X+(10+x)+(4x-8)+(2x+1)+(3x-3)=132

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Solution for X+(10+x)+(4x-8)+(2x+1)+(3x-3)=132 equation: 



X+(10+X)+(4X-8)+(2X+1)+(3X-3)=132

We move all terms to the left:

X+(10+X)+(4X-8)+(2X+1)+(3X-3)-(132)=0

We add all the numbers together, and all the variables

X+(X+10)+(4X-8)+(2X+1)+(3X-3)-132=0

We get rid of parentheses

X+X+4X+2X+3X+10-8+1-3-132=0

We add all the numbers together, and all the variables

11X-132=0

We move all terms containing X to the left, all other terms to the right

11X=132

X=132/11

X=12

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