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

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



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

We move all terms to the left:

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

We multiply parentheses

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



We calculate terms in parentheses: -(5(y-1)+2y+3), so:

5(y-1)+2y+3

We add all the numbers together, and all the variables

2y+5(y-1)+3

We multiply parentheses

2y+5y-5+3

We add all the numbers together, and all the variables

7y-2

Back to the equation:

-(7y-2)



We add all the numbers together, and all the variables

5y-(7y-2)-8=0

We get rid of parentheses

5y-7y+2-8=0

We add all the numbers together, and all the variables

-2y-6=0

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

-2y=6

y=6/-2

y=-3

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