Chapter 3 - Inequalities Algebra I. Table of Contents 3.1 - Graphing and Writing Inequalities 3.1...

Chapter 3 - Inequalities

Algebra I

Table of Contents

• 3.1 - Graphing and Writing Inequalities• 3.2 - Solving Inequalities by Adding or Subtracting• 3.3 - Solving inequalities by multiplying or dividing• 3.4 - Solving Two Step and Multi-Step Inequalities• 3.5 - Solving Inequalities with Variables on Both Sides• 3.6 - Solving Compound Inequalities• 3.7- Solving Absolute-Value Inequalities

3.1- Graphing and Writing Inequalities

Algebra I

An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs:

≤A ≤ BA is less than or

equal to B.

<A < BA is lessthan B.

>A > B

A is greaterthan B.

≥A ≥ B

A is greaterthan or

equal to B.

≠A ≠ B

A is notequal to B.

A solution of an inequality is any value that makes the inequality true.

3.1 Algebra 1 (bell work)

Describe the solutions of x – 6 ≥ 4 in words.

It appears that the solutions of x – 6 ≥ 4 are all real numbers greater than or equal to 10.

Describe the solutions of 2p > 8 in words.

It appears that the solutions of 2p > 8 are all real numbers greater than 4.

3.1 Example 1 Identifying Solutions of Inequalities

+6 +6

X ≥ 10

2 2

p > 4

3.1Do not Copy

Graph each inequality

A. m ≥

0 1– 2 3 3

B. t < 5(–1 + 3) t < 5(–1 + 3)

t < 5(2) t < 10

–4 –2 0 2 4 6 8 10 12 –6 –8

3.1 Example 2 Graphing Inequalities

Write the inequality shown by each graph

x < 2 x ≥ –0.5

3.1 Example 3 Writing an Inequality from a Graph

Math Joke

• Q: Why did the Moore family name their son Lester?

• A: So he could be called either ‘Moore’ or ‘les’

3.1

Ray’s dad told him not to turn on the air conditioner unless the temperature is at least 85°F.

Define a variable and write an inequality for the temperatures at which Ray can turn on the air conditioner. Graph the solutions.

Let t represent the temperatures at which Ray can turn on the air conditioner.

75 80 85 9070

Turn on the AC when temperature is at least 85°F

t ≥ 85

t 85

3.1 Example 4 Application

A store’s employees earn at least $8.50 per hour

Define a variable and write an inequality for the amount the employees may earn per hour. Graph the solutions.

Let w represent an employee’s wages.

An employee earns at least $8.50

w ≥ 8.50

4 6 8 10 12−2 0 2 14 16 18

8.5

w ≥ 8.5

3.1

HW pg. 173

• 3.1-– 3-15 (Odd), 16, 17, 18-32 (Even), 33, 42-47, 55, 74-

81

– Follow All HW Guidelines or ½ off

3.2 - Solving Inequalities by Adding or Subtracting

Algebra I

3.2 Algebra 1 (bell work)

Just Read

Solve the inequality and graph the solutions.

x + 12 < 20–12 –12

x + 0 < 8

x < 8

–10 –8 –6 –4 –2 0 2 4 6 8 10

3.2 Example 1 Using Addition and Subtraction to Solve Inequalities

+5 +5

d + 0 > –2

d > –2

d – 5 > –7

–10 –8 –6 –4 –2 0 2 4 6 8 10

d – 5 > –7

Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions.

You can check the endpoint and the direction of the inequality symbol.

The solutions of x + 9 < 15 are given by x < 6.

3.2

Math Joke

• Q: Why did the parents think their little variable was sick?

• A: The nurse said he had to be isolated

3.2

Sami has a gift card. She has already used $14 of the of the total value, which was $30.

Write, solve, and graph an inequality to show how much more she can spend.

Amount remaining

plus $30.is at mostamount used

g + 14 ≤ 30

g + 14 ≤ 30

– 14 – 14

g + 0 ≤ 16

g ≤ 16

0 2 4 6 8 10 12 14 16 18 10

3.2 Example 2/3 Application

Mrs. Lawrence wants to buy an antique bracelet at an auction. She is willing to bid no more than $550. So far, the highest bid is $475.

Write and solve an inequality to determine the amount Mrs. Lawrence can add to the bid. Check your answer.

Let x represent the amount Mrs. Lawrence can add to the bid.

$475 plus amount can add

is at most $550.

x+475 ≤ 550

475 + x ≤ 550

–475 – 475

x ≤ 75

0 + x ≤ 75

3.2

Josh wants to try to break the school bench press record of 282 pounds. He currently can bench press 250 pounds. Write and solve an inequality to determine how many more pounds Josh needs to lift to break the school record. Check your answer.

Let p represent the number of additional pounds Josh needs to lift.

250 pounds plus additional pounds is greater than

282 pounds.

250 + p > 282

250 + p > 282

–250 –250

p > 32

3.2

HW pg. 179

• 3.2-– 1-15, 25-31, 47-56– Ch: 35

– Follow All HW Guidelines or ½ off

3.3 - Solving Inequalities by Multiplying or Dividing

Algebra I

3.3 Algebra 1 (bell work)

Just Read

Solve the inequality and graph the solutions.

7x > –42

7x > –42

>

1x > –6 x > –6

–10 –8 –6 –4 –2 0 2 4 6 8 10

3.3 Example 1 Multiplying or Dividing by a Positive Number

3(2.4) ≤ 3

m ≥ 7.2)

0 2 4 6 8 10 12 14 16 18 20

r < 16

0 2 4 6 8 10 12 14 16 18 20

Solve the inequality and graph the solutions3.3

If you multiply or divide both sides of an inequality by a negative number, the resulting inequality is not a true statement.

You need to reverse the inequality symbol to make the statement true.

3.3

Math Joke

• Q: What did the teacher do to prepare for class?

• A: She made a “less-than” plan (lesson plan)

3.3

Solve the inequality and graph the solutions.

–12x > 84

x < –7

–10 –8 –6 –4 –2 0 2 4 6–12–14

–7

3.3 Example 2 Multiplying or Dividing by a Negative Number

16 18 20 22 2410 14 26 28 3012

x 24)

Solve each inequality and graph the solutions

a. 10 ≥ –x

–1(10) ≤ –1(–x)

–10 ≤ x

b. 4.25 > –0.25h

–17 < h

–20 –16 –12 –8 –4 0 4 8 12 16 20

–17–10 –8 –6 –4 –2 0 2 4 6 8 10

3.3

$4.30 times number of tubes is at most $20.00.

4.30 • p ≤ 20.00

Jill has a $20 gift card to an art supply store where 4 oz tubes of paint are $4.30 each after tax. What are the possible numbers of tubes that Jill can buy?Let p represent the number of tubes of paint that Jill can buy.

4.30p ≤ 20.00

p ≤ 4.65

Since Jill can buy only whole numbers of tubes, she can buy 0, 1, 2, 3, or 4 tubes of paint.

3.3 Example 3 Application

A pitcher holds 128 ounces of juice. What are the possible numbers of 10-ounce servings that one pitcher can fill?

10 oz timesnumber of servings

is at most 128 oz

10 • x ≤ 128

Let x represent the number of servings of juice the pitcher can contain.

10x ≤ 128

x ≤ 12.8

The pitcher can fill 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12 servings.

3.3

HW pg. 185

• 3.3-– 19-41 (Odd), 17, 42, 51-54, 66, 78-87– Ch: 62

– Follow All HW Guidelines or ½ off

3.4 - Solving Two-Step and Multi-Step Inequalities

Algebra I

Solve the inequality and graph the solutions.

45 + 2b > 61

45 + 2b > 61–45 –45

2b > 16

b > 8

0 2 4 6 8 10 12 14 16 18 20

3.4 Example 1 Solving Multi-Step Inequalities

–12 ≥ 3x + 6

–12 ≥ 3x + 6– 6 – 6

–18 ≥ 3x

–6 ≥ x

–10 –8 –6 –4 –2 0 2 4 6 8 10

Solve the inequality and graph the solutions.

x < –11

–5 –5x + 5 < –6

–20 –12 –8 –4–16 0

–11

3.4 Example 2 Simplifying Before Solving Inequalities

–4(2 – x) ≤ 8

−4(2 – x) ≤ 8−4(2) − 4(−x) ≤ 8

–8 + 4x ≤ 8+8 +8

4x ≤ 16

x ≤ 4

–10 –8 –6 –4 –2 0 2 4 6 8 10

Solve the inequality and graph the solutions.

4f + 3 > 2–3 –3

4f > –1

4f > –1

0

3.4

Math Joke

• Q: What did the doctor say to the multi-step inequality?

• A: I can solve your problems with a few operations

3.4

To rent a certain vehicle, Rent-A-Ride charges $55.00 per day with unlimited miles. The cost of renting a similar vehicle at We Got Wheels is $38.00 per day plus $0.20 per mile.

For what number of miles in the cost at Rent-A-Ride less than the cost at We Got Wheels?

Let m represent the number of miles. The cost for Rent-A-Ride should be less than that of We Got Wheels.

Cost at Rent-A-Ride

must be less than

daily cost at We Got

Wheelsplus

$0.20 per mile times # of

miles.

55 < 38 + 0.20 m

3.4 Example 3 Application

85 < m

Rent-A-Ride costs less when the number of miles is more than 85.

55 < 38 + 0.20m

–38 –3855 < 38 + 0.20m

17 < 0.20m

3.4

The average of Jim’s two test scores must be at least 90 to make an A in the class. Jim got a 95 on his first test.

What grades can Jim get on his second test to make an A in the class?

Let x represent the test score needed. The average score is the sum of each score divided by 2.

First test score plus

second test score

divided by

number of scores

is greater than or equal

to

total score

(95 + x) 2 ≥ 90

3.4

The score on the second test must be 85 or higher.

95 + x ≥ 180

–95 –95

x ≥ 85

3.4

HW pg.193

• 3.4-– 15-37 (Odd), 44, 49,51-54, 59, 77-86– Ch: 61

– Follow All HW Guidelines or ½ off

3.5 - Solving Inequalities with Variables on Both Sides

Algebra I

Solve the inequality and graph the solutions.

y ≤ 4y + 18

y ≤ 4y + 18–y –y

0 ≤ 3y + 18

–18 – 18

–18 ≤ 3y

y –6

–10 –8 –6 –4 –2 0 2 4 6 8 10

3.5 Example 1 Solving Inequalities with Variables on Both Sides

4m – 3 < 2m + 6

–2m – 2m

2m – 3 < + 6

+ 3 + 3

2m < 9

4 5 6

Solve the inequality and graph the solutions

5t + 1 < –2t – 6

5t + 1 < –2t – 6+2t +2t

7t + 1 < –6

– 1 < –1

7t < –7

7t < –7

7 7t < –1

–5 –4 –3 –2 –1 0 1 2 3 4 5

3.5

The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding.How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean?

Let w be the number of windows.

312 + 12 • w < 36 • w312 + 12w < 36w

312 < 24w

13 < w

The Home Cleaning Company is less expensive for houses with more than 13 windows.

HomeCleaningCompany

siding charge

plus$12 per window

# of windows

is lessthan

PowerClean

cost per window

# ofwindows.

timestimes

3.5 Example 2 Application

Solve the inequality and graph the solutions.

2(k – 3) > 6 + 3k – 3

2(k – 3) > 3 + 3k

2k + 2(–3) > 3 + 3k

2k – 6 > 3 + 3k–2k – 2k

–6 > 3 + k

–3 –3

–9 > k

3.5 Example 3 Simplifying Each Side Before Solving

3.5 Just Read

Math Joke

• Q: What did Miss Manners say to the inequality symbol?

• A: It’s not polite to point

3.5

Solve the inequality.

2x – 7 ≤ 5 + 2x

2x – 7 ≤ 5 + 2x–2x –2x

–7 ≤ 5

The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

3.5 Example 4 All Real Numbers as Solutions or No Solutions

2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)

6y – 4 – 4 ≥ 6y + 21

6y – 8 ≥ 6y + 21

–6y –6y

–8 ≥ 21 No values of y make the inequality true. There are no solutions.

4(y – 1) ≥ 4y + 2

4(y – 1) ≥ 4y + 2

4(y) + 4(–1) ≥ 4y + 2

4y – 4 ≥ 4y + 2

Solve the inequality.

–4y –4y

–4 ≥ 2 No values of y make the inequality true. There are no solutions.

3.5 Optional

HW pg. 199

• 3.5-– 1-17, 26, 45-49– Ch: 50-56

– Follow All HW Guidelines or ½ off

3.6 - Solving Compound Inequalities

Algebra I

3.6 Algebra 1 (bell work)

Copy Boxed Parts

Pg. 204

The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions.

Let p be the pH level of the shampoo.

6.0 is less than or equal to

pH level is less than or equal to

6.5

6.0 ≤ p ≤ 6.5

6.0 ≤ p ≤ 6.5

5.9 6.1 6.2 6.36.0 6.4 6.5

3.6 Example 1 Application

The free chlorine in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.

Let c be the chlorine level of the pool.

1.0 is less than or equal to

chlorine is less than or equal to

3.0

1.0 ≤ c ≤ 3.0

1.0 ≤ c ≤ 3.0

0 2 3 4 1 5 6

3.6

In this diagram, oval A represents some integer solutions of x < 10 and oval B represents some integer solutions of x > 0.

The overlapping region represents numbers that belong in both ovals. Those numbers are solutions of both x < 10 and x > 0.

3.6

You can graph the solutions of a compound inequality involving AND by using the idea of an overlapping region.

The overlapping region is called the intersection and shows the numbers that are solutions of both inequalities.

3.6

Solve the compound inequality and graph the solutions.

–5 < x + 1 < 2

–5 < x + 1 < 2

–1 – 1 – 1

–6 < x < 1

–10 –8 –6 –4 –2 0 2 4 6 8 10

3.6 Example 2 Solving Compound Inequalities Involving AND

8 < 3x – 1 ≤ 11

8 < 3x – 1 ≤ 11+1 +1 +1

9 < 3x ≤ 12

3 < x ≤ 4

In this diagram, circle A represents some integer solutions of x < 0, and circle B represents some integer solutions of x > 10.

The combined shaded regions represent numbers that are solutions of either x < 0 or x >10.

3.6

Math Joke

• Q: How does a math teacher get a compound fracture?

• A: She breaks her hAND

3.6

You can graph the solutions of a compound inequality involving OR by using the idea of combining regions.

The combine regions are called the union and show the numbers that are solutions of either inequality.

3.6 Day 2

Solve the inequality and graph the solutions.

8 + t ≥ 7 OR 8 + t < 2

8 + t ≥ 7 OR 8 + t < 2

–8 –8 –8 −8

t ≥ –1 OR t < –6

–10 –8 –6 –4 –2 0 2 4 6 8 10

3.6 Example 3 Solving Compound Inequalities Involving OR

4x ≤ 20 OR 3x > 21

4x ≤ 20 OR 3x > 21

x ≤ 5 OR x > 7

0 2 4 6 8 10–10 –8 –6 –4 –2

Every solution of a compound inequality involving AND must be a solution of both parts of the compound inequality.

If no numbers are solutions of both simple inequalities, then the compound inequality has no solutions.

The solutions of a compound inequality involving OR are not always two separate sets of numbers.

There may be numbers that are solutions of both parts of the compound inequality.

3-6

Write the compound inequality shown by the graph.

The shaded portion of the graph is not between two values, so the compound inequality involves OR.

The compound inequality is x ≤ –8 OR x > 0.

3.6 Example 4 Writing a Compound Inequality from a Graph

The shaded portion of the graph is between the values –2 and 5, so the compound inequality involves AND.

The compound inequality is m > –2 AND m < 5 (or -2 < m < 5).

The shaded portion of the graph is between the values –9 and –2, so the compound inequality involves AND.

The compound inequality is –9 < x AND x < –2 (or –9 < x < –2).

Write the compound inequality shown by the graph.3.6

The shaded portion of the graph is not between two values, so the compound inequality involves OR.

The compound inequality is x ≤ –3 OR x ≥ 2.

HW pg. 208

• 3.6-– Day 1: 1-6, 15-19, 57-65– Day 2: 7-14, 24-28, 34, 44– Ch: 29

– Follow All HW Guidelines or ½ off

3.7- Solving Absolute Value Inequalities

Algebra I

3.7

Solve the inequality and graph the solutions. Then write the solutions as a compound inequality.

|x| + 2 ≤ 6

|x| + 2 ≤ 6–2 –2

|x| ≤ 4

x ≥ –4 AND x ≤ 4

–5 –4 –3 –2 –1 0 1 2 3 4 5

4 units 4 units

–4 ≤ x ≤ 4

3.7

|x| – 5 < –4

|x| – 5 < –4+5 +5|x| < 1

–1 < x AND x < 1

–1 < x < 1

–2 –1 0 1 2

unit1

unit1

Solve the inequality and graph the solutions. Then write the solutions as a compound inequality.

3.7

3.7

Math Joke

• Q: What does an absolute-value expression work on when it goes to the gym?

• A: Its “abs”!

3.7

Solve the inequality and graph the solutions. Then write the solutions as a compound inequality.

|x| + 2 > 7|x| + 2 > 7

– 2 –2

|x| > 5

x < –5 OR x > 5

–10 –8 –6 –4 –2 0 2 4 6 8 10

5 units 5 units

3.7

Solve the inequality and graph the solutions. Then write the solutions as a compound inequality.

|x + 3| – 5 > 9

|x + 3| – 5 > 9+ 5 +5

|x + 3| > 14

–16 –12 –8 –4 0 4 8 12 16

x + 3 < –14 OR x + 3 > 14

14 units 14 units

3.7

|x| – 7 > –1|x| – 7 > –1

+7 +7|x| > 6

–10 –8 –6 –4 –2 0 2 4 6 8 10

x < –6 OR x > 6

Solve the inequality and graph the solutions. Then write the solutions as a compound inequality.

6 units 6 units

3.7

HW pg. 215

• 3.7-– 1-12, 14-22, 61-64

– Follow All HW Guidelines or ½ off