8(y-4)-y=128-3(2+5y)

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Solution for 8(y-4)-y=128-3(2+5y) equation: 



8(y-4)-y=128-3(2+5y)

We move all terms to the left:

8(y-4)-y-(128-3(2+5y))=0

We add all the numbers together, and all the variables

8(y-4)-y-(128-3(5y+2))=0

We add all the numbers together, and all the variables

-1y+8(y-4)-(128-3(5y+2))=0

We multiply parentheses

-1y+8y-(128-3(5y+2))-32=0



We calculate terms in parentheses: -(128-3(5y+2)), so:

128-3(5y+2)

determiningTheFunctionDomain
-3(5y+2)+128

We multiply parentheses

-15y-6+128

We add all the numbers together, and all the variables

-15y+122

Back to the equation:

-(-15y+122)



We add all the numbers together, and all the variables

7y-(-15y+122)-32=0

We get rid of parentheses

7y+15y-122-32=0

We add all the numbers together, and all the variables

22y-154=0

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

22y=154

y=154/22

y=7

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