How do you solve #7abs(x + 3 ) = 42#?

1 Answer
Alan P.
Jul 22, 2015

#x=3# or #x=-9#

Explanation:

If #7abs(x+3) = 42#
then #abs(x+3) = 6#
#color(white)("XXXX")##color(white)("XXXX")#(after dividing both sides by 7)

If #(x+3) < 0#
then #abs(x+3) = -x-3 = 6#
#color(white)("XXXX")##rarr x = -9#

if #(x+3) >= 0#
then #abs(x+3) = x+3 = 6#
#color(white)("XXXX")##rarr x= 3#

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