How do you solve #2[1-3(x+2)]= (-x)#?

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Answer 1 · 2018-05-13T14:46:09

#x = -2#

#2(1−3(x+2))= −x #

Let's start with the left side. First we distribute the #-3#.

#2(1- 3x -6 )= -x #

We combine the integers.

#2(-3x - 5)= -x #

Let's distribute the two.

#-6x -10 = -x #

Now we add #6x# to both sides to move the #-6x# to the right.

#-10 = 5x #

Dividing by five,

#x = -2#

Check:

# 2(1- 3(-2+2))=2= -x quad sqrt#