Holt Algebra 2 2-5 Linear Inequalities in Two Variables 2-5 Linear Inequalities in Two Variables...

Holt Algebra 2

2-5 Linear Inequalities in Two Variables 2-5 Linear Inequalities in Two Variables

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 2

2-5 Linear Inequalities in Two Variables

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


Graph linear inequalities on the coordinate plane.Solve problems using linear inequalities.

Objectives

Holt Algebra 2


linear inequalityboundary line

Vocabulary

Holt Algebra 2


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.

Holt Algebra 2


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 where y > 2x – 4 is above the boundary line where y = 2x – 4.

Holt Algebra 2


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

Holt Algebra 2


Think of the underlines in the symbols ≤ and ≥ as representing solid lines on the graph.

Helpful Hint

Holt Algebra 2


Example 1A: Graphing Linear Inequalities

Graph the inequality .

The boundary line is which has a

y-intercept of 2 and a slope of .

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

Then shade the region above the boundary line to show

.

Holt Algebra 2


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.

?

2 > 1 ?

Holt Algebra 2


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.

Example 1B: Graphing Linear Inequalities

Holt Algebra 2


Check The point (0, –2) is a solution because –2 ≤ –1. Note that any point on or below y = –1 is a solution, regardless of the value of x.

Example 1B Continued

Holt Algebra 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.

Check It Out! Example 1a

Graph the inequality y ≥ 3x –2.

Holt Algebra 2


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.

y ≥ 3x –2

Check It Out! Example 1a Continued

2 ≥ 3(–3) –2?

2 ≥ (–9) –2?

2 > –11 ?

Holt Algebra 2


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.

Check It Out! Example 1b

Holt Algebra 2


Check It Out! Example 1b Continued

Check The point (0, –4) is a solution because –4 < –3. Note that any point below y < –4 is a solution, regardless of the value of x.

Holt Algebra 2


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.

Helpful Hint

Holt Algebra 2


Graph 3x + 4y ≤ 12 using intercepts.

Example 2: Graphing Linear Inequalities 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 x-intercept

3x + 4y = 12

3(0) + 4y = 12 3x + 4(0) = 12

4y = 12

3x + 4y = 12

y = 3

3x = 12

x = 4

Holt Algebra 2


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 the inequality. Because 0 + 0 ≤ 12 is true, shade the region that contains (0, 0).

(0, 3)

(4, 0)

Holt Algebra 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 x-intercept

3x – 4y = 12

3(0) – 4y = 12 3x – 4(0) = 12

– 4y = 12

3x – 4y = 12

y = – 3

3x = 12

x = 4

Check It Out! Example 2

Holt Algebra 2


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 region to shade.Substitute (0, 0) into the inequality. Because 0 + 0 >12 is false, shade the region that does not contain (0, 0).

(4, 0)


(0, –3)

Holt Algebra 2


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.

Caution

Holt Algebra 2


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?

Holt Algebra 2


1 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.

Holt Algebra 2


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.

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

2 Make a Plan

135y3.00+x4.50

total.is at least

number of child tickets

timeschild price

plusnumber of

adult tickets

timesAdult price

• •

Holt Algebra 2


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.

Solve3

4.5(0) + 3y = 135 4.5x + 3(0) = 135

y = 45 x = 30

Holt Algebra 2


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.

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

Look Back4

4.5x + 3(25) ≥ 135

4.5x + 75 ≥ 135

4.5x ≥ 60, so x ≥ 13.3_

Holt Algebra 2



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


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.

1 Understand the Problem

Holt Algebra 2


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.

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

2 Make a Plan

1500y125+x40

total.is less than

number awarded

timeslarge prize

plusnumber awarded

timesSmall prize

≤

Holt Algebra 2


Find the intercepts of the boundary line.

Graph the boundary line through (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.

Solve3

40(0) + 125y = 1500 40x + 125(0) = 1500

y = 12 x = 37.5

Holt Algebra 2


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.

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

Look Back4

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. Press and use the left arrow key to move to the left side.

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

Holt Algebra 2


Solve for y. Graph the solution.

Subtract 8x from both sides.

Divide by –2, and reverse the inequality symbol.

8x – 2y > 8

–2y > –8x + 8

Example 4: Solving and Graphing Linear Inequalities

y < 4x – 4

Multiply both sides by

Holt Algebra 2


Note that the graph is shown in the STANDARD SQUARE window. ( 6:ZStandard followed by 5:ZSquare).

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

Example 4 Continued

Holt Algebra 2


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

Subtract 3x from both sides.

Divide by –4, and reverse the inequality symbol.

–4y > –3x + 12


Divide both sides by 2.3x – 4y > 12

Holt Algebra 2


Note that the graph is shown in the STANDARD SQUARE window. ( 6:ZStandard followed by 5:ZSquare).

Use the calculator option to shade below the line

.

Check It Out! Example 4 Continued

Holt Algebra 2


Lesson Quiz: Part I

1. Graph 2x –5y 10 using intercepts.

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

y > –3x + 2

Holt Algebra 2


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 number of boxes the chef can buy.

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

18x + 12y ≤ 144

no more than 5

Lesson Quiz: Part II

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Holt Algebra 2 2-5 Linear Inequalities in Two Variables 2-5 Linear Inequalities in Two Variables...

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