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.