(2x+10)+(7x-5)+(3x+4)=129

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Solution for (2x+10)+(7x-5)+(3x+4)=129 equation: 



(2x+10)+(7x-5)+(3x+4)=129

We move all terms to the left:

(2x+10)+(7x-5)+(3x+4)-(129)=0

We get rid of parentheses

2x+7x+3x+10-5+4-129=0

We add all the numbers together, and all the variables

12x-120=0

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

12x=120

x=120/12

x=10

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