4(y+2)-8+4y=32+2(2y-4)

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Solution for 4(y+2)-8+4y=32+2(2y-4) equation: 



4(y+2)-8+4y=32+2(2y-4)

We move all terms to the left:

4(y+2)-8+4y-(32+2(2y-4))=0

We add all the numbers together, and all the variables

4y+4(y+2)-(32+2(2y-4))-8=0

We multiply parentheses

4y+4y-(32+2(2y-4))+8-8=0



We calculate terms in parentheses: -(32+2(2y-4)), so:

32+2(2y-4)

determiningTheFunctionDomain
2(2y-4)+32

We multiply parentheses

4y-8+32

We add all the numbers together, and all the variables

4y+24

Back to the equation:

-(4y+24)



We add all the numbers together, and all the variables

8y-(4y+24)=0

We get rid of parentheses

8y-4y-24=0

We add all the numbers together, and all the variables

4y-24=0

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

4y=24

y=24/4

y=6

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