Holt Algebra 2 2-5 Linear Inequalities in Two Variables 6-6 Linear Inequalities in Two Variables 1. Warm Up (Slide #2) 1. Warm Up (Slide #2) 3. Lesson

Holt Algebra 2

2-5 Linear Inequalities in Two Variables 6-6 Linear Inequalities in Two Variables

1. Warm Up 1. Warm Up (Slide #2)(Slide #2)

3. Lesson Presentation 3. Lesson Presentation (Slide #6 – 33)(Slide #6 – 33)

6. Lesson Quiz 6. Lesson Quiz (Slide #36,37)(Slide #36,37)

4. Text Questions 4. Text Questions (Slide #34)(Slide #34)

LESSON PLAN

2. Vocab 2. Vocab (Slides #4 - 5)(Slides #4 - 5)

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

Holt Algebra 2


1. Graph linear inequalities on the coordinate plane

2. Solve problems using linear inequalities

Objectives

Holt Algebra 2


linear inequality

Vocabulary

Holt Algebra 2


A linear inequality relates two variables using an inequality symbol such as y > 2x – 4.

Its graph is a regionregion (…or a wedge-shaped areaa wedge-shaped area) of the coordinate plane bounded by a line.

Holt Algebra 2


For example:

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 has a y-int. +2 and slope of .

Draw the line dashed since it is not part of the solution.

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.

?

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

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The boundary line has a y–intercept of ____

and a slope of _____.

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

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

Check It Out! Example 1a

Graph the inequality y ≥ 3x –2.

-2-2

+3+3

solidsolid

aboveabove

Holt Algebra 2


Check Is (–3, 2) a solution to 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 ________________line.

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

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

Check It Out! Example 1b

horizontalhorizontal

dasheddashed

belowbelow

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Check It Out! Example 1b Continued

Check Is (0, –4) a solution? Yes!Yes!

Holt Algebra 2


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.

y-intercept (is where x = 0) x-intercept (is where y = 0)

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)

Graphing 3x + 4y ≤ 12 using intercepts:

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)

Graph 3x – 4y > 12 using intercepts.

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 number of adult tickets and y as the number of child tickets, 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


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.

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.

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

y = 45 x = 30

4.5x + 3y ≥ 135.

Holt Algebra 2


If 25 child tickets are sold,

At least 14 adult tickets must be sold.

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


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 ≤ 1500

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.

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

y = 12 x = 37.5

40x + 125y ≤ 1500

Holt Algebra 2


If 4 large prizes are awarded,

No more than 25 small prizes can be awarded.

40x + 125(4) ≤ 1500

40x + 500 ≤ 1500

40x ≤ 1000

x ≤ 25

40x + 125y ≤ 1500

Holt Algebra 2


Graphing a linear inequality with a graphing calculator.

Press the key.

Press the left arrow key to move to the left side of

Each time you press you will see one of the graph styles shown here.

>

<

Holt Algebra 2


1. Key “Y=“2. Key “<“ twice3. Key “ENTER” three times to get to the “less than” graphic 4. Key “>” twice to return to the equation area5. Key “4X-4”6. Key “GRAPH”

Use the calculator graph the solution to y < 4x - 4

Check It Out! Example 4 Continued

Holt Algebra 2


1. Key “Y=“2. Key “<“ twice3. Key “ENTER” three times4. Key “>” twice5. Key “(3÷4)X-3”6. Key “GRAPH”

Use the calculator graph the solution to

Check It Out! Example 4 Continued

Holt Algebra 2


Graph this inequality on a calculator: 2x + 3y < -5

2x + 3y < -5-2x

3y < -2x – 5 /3

y < -(2/3)x - 5/3

-2x

/3 /3

To use the graphing calculator, you must first rewrite the inequality so you are “solving for ‘y’”)

Documents

Holt Algebra 2 2-5 Linear Inequalities in Two Variables 6-6 Linear Inequalities in Two Variables 1. Warm Up (Slide #2) 1. Warm Up (Slide #2) 3. Lesson