3(x+1)+2(x-4)+6(x+1)=2

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Solution for 3(x+1)+2(x-4)+6(x+1)=2 equation: 



3(x+1)+2(x-4)+6(x+1)=2

We move all terms to the left:

3(x+1)+2(x-4)+6(x+1)-(2)=0

We multiply parentheses

3x+2x+6x+3-8+6-2=0

We add all the numbers together, and all the variables

11x-1=0

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

11x=1

x=1/11

x=1/11

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