2y+3(y-4)=7y-4(y-10)

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Solution for 2y+3(y-4)=7y-4(y-10) equation: 



2y+3(y-4)=7y-4(y-10)

We move all terms to the left:

2y+3(y-4)-(7y-4(y-10))=0

We multiply parentheses

2y+3y-(7y-4(y-10))-12=0



We calculate terms in parentheses: -(7y-4(y-10)), so:

7y-4(y-10)

We multiply parentheses

7y-4y+40

We add all the numbers together, and all the variables

3y+40

Back to the equation:

-(3y+40)



We add all the numbers together, and all the variables

5y-(3y+40)-12=0

We get rid of parentheses

5y-3y-40-12=0

We add all the numbers together, and all the variables

2y-52=0

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

2y=52

y=52/2

y=26

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