Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Lesson Quiz 3.11. Describe the solutions of 7 < x + 4.

all real numbers greater than 3

2. Graph h ≥ –4.75

–5 –4.75 –4.5

Write the inequality shown by each graph.

3. x ≥ 3

4. x < –5.5

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

4 – 6a + 4a = –1 – 5(7 – 2a)

4 – 6a + 4a = -1 + (-5)(7) + (-5)(-2a)

4 – 6a + 4a = -1 + (-35) + (10a)

4 – 2a = -36 + 10a

+36 +2a +36 + 2a

40 = 12a

12 12

10/3 = a

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

own!your on Finish

236

)12

3(2)6(

2

12

12

3)6(

2

1

bb

bb

bb

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Students will be able to: Solve one-step inequalities by using addition and solve one-step inequalities by using subtraction.

Learning Target

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Solve the inequality and graph the solutions.

x + 12 < 20

–10 –8 –6 –4 –2 0 2 4 6 8 10

12 128x

d – 5 > –7

–10 –8 –6 –4 –2 0 2 4 6 8 10

5 52d

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

0.9 ≥ n – 0.3

–10 –8 –6 –4 –2 0 2 4 6 8 10

> –3 + t

–10 –8 –6 –4 –2 0 2 4 6 8 10

0.3 0.31.2 n

1.2

3 31

52t

15

2

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

14 30x

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 left on gift cardx

14 1416x

1614 18

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

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.

475 550x amount she can add to the highest bidx

475 475$75x

7573 77

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Students will be able to: Solve one-step inequalities by using multiplication and solve

one-step inequalities by using division.

Learning Target

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Solve the inequality and graph the solutions.

7x > –42

–10

–8

–6–4 –2 0 2 4 6 8 10

–10

–8

–6–4 –2 0 2 4 6 8 107 76x

3 3

7.2 m7.2

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Solve the inequality and graph the solutions.4

3

4

3

16r

4

1614 18

–50 ≥ 5q

5 510 q –

10–8

–6–4 –2 0 2 4 6 8 10

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Solve the inequality and graph the solutions.

4

3

4

3

36g

9

3634 38

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

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.

Solve the inequality and graph the solutions.

–12x > 84

–10

–8

–6–4 –2 0 2 4 6 8 1012 12

7x

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Solve the inequality and graph the solutions.

3 3

24 x2422 26

4.25 > –0.25h0.25 0.25

17 h 1719 15

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

4.30 20T

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?

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

4.30 4.304.65T

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

10 128S

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

pitcher can fill?

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

HW: pp.183-185/1-17,18-42

10 1012.8S

Holt Algebra 1

3-2Solving Inequalities by Adding or Subtracting

Warm UpGraph each inequality. Write an inequality for each situation. 1. The temperature must be at least –10°F.

2. The temperature must be no more than 90°F.

x ≥ –10

x ≤ 90

Solve each equation.

3. x – 4 = 10 14

4. 15 = x + 1.1 13.9

–10 0 10

–90 0 90