Warm Up

Lesson Presentation

Lesson Quiz

UNIT 2.8 TWO-VARIABLE INEQUALITIES

Warm UpFind the intercepts of each line.1. 3x + 2y = 18

2. 4x – y = 8

3. 5x + 10 = 2y

Write the function in slope-intercept form. Then graph.

4. 2x + 3y = –3

(0, 9), (6, 0)

(0, –8), (2, 0)

(0, 5), (–2, 0)

Holt Algebra 2

ObjectivesGraph linear inequalities on the coordinate plane.

Solve problems using linear inequalities.

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2-5 Linear Inequalities in Two Variables

Vocabulary

linear inequality boundary line

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2-5 Linear Inequalities in Two Variables

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Linear functions form the basis of linear inequalities. A linear inequality in two variables relates two variables using an inequality symbol, such as y > 2x – 4. Its graph is a region of the coordinate plane bounded by a line. The line is a boundary line, which divides the coordinate plane into two regions.

For example, the line y = 2x – 4, shown at right, divides the coordinate plane into two parts: one where y > 2x – 4 and one where y < 2x – 4. In the coordinate plane higher points have larger y values, so the region wherey > 2x – 4 is above the boundary line where y = 2x – 4.

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To graph y � 2x – 4, make the boundary line solid, and shade the region above the line. To graphy > 2x – 4, make theboundary line dashed because y-values equal to 2x – 4 are not included.

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Think of the underlines in the symbols ::: and �as representing solid lines on the graph.

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Helpful Hint

Example 1A: Graphing Linear Inequalities

Graph the inequality

.

The boundary line is

y-intercept of 2 and a slope of

which has a

.

Draw the boundary line dashed because it is not part of the solution.

Then shade the region above the boundary line to show

.

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Example 1A Continued

Check Choose a point in the solution region, such as (3, 2) and test it in the inequality.

The test point satisfies the inequality, so the solution region appears to be correct.

?

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2 > 1 ?

Example 1B: Graphing Linear Inequalities

Graph the inequality y � –1.

Recall that y= –1 is a horizontal line.

Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph.

Step 2 Shade the region below the boundary line to show where y < –1.Holt Algebra 2

Example 1B Continued

Check The point (0, –2) is a solution because –2 :: –1.

Note that any point on orbelow y = –1 is a solution,

regardless of the value of x.

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Check It Out! Example 1a

Graph the inequality y =:? 3x –2.

The boundary line is y = 3x – 2 which has a

y–intercept of –2 and a slope of 3.

Draw a solid line because it is part of the solution.

Then shade the region above the boundary line to show y > 3x – 2.

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Check Choose a point in the solution region, such as (–3, 2) and test it in the inequality.

y � 3x –2

?

2 � 3(–3) –2?

2 � (–9) –2

2 > –11

The test point satisfies the inequality, so the solution region appears to be correct.

Check It Out! Example 1a Continued

?

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Check It Out! Example 1b

Graph the inequality y < –3.

Recall that y = –3 is a horizontal line.

Step 1 Draw the boundary line dashed because it is not part of the solution.

Step 2 Shade the region below the boundary line to show where y < –3.

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Check It Out! Example 1b Continued

Check The point (0, –4) is a solution because –4 < –3.

Note that any point belowy < –4 is a solution, regardless

of the value of x.

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If the equation of the boundary line is not in slope-intercept form, you can choose a test point that is not on the line to determine which region to shade. If the point satisfies the inequality, then shade the region containing that point.Otherwise, shade the other region.

The point (0, 0) is the easiest point to test if it is not on the boundary line.

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Helpful Hint

Example 2: Graphing Linear Inequalities Using Intercepts

Graph 3x + 4y � 12 using intercepts.

Step 1 Find the intercepts.

Substitute x = 0 and y = 0 into 3x + 4y = 12 to find the intercepts of the boundary line.

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y-intercept

3x + 4y = 12

3(0) + 4y = 12

4y = 12

y = 3

x-intercept

3x + 4y = 12

3x + 4(0) = 12

3x = 12

x = 4

Example 2 Continued

Step 2 Draw the boundary line. The line goes through (0, 3) and(4, 0). Draw a solid line for the boundary line because it is part of the graph.

Step 3 Find the correct region to shade.Substitute (0, 0) into theinequality. Because 0 + 0 ::: 12 is true, shade the region that contains (0, 0).

(0, 3)

(4, 0)

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Holt Algebra 2

Check It Out! Example 2

Graph 3x – 4y > 12

using intercepts. Step 1 Find the

intercepts.

Substitute x = 0 and y = 0 into 3x – 4y = 12 to find the intercepts of the boundary line.

y-intercept

3x – 4y = 12

3(0) – 4y = 12

– 4y = 12

y = – 3

x-intercept

3x – 4y = 12

3x – 4(0) = 12

3x = 12

x = 4

Step 2 Draw the boundary line. The line goes through (0, –3) and(4, 0). Draw the boundary line dashed because it is not part of the solution.

Step 3 Find the correct regionto shade.Substitute (0, 0) into theinequality. Because 0 + 0 >12 is false, shade the region that does not contain (0, 0).

(4, 0)

Check It Out! Example 2

(0, –3)

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Many applications of inequalities in two variables use only nonnegative values for the variables.Graph only the part of the plane that includes realistic solutions.

Don’t forget which variable represents which quantity.

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Caution

Example 3: Problem-Solving Application

A school carnival charges $4.50 for adults and $3.00 for children. The school needs to make at least $135 to cover expenses.

A. Using x as the adult ticket price and y as the child ticket price, write and graph an inequality for the amount the school makes on ticket sales.

B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses?

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Understand the Problem

The answer will be in two parts: (1) an inequality graph showing the number of each type of ticket that must be sold to cover expenses (2) the number of adult tickets that must be sold to make at least $135 if 25 child tickets are sold.

List the important information:• The school sells tickets at $4.50 for adults

and $3.00 for children.• The school needs to make at least $135.

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Make a Plan

Let x represent the number of adult tickets and y represent the number of child tickets that must be sold. Write an inequality to represent the situation.

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An inequality that models the problem is 4.5x + 3y � 135.

Adult price

times number of adult

tickets

plus child price

times number of child tickets

is at least

total.

4.50 • x + 3.00 • y 135

Solve

Find the intercepts of the boundary line.

Graph the boundary line through (0, 45) and (30, 0) as a solid line.Shade the region above the line that is in the first quadrant, as ticket sales cannot be negative.

4.5(0) + 3y = 135

y = 45

4.5x + 3(0) = 135

x = 30

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If 25 child tickets are sold,

Substitute 25 for y in 4.5x + 3y �

135. Multiply 3 by 25.

A whole number of tickets must be sold.

At least 14 adult tickets must be sold.

Look Back

14($4.50) + 25($3.00) = $138.00, so the answer is reasonable.

4.5x + 3(25) ;: 135

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4.5x + 75 ;: 135_4

.5x ;: 60, so x ;: 13.3

Check It Out! Example 3

A café gives away prizes. A large prize costs the café $125, and the small prize costs $40. The café will not spend more than $1500. How many of each prize can be awarded? How many small prizes can be awarded if 4 large prizes are given away?

Holt Algebra 2

Understand the Problem

The answer will be in two parts: (1) an inequality graph showing the number of each type of prize awarded not too exceed a certain amount (2) the number of small prizes awarded if 4 large prizes are awarded.

List the important information:• The café awarded large prizes valued at

$125 and $40 for small prizes.• The café will not spend over $1500.

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Make a Plan

Let x represent the number of small prizes and y represent the number of large prizes, the total not too exceed $1500. Write an inequality to represent the situation.

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An inequality that models the problem is 40x + 125y ::: 135.

Small prize

times number awarded

plus large prize

times number awarded

is less than

total.

40 x + 125 y :: 1500

Solve

Find the intercepts of the boundary line.

40(0) + 125y = 1500

y = 12

Graph the boundary linethrough (0, 12) and (37.5, 0) as a solid line.

Shade the region below the line that is in the first quadrant, as prizes awarded cannot be negative.

40x + 125(0) = 1500

x = 37.5

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If 4 large prizes are awarded,

Substitute 4 for y in 40x + 125y ::

135. Multiply 125 by 4.

A whole number of small prizes must be awarded.

No more than 25 small prizes can be awarded.

Look Back

$40(25) + $125(4) = $1500, so the answer is reasonable.

40x + 125(4) :: 1500

40x + 500 :: 1500

40x ;: 1000, so x :: 25

Holt Algebra 2

You can graph a linear inequality that is solved for y with a graphing calculator. Pressand use the left arrow key to move to the left side.

Each time you press you will see one of thegraph styles shown here. You are already familiar with the line style.

Holt Algebra 2

y < 4x – 4

Example 4: Solving and Graphing Linear Inequalities

Solve for y. Graph the solution.

Multiply both sides by

8x – 2y > 8 –2y > –8x + 8 Subtract 8x from both sides.

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Divide by –2, and reverse the inequality symbol.

Example 4 Continued

Use the calculator option to shade below the liney < 4x – 4.

Note that the graph is shown in the STANDARD SQUAREwindow.

(by

6:ZStandard followed5:ZSquare).

Check It Out! Example 4

Solve 2(3x – 4y) > 24 for y. Graph the solution.

Subtract 3x from both sides.

Divide by –4, and reverse the inequality symbol.

3x – 4y > 12

–4y > –3x + 12

Divide both sides by 2.

6:ZStandard followed5:ZSquare).

Check It Out! Example 4 Continued

Use the calculator option to shade below the line

.

Note that the graph is shown in the STANDARD SQUAREwindow.

(by

Lesson Quiz: Part I

1. Graph 2x –5y 10 using intercepts.

2. Solve –6y < 18x – 12 for y. Graph the solution.

y > –3x + 2

Lesson Quiz: Part II

3. Potatoes cost a chef $18 a box, and carrots cost $12 a box. The chef wants to spend no more than

$144. Use x as the number of boxes of potatoes and y as the number of boxes of carrots.

a. Write an inequality for the chef can buy.

18x + 12y ::: 144

the number of boxes

b. How many boxes of potatoes can the chef order if she orders 4 boxes of carrot?

no more than 5

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