How do you solve #1/2abs(4x – 8) - 7 = 3#?

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Answer 1 · 2015-07-05T12:11:42

Rearrange and split into cases to find: #x = 7# or #x = -3#

First add #7# to both sides to get:

#1/2abs(4x-8)=10#

Divide both sides by #2# to get:

#5 = 1/4abs(4x-8) = abs(x-2)#

There are now two cases:

Case 1 : (x - 2) >= 0

If #(x - 2) >= 0# then #abs(x-2) = (x-2)# and we have:

#5 = x-2#, hence #x = 7#

This is a valid solution since #x - 2 = 7 - 2 > 0# thus satisfying the condition of the case.

Case 2 : (x - 2) < 0

If #(x - 2) < 0# then #abs(x-2) = -(x-2) = 2 - x# and we have:

#5 = 2-x#, hence #x = -3#

This is a valid solution since #x - 2 = -3 -2 < 0# thus satisfying the condition of the case.