6y+2(y-3)=5(y+1)-4

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Solution for 6y+2(y-3)=5(y+1)-4 equation: 



6y+2(y-3)=5(y+1)-4

We move all terms to the left:

6y+2(y-3)-(5(y+1)-4)=0

We multiply parentheses

6y+2y-(5(y+1)-4)-6=0



We calculate terms in parentheses: -(5(y+1)-4), so:

5(y+1)-4

We multiply parentheses

5y+5-4

We add all the numbers together, and all the variables

5y+1

Back to the equation:

-(5y+1)



We add all the numbers together, and all the variables

8y-(5y+1)-6=0

We get rid of parentheses

8y-5y-1-6=0

We add all the numbers together, and all the variables

3y-7=0

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

3y=7

y=7/3

y=2+1/3

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