(2x+3)+(4x+5)+(3x+1)=108

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Solution for (2x+3)+(4x+5)+(3x+1)=108 equation: 



(2x+3)+(4x+5)+(3x+1)=108

We move all terms to the left:

(2x+3)+(4x+5)+(3x+1)-(108)=0

We get rid of parentheses

2x+4x+3x+3+5+1-108=0

We add all the numbers together, and all the variables

9x-99=0

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

9x=99

x=99/9

x=11

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