4(2x-4)-5=4(x-4)+27

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Solution for 4(2x-4)-5=4(x-4)+27 equation:



4(2x-4)-5=4(x-4)+27
We move all terms to the left:
4(2x-4)-5-(4(x-4)+27)=0
We multiply parentheses
8x-(4(x-4)+27)-16-5=0
We calculate terms in parentheses: -(4(x-4)+27), so:
4(x-4)+27
We multiply parentheses
4x-16+27
We add all the numbers together, and all the variables
4x+11
Back to the equation:
-(4x+11)
We add all the numbers together, and all the variables
8x-(4x+11)-21=0
We get rid of parentheses
8x-4x-11-21=0
We add all the numbers together, and all the variables
4x-32=0
We move all terms containing x to the left, all other terms to the right
4x=32
x=32/4
x=8

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