7(y-4)=3y-32

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Solution for 7(y-4)=3y-32 equation: 



7(y-4)=3y-32

We move all terms to the left:

7(y-4)-(3y-32)=0

We multiply parentheses

7y-(3y-32)-28=0

We get rid of parentheses

7y-3y+32-28=0

We add all the numbers together, and all the variables

4y+4=0

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

4y=-4

y=-4/4

y=-1

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