3(5y+3)+20=5(4y+2)-12

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Solution for 3(5y+3)+20=5(4y+2)-12 equation: 



3(5y+3)+20=5(4y+2)-12

We move all terms to the left:

3(5y+3)+20-(5(4y+2)-12)=0

We multiply parentheses

15y-(5(4y+2)-12)+9+20=0



We calculate terms in parentheses: -(5(4y+2)-12), so:

5(4y+2)-12

We multiply parentheses

20y+10-12

We add all the numbers together, and all the variables

20y-2

Back to the equation:

-(20y-2)



We add all the numbers together, and all the variables

15y-(20y-2)+29=0

We get rid of parentheses

15y-20y+2+29=0

We add all the numbers together, and all the variables

-5y+31=0

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

-5y=-31

y=-31/-5

y=6+1/5

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