((3y+3)+(4y-1))=130

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Solution for ((3y+3)+(4y-1))=130 equation: 



((3y+3)+(4y-1))=130

We move all terms to the left:

((3y+3)+(4y-1))-(130)=0



We calculate terms in parentheses: +((3y+3)+(4y-1)), so:

(3y+3)+(4y-1)

We get rid of parentheses

3y+4y+3-1

We add all the numbers together, and all the variables

7y+2

Back to the equation:

+(7y+2)



We get rid of parentheses

7y+2-130=0

We add all the numbers together, and all the variables

7y-128=0

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

7y=128

y=128/7

y=18+2/7

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