How do you solve the compound inequalities $3 x \geq - 12$ $3x≥-12$ and $8 x \leq 16$ $8x≤16$ ?

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Answer 1 · 2015-05-09T15:25:32

Taking the two inequalities separately
$3 x \geq - 12$ $3x>= -12$
$\to x \geq - 4$ $rarr x>= -4$

$8 x \leq 16$ $8x<=16$
$\to x \leq 2$ $rarr x<=2$

Combining:
$3 x \geq - 12 and 8 x \leq 16$ $3x>=-12" and " 8x<=16$
$\to - 4 \leq x \leq 2$ $rarr -4 <= x <= 2$

Answer 2 · 2015-05-09T15:27:24

(1) $3 x \geq - 12 \to x \geq - 4$ $3x >= - 12 -> x >= -4$

(2) $8 x \leq 16 \to x \leq 2$ $8x <= 16 -> x <= 2$

Compound solution set : $- 4 \leq x \leq 2$ $-4 <= x <= 2$

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

How do you solve the compound inequalities 3x≥-123x≥−123x≥-12 and 8x≤168x≤168x≤16?