Question

solve the following inequalities:
a. $3 - 5x \geq -22$
b. $3 < \frac{x}{6} + 5$
c. $3x + 6 \leq 12$

Explanation:

Part a:

Step1: Subtract 3 from both sides

$3 - 5x - 3 \geq -22 - 3$
$-5x \geq -25$

Step2: Divide by -5, flip inequality

$\frac{-5x}{-5} \leq \frac{-25}{-5}$
$x \leq 5$

Part b:

Step1: Subtract 5 from both sides

$3 - 5 < \frac{x}{6} + 5 - 5$
$-2 < \frac{x}{6}$

Step2: Multiply both sides by 6

$-2 \times 6 < \frac{x}{6} \times 6$
$x > -12$

Part c:

Step1: Subtract 6 from both sides

$3x + 6 - 6 \leq 12 - 6$
$3x \leq 6$

Step2: Divide both sides by 3

$\frac{3x}{3} \leq \frac{6}{3}$
$x \leq 2$

Answer:

a. $x \leq 5$
b. $x > -12$
c. $x \leq 2$