How do you solve #-2(x-4)=2#?

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Answer 1 · 2018-03-25T00:10:39

See a solution process below..

#-2(x - 4) = 2#

First remove the brackets by multiplying it with the values outside the bracket;

#-2x + 8 = 2#

Recall; #- xx - = +#

That is why #-2 xx -4 = 8#

Hence;

#-2x + 8 = 2#

Subtract #8# from both sides..

#-2x + 8 - 8 = 2 - 8#

#-2x = -6#

Divide both sides by #-2#

#(-2x)/(-2) = (-6)/(-2)#

#(cancel(-2)x)/cancel(-2) = (-6)/(-2)#

#x = (-6)/(-2)#

#x = 3#