Question

12x +7 < −11 OR 5x −8 > 40

Answer 1

For this case we must resolve each of the inequalities and find the solution set.

Inequality 1:

$12x + 7$

We subtract 7 from both sides of the inequality:

$12x$

We divide between 12 on both sides of the inequality:

$x$

Thus, the solution is given by all values of x less than$- \frac {3} {2}.$

Inequality 2:

$5x-8> 40$

We add 8 to both sides of the inequality:

$5x> 40 + 8\\5x> 48$

We divide between 5 on both sides of the inequality:

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

Thus, the solution is given by all values of x greater than$\frac {48} {5}.$

The solution set is given by:

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

Answer:

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