37-5+5y+3(y+5)=-5(2y-7)-8(y+1)+12y+28

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Solution for 37-5+5y+3(y+5)=-5(2y-7)-8(y+1)+12y+28 equation: 



37-5+5y+3(y+5)=-5(2y-7)-8(y+1)+12y+28

We move all terms to the left:

37-5+5y+3(y+5)-(-5(2y-7)-8(y+1)+12y+28)=0

We add all the numbers together, and all the variables

5y+3(y+5)-(-5(2y-7)-8(y+1)+12y+28)+32=0

We multiply parentheses

5y+3y-(-5(2y-7)-8(y+1)+12y+28)+15+32=0



We calculate terms in parentheses: -(-5(2y-7)-8(y+1)+12y+28), so:

-5(2y-7)-8(y+1)+12y+28

We add all the numbers together, and all the variables

12y-5(2y-7)-8(y+1)+28

We multiply parentheses

12y-10y-8y+35-8+28

We add all the numbers together, and all the variables

-6y+55

Back to the equation:

-(-6y+55)



We add all the numbers together, and all the variables

8y-(-6y+55)+47=0

We get rid of parentheses

8y+6y-55+47=0

We add all the numbers together, and all the variables

14y-8=0

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

14y=8

y=8/14

y=4/7

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