8(y-4)=5(y+2)-y

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Solution for 8(y-4)=5(y+2)-y equation: 



8(y-4)=5(y+2)-y

We move all terms to the left:

8(y-4)-(5(y+2)-y)=0

We multiply parentheses

8y-(5(y+2)-y)-32=0



We calculate terms in parentheses: -(5(y+2)-y), so:

5(y+2)-y

We add all the numbers together, and all the variables

-1y+5(y+2)

We multiply parentheses

-1y+5y+10

We add all the numbers together, and all the variables

4y+10

Back to the equation:

-(4y+10)



We get rid of parentheses

8y-4y-10-32=0

We add all the numbers together, and all the variables

4y-42=0

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

4y=42

y=42/4

y=10+1/2

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