3(3x+2)+68=13x+3-4x+71

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Solution for 3(3x+2)+68=13x+3-4x+71 equation:



3(3x+2)+68=13x+3-4x+71
We move all terms to the left:
3(3x+2)+68-(13x+3-4x+71)=0
We add all the numbers together, and all the variables
3(3x+2)-(9x+74)+68=0
We multiply parentheses
9x-(9x+74)+6+68=0
We get rid of parentheses
9x-9x-74+6+68=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0

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