How do you solve #-4(x-1)<12#?

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Answer 1 · 2016-07-10T17:38:51

#x > -2#

Remember when dividing/multiplying a negative number when dealing with inequalities, the symbol must be switched.

#-4(x-1)<12#

Divide out the #-4#.

#(x-1) > -3#

Use additive inverse to get #x# by itself (add #1#).

#x > -2#

We found that #x# is a solution when it accounts for all values greater than #-2#.

Answer 2 · 2016-07-10T17:40:58

#x > -2#

#-4*(x-1) < 12#

# 4*x-4 > -12#

#4*x > -8#

#x > -2#

graph{x> -2 [-10, 10, -5, 5]}