68=2(2x+1)+2(3x+3)

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Solution for 68=2(2x+1)+2(3x+3) equation: 



68=2(2x+1)+2(3x+3)

We move all terms to the left:

68-(2(2x+1)+2(3x+3))=0



We calculate terms in parentheses: -(2(2x+1)+2(3x+3)), so:

2(2x+1)+2(3x+3)

We multiply parentheses

4x+6x+2+6

We add all the numbers together, and all the variables

10x+8

Back to the equation:

-(10x+8)



We get rid of parentheses

-10x-8+68=0

We add all the numbers together, and all the variables

-10x+60=0

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

-10x=-60

x=-60/-10

x=+6

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