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

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Solution for 2(y-3)+4(2y+1)=8-5(y-4) equation: 



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

We move all terms to the left:

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

We multiply parentheses

2y+8y-(8-5(y-4))-6+4=0



We calculate terms in parentheses: -(8-5(y-4)), so:

8-5(y-4)

determiningTheFunctionDomain
-5(y-4)+8

We multiply parentheses

-5y+20+8

We add all the numbers together, and all the variables

-5y+28

Back to the equation:

-(-5y+28)



We add all the numbers together, and all the variables

10y-(-5y+28)-2=0

We get rid of parentheses

10y+5y-28-2=0

We add all the numbers together, and all the variables

15y-30=0

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

15y=30

y=30/15

y=2

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