7(y+3)+y=3(y+3)-8

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Solution for 7(y+3)+y=3(y+3)-8 equation: 



7(y+3)+y=3(y+3)-8

We move all terms to the left:

7(y+3)+y-(3(y+3)-8)=0

We add all the numbers together, and all the variables

y+7(y+3)-(3(y+3)-8)=0

We multiply parentheses

y+7y-(3(y+3)-8)+21=0



We calculate terms in parentheses: -(3(y+3)-8), so:

3(y+3)-8

We multiply parentheses

3y+9-8

We add all the numbers together, and all the variables

3y+1

Back to the equation:

-(3y+1)



We add all the numbers together, and all the variables

8y-(3y+1)+21=0

We get rid of parentheses

8y-3y-1+21=0

We add all the numbers together, and all the variables

5y+20=0

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

5y=-20

y=-20/5

y=-4

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