5y+3(y-4)=6(y+1)-3

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Solution for 5y+3(y-4)=6(y+1)-3 equation: 



5y+3(y-4)=6(y+1)-3

We move all terms to the left:

5y+3(y-4)-(6(y+1)-3)=0

We multiply parentheses

5y+3y-(6(y+1)-3)-12=0



We calculate terms in parentheses: -(6(y+1)-3), so:

6(y+1)-3

We multiply parentheses

6y+6-3

We add all the numbers together, and all the variables

6y+3

Back to the equation:

-(6y+3)



We add all the numbers together, and all the variables

8y-(6y+3)-12=0

We get rid of parentheses

8y-6y-3-12=0

We add all the numbers together, and all the variables

2y-15=0

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

2y=15

y=15/2

y=7+1/2

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