# How do you solve the compound inequalities 3x≥-12 and 8x≤16?

May 9, 2015

Taking the two inequalities separately
$3 x \ge - 12$
$\rightarrow x \ge - 4$

$8 x \le 16$
$\rightarrow x \le 2$

Combining:
$3 x \ge - 12 \text{ and } 8 x \le 16$
$\rightarrow - 4 \le x \le 2$

May 9, 2015

(1) $3 x \ge - 12 \to x \ge - 4$

(2) $8 x \le 16 \to x \le 2$

Compound solution set : $- 4 \le x \le 2$

The solution set is the close interval: [-4, 2]. The 2 end points are included in the solution set.