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

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



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

We move all terms to the left:

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

We multiply parentheses

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



We calculate terms in parentheses: -(7y-2(y-10)), so:

7y-2(y-10)

We multiply parentheses

7y-2y+20

We add all the numbers together, and all the variables

5y+20

Back to the equation:

-(5y+20)



We add all the numbers together, and all the variables

7y-(5y+20)-12=0

We get rid of parentheses

7y-5y-20-12=0

We add all the numbers together, and all the variables

2y-32=0

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

2y=32

y=32/2

y=16

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