Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve each inequality. Graph the solutions. 1.w + 4 –5 2.7 < z – 3 3.4 > a + 64.x – 5 –6 > < (For

Solving Two-Step InequalitiesSolving Two-Step InequalitiesPRE-ALGEBRA LESSON 7-6PRE-ALGEBRA LESSON 7-6

Solve each inequality. Graph the solutions.

1. w + 4 –5 2. 7 < z – 3

3. 4 > a + 6 4. x – 5 –6

>

<

(For help, go to Lesson 2-9.)

Check Skills You’ll Need

7-6


Solutions

1. w + 4 –5 2. 7 < z – 3w + 4 – 4 –5 – 4 7 + 3 < z – 3 + 3 w –9 10 < z

z > 10

>>

>

3. 4 > a + 6 4. x – 5 –64 – 6 > a + 6 – 6 x – 5 + 5 –6 + 5 –2 > a x –1 a < –2

<<<

7-6

Solving Two-Step InequalitiesSolving Two-Step Inequalities

Solve and graph 7g + 11 > 67.

PRE-ALGEBRA LESSON 7-6PRE-ALGEBRA LESSON 7-6

7g + 11 > 67

7g + 11 – 11 > 67 – 11 Subtract 11 from each side.

7g > 56 Simplify.

g > 8 Simplify.

Divide each side by 7.>7g7

567

Quick Check

7-6


Solve 6 – r – 6.


< 23

6 – r – 6<23

6 + 6 – r – 6 + 6< 23

Add 6 to each side.

Simplify.12 – r< 23

Simplify.>–18 r, or r –18<

32– (12)

32–

23– r Multiply each side by . Reverse the direction of

the inequality symbol.

32

–>

Quick Check

7-6


Dale has $25 to spend at a carnival. If the admission to

the carnival is $4 and the rides cost $1.50 each, what is the

greatest number of rides Dale can go on?


25Inequality 4 + •1.5 r <

Let = number of rides Dale goes on.

Words $4 admission + $1.50/ridenumberof rides

is less thanor equal to

$25•

r

7-6


(continued)


The greatest number of rides Dale can go on is 14.

4 + 1.5r 25<

Subtract 4 from each side.4 + 1.5r – 4 25 – 4<

Simplify.1.5r 21<

Divide each side by 1.5.1.5r1.5

211.5<

Simplify.r 14<

Quick Check

7-6


Solve each inequality.

1. -2m + 4 34

2. 5a - 9 > 11

a > 4

7-6

<

m 15>

Documents

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve each inequality. Graph the solutions. 1.w + 4 –5 2.7 < z – 3 3.4 > a + 64.x – 5 –6 > < (For