3y+2y-3(y-4)=32

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Solution for 3y+2y-3(y-4)=32 equation: 



3y+2y-3(y-4)=32

We move all terms to the left:

3y+2y-3(y-4)-(32)=0

We add all the numbers together, and all the variables

5y-3(y-4)-32=0

We multiply parentheses

5y-3y+12-32=0

We add all the numbers together, and all the variables

2y-20=0

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

2y=20

y=20/2

y=10

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