2(3y-4)-5=7y+25-(4y+8)

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Solution for 2(3y-4)-5=7y+25-(4y+8) equation: 



2(3y-4)-5=7y+25-(4y+8)

We move all terms to the left:

2(3y-4)-5-(7y+25-(4y+8))=0

We multiply parentheses

6y-(7y+25-(4y+8))-8-5=0



We calculate terms in parentheses: -(7y+25-(4y+8)), so:

7y+25-(4y+8)

determiningTheFunctionDomain
7y-(4y+8)+25

We get rid of parentheses

7y-4y-8+25

We add all the numbers together, and all the variables

3y+17

Back to the equation:

-(3y+17)



We add all the numbers together, and all the variables

6y-(3y+17)-13=0

We get rid of parentheses

6y-3y-17-13=0

We add all the numbers together, and all the variables

3y-30=0

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

3y=30

y=30/3

y=10

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