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12x+7<-11 or 5x-8>40

12x+7<-11 or 5x-8>40-example-1

2 Answers

4 votes

Answer:

A

Step-by-step explanation:

Given

12x + 7 < - 11 or 5x - 8 > 40

Solve each inequality

12x + 7 < - 11 ( subtract 7 from both sides )

12x < - 18 ( divide both sides by 12 )

x < -
(3)/(2)

OR

5x - 8 > 40 ( add 8 to both sides )

5x > 48 ( divide both sides by 5 )

x >
(48)/(5)

Solution is

x < -
(3)/(2) or x >
(48)/(5) → A

User Hynekcer
by
4.4k points
2 votes

To solve the compound inequality 12x+7<-11 OR 5x-8>40, first solve 12x+7<-11 and then solve 5x-8>40. The final solution is x < -3/2 or x > 48/5.

Step-by-step explanation:

To solve the given compound inequality, we will solve each inequality separately and then combine the solutions.

For the first inequality, 12x + 7 < -11, we will subtract 7 from both sides and divide by 12 to solve for x:

12x + 7 < -11

12x < -18

x < -18/12

x < -3/2

For the second inequality, 5x - 8 > 40, we will add 8 to both sides and divide by 5 to solve for x:

5x - 8 > 40

5x > 48

x > 48/5

Therefore, our final solution is x < -3/2 or x > 48/5.

User Obecker
by
4.9k points