How do you solve #2x-7+19=6x-x+12#?

1 Answer
Dec 23, 2016

You first get the #x#'s and the numbers on opposite sides of #=#

Explanation:

Subtract #2x# on both sides:
#cancel(2x)-cancel(2x)-7+19=6x-x-2x+12#
Now subtract #12#:
#-7+19-12=6x-x-2x+cancel12-cancel12#
Do the adding and subtracting:
#0=3x->x=0#

Of course, if you had started by subtracting the #12#, you would gotten:
#2x=5x# which is only true is #x=0#