116=2(2y)+2(3y+3)

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Solution for 116=2(2y)+2(3y+3) equation: 



116=2(2y)+2(3y+3)

We move all terms to the left:

116-(2(2y)+2(3y+3))=0



We calculate terms in parentheses: -(22y+2(3y+3)), so:

22y+2(3y+3)

We multiply parentheses

22y+6y+6

We add all the numbers together, and all the variables

28y+6

Back to the equation:

-(28y+6)



We get rid of parentheses

-28y-6+116=0

We add all the numbers together, and all the variables

-28y+110=0

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

-28y=-110

y=-110/-28

y=3+13/14

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