Question

12x+7<−11 OR 5x−8>40. Solve for the inequality

Answer 1

Answer:

$(-\infty,-\frac{3}{2})\text{U}(\frac{48}{5},\infty)$

Step-by-step explanation:

We have been given a compound inequality $12x+7<-11\text{ (or) } 5x-8>40$. We are asked to solve the given inequality.

First of all, we will solve both parts of inequality separately and then we will combine the solution.

$12x+7<-11$

$12x+7-7<-11-7$

$12x<-18$

$\frac{12x}{12}<-\frac{18}{12}$

$x<-\frac{3}{2}$

$x<-1.5$

$5x-8>40$

$5x-8+8>40+8$

$5x>48$

$\frac{5x}{5}>\frac{48}{5}$

$x>\frac{48}{5}$

$x>9.6$

Therefore, the solution of the given compound inequality would be $x<-\frac{3}{2}\text{ (or) } x>\frac{48}{5}$ that is $(-\infty,-\frac{3}{2})\text{U}(\frac{48}{5},\infty)$ in interval notation.

Answer 2

The solution for the inequality 12x+7<−11 is x < -3/2

The solution for the inequality  5x−8>40 is x < 48/5

The given inequalities are:

12x + 7 < −11

5x − 8 > 40

First, solve 12x + 7 < −11

Subtract 7 from both sides

12x  +  7  -  7   <  -11 - 7

12x    <   -18

Divide both sides by 12

$\frac{12x}{12} < \frac{-18}{12} \\\\x < \frac{-3}{2}$

First, solve 5x - 8 < 40

Add 8 to both sides

5x - 8  + 8   <  40  +  8

5x    <   48

Divide both sides by 5

$\frac{5x}{5} < \frac{48}{5} \\\\x < \frac{48}{5}$

The solution for the inequality 12x+7<−11 is x < -3/2

The solution for the inequality  5x−8>40 is x < 48/5

Learn more here: https://brainly.com/question/17448505