31-5+8y-2(y+3)=-3(4y-5)-7(y-1)-9y+10

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Solution for 31-5+8y-2(y+3)=-3(4y-5)-7(y-1)-9y+10 equation: 



31-5+8y-2(y+3)=-3(4y-5)-7(y-1)-9y+10

We move all terms to the left:

31-5+8y-2(y+3)-(-3(4y-5)-7(y-1)-9y+10)=0

We add all the numbers together, and all the variables

8y-2(y+3)-(-3(4y-5)-7(y-1)-9y+10)+26=0

We multiply parentheses

8y-2y-(-3(4y-5)-7(y-1)-9y+10)-6+26=0



We calculate terms in parentheses: -(-3(4y-5)-7(y-1)-9y+10), so:

-3(4y-5)-7(y-1)-9y+10

We add all the numbers together, and all the variables

-9y-3(4y-5)-7(y-1)+10

We multiply parentheses

-9y-12y-7y+15+7+10

We add all the numbers together, and all the variables

-28y+32

Back to the equation:

-(-28y+32)



We add all the numbers together, and all the variables

6y-(-28y+32)+20=0

We get rid of parentheses

6y+28y-32+20=0

We add all the numbers together, and all the variables

34y-12=0

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

34y=12

y=12/34

y=6/17

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