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

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Answer 1 · 2015-05-09T15:25:32

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

$8x<=16$
$rarr x<=2$

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

Answer 2 · 2015-05-09T15:27:24

(1) $3x >= - 12 -> x >= -4$

(2) $8x <= 16 -> x <= 2$

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

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≥-12 and 8x≤168x≤16?