This system of inequalities shares no points in common.

In the following examples, we will continue to practice graphing the solution region for systems of linear inequalities.  We will also graph the solutions to a system that includes a compound inequality.

Shade the region of the graph that represents solutions for both inequalities.  [latex]x+y\geq1[/latex] and [latex]y–x\geq5[/latex].

Find an ordered pair on either side of the boundary line. Insert the x – and y -values into the inequality [latex]x+y\geq1[/latex] and see which ordered pair results in a true statement.

[latex]\begin{array}{r}\text{Test }1:\left(−3,0\right)\\x+y\geq1\\−3+0\geq1\\−3\geq1\\\text{FALSE}\\\\\text{Test }2:\left(4,1\right)\\x+y\geq1\\4+1\geq1\\5\geq1\\\text{TRUE}\end{array}[/latex]

Since (4, 1) results in a true statement, the region that includes (4, 1) should be shaded.

Do the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y–x=5\left(\text{or }y=x+5\right)[/latex] and is solid. Test point (−3, 0) is not a solution of [latex]y–x\geq5[/latex], and test point (0, 6) is a solution.

The purple region in this graph shows the set of all solutions of the system.

The videos that follow show more examples of graphing the solution set of a system of linear inequalities.

The system in our last example includes a compound inequality.  We will see that you can treat a compound inequality like two lines when you are graphing them.

Find the solution to the system 3 x + 2 y < 12 and −1 ≤ y ≤ 5.

Graph one inequality. First graph the boundary line, then test points.

Remember, because the inequality 3 x + 2 y < 12 does not include the equal sign, draw a dashed border line.

Testing a point (like (0, 0) will show that the area below the line is the solution to this inequality.

The inequality −1 ≤ y ≤ 5 is actually two inequalities: −1 ≤ y , and y ≤ 5. Another way to think of this is y must be between −1 and 5. The border lines for both are horizontal. The region between those two lines contains the solutions of −1 ≤ y ≤ 5. We make the lines solid because we also want to include y = −1 and y = 5.

Graph this region on the same axes as the other inequality.

In the video that follows, we show how to solve another system of inequalities.

Applications

In our first example we will show how to write and graph a system of linear inequalities that models the amount of sales needed to obtain a specific amount of money.

Cathy is selling ice cream cones at a school fundraiser. She is selling two sizes: small (which has 1 scoop) and large (which has 2 scoops). She knows that she can get a maximum of 70 scoops of ice cream out of her supply. She charges $3 for a small cone and $5 for a large cone.

Cathy wants to earn at least $120 to give back to the school. Write and graph a system of inequalities that models this situation.

First, identify the variables. There are two variables: the number of small cones and the number of large cones.

s = small cone

l = large cone

Write the first equation: the maximum number of scoops she can give out. The scoops she has available (70) must be greater than or equal to the number of scoops for the small cones ( s ) and the large cones (2 l ) she sells.

[latex]s+2l\le70[/latex]

Write the second equation: the amount of money she raises. She wants the total amount of money earned from small cones (3 s ) and large cones (5 l ) to be at least $120.

[latex]3s+5l\ge120[/latex]

Write the system.

[latex]\begin{cases}s+2l\le70\\3s+5l\ge120\end{cases}[/latex]

Now graph the system. The variables x and y have been replaced by s and l ; graph s along the x -axis, and l along the y -axis.

Now graph the region [latex]3s+5l\ge120[/latex] Graph the boundary line and then test individual points to see which region to shade. The graph is shown below.

Graphing the regions together, you find the following:

The region in purple is the solution. As long as the combination of small cones and large cones that Cathy sells can be mapped in the purple region, she will have earned at least $120 and not used more than 70 scoops of ice cream.

In a previous example for finding a solution to a system of linear equations, we introduced a manufacturer’s cost and revenue equations:

Cost: [latex]y=0.85x+35,000[/latex]

Revenue: [latex]y=1.55x[/latex]

The cost equation is shown in blue in the graph below, and the revenue equation is graphed in orange.The point at which the two lines intersect is called the break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The region to the left represents quantities for which the company suffers a loss.

In the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue.

Note how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the “sweet spot” that the company wants to achieve where they produce enough bike frames at a minimal enough cost to make money. They don’t want more money going out than coming in!

Define the profit region for the skateboard manufacturing business using inequalities, given the system of linear equations:

We know that graphically,  solutions to linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the skateboard manufacturer will make a profit.

Let’s start with the revenue equation.  We know that the break even point is at (50,000, 77,500) and the profit region is the blue area.  If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use.

Let’s test the point [latex]\left(65,00,100,000\right)[/latex] in both equations to determine which inequality sign to use.

[latex]\begin{array}{l}y=0.85x+{35,000}\\{100,000}\text{ ? }0.85\left(65,000\right)+35,000\\100,000\text{ ? }90,250\end{array}[/latex]

We need to use > because 100,000 is greater than 90,250

The cost inequality that will ensure the company makes profit – not just break even – is [latex]y>0.85x+35,000[/latex]

Now test the point in the revenue equation:

[latex]\begin{array}{l}y=1.55x\\100,000\text{ ? }1.55\left(65,000\right)\\100,000\text{ ? }100,750\end{array}[/latex]

We need to use < because 100,000 is less than 100,750

The revenue inequality that will ensure the company makes profit – not just break even – is [latex]y<1.55x[/latex]

The systems of inequalities that defines the profit region for the bike manufacturer:

[latex]\begin{array}{l}y>0.85x+35,000\\y<1.55x\end{array}[/latex]

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is:

In the following video you will see an example of how to find the break even point for a small sno-cone business.

And here is one more video example of solving an application using a sustem of linear inequalities.

We have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses.  The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.

• Ex 1: Graphing Linear Inequalities in Two Variables (Slope Intercept Form). Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/Hzxc4HASygU . License : CC BY: Attribution
• System of Equations App: Break-Even Point.. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : . License : CC BY: Attribution
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Ex 2: Graphing Linear Inequalities in Two Variables (Standard Form). Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/2VgFg2ztspI . License : CC BY: Attribution
• Unit 13: Graphing, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
• Use a Graph Determine Ordered Pair Solutions of a Linear Inequalty in Two Variable. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/GQVdDRVq5_o . License : CC BY: Attribution
• Ex: Determine if Ordered Pairs Satisfy a Linear Inequality. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/-x-zt_yM0RM . License : CC BY: Attribution
• Ex 1: Graph a System of Linear Inequalities. . Authored by : James Sousa (Mathispower4u.com) .. Located at : . License : CC BY: Attribution
• Ex 2: Graph a System of Linear Inequalities.. Authored by : James Sousa (Mathispower4u.com). Located at : . License : CC BY: Attribution
• Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program.. Provided by : Monterey Institute of Technology. . Located at : http://nrocnetwork.org/dm-opentext. . License : CC BY: Attribution
• Determine the Solution to a System of Inequalities (Compound).. Authored by : James Sousa (Mathispower4u.com) . Located at : . License : CC BY: Attribution
• College Algebra. Authored by : Jay Abrams, et al.. Provided by : OpenStax. Located at : https://openstaxcollege.org/textbooks/college-algebra. . License : CC BY: Attribution

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