7y+4=3(y-1)+35

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Solution for 7y+4=3(y-1)+35 equation: 



7y+4=3(y-1)+35

We move all terms to the left:

7y+4-(3(y-1)+35)=0



We calculate terms in parentheses: -(3(y-1)+35), so:

3(y-1)+35

We multiply parentheses

3y-3+35

We add all the numbers together, and all the variables

3y+32

Back to the equation:

-(3y+32)



We get rid of parentheses

7y-3y-32+4=0

We add all the numbers together, and all the variables

4y-28=0

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

4y=28

y=28/4

y=7

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