Question

Solve the following inequality. 21<_-3(x - 4) < 30 A. 28<_x < 37 B. -3 <_ x < 6 C. -6 < x <_ -3 D. -10 &lt; x <_ -3

Answer 1

Answer:

Option C $-6$

Step-by-step explanation:

we have

$21\leq -3(x-4)$

The compound inequality can be divided into two inequality

$21\leq -3(x-4)$ -----> inequality A

$-3(x-4)$ ----> inequality B

Solve inequality A

$21\leq -3x+12$

$9\leq -3x$

Divide by -3 both sides

when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$-3\geq x$

Rewrite

$x\leq -3$

The solution of the inequality A is the interval (-∞,-3]

Solve the inequality B

$-3x+12$

$-3x$

Divide by -3 both sides

when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$x>-6$

The solution of the inequality B is the interval [-6,∞)

The solution of the compound inequality is

[-6,∞) ∩ (-∞,-3]=(-6,-3]

$-6$