3(2y-3)(3y+1)=2(9y-4)(y+1)-7

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



3(2y-3)(3y+1)=2(9y-4)(y+1)-7

We move all terms to the left:

3(2y-3)(3y+1)-(2(9y-4)(y+1)-7)=0

We multiply parentheses ..

3(+6y^2+2y-9y-3)-(2(9y-4)(y+1)-7)=0



We calculate terms in parentheses: -(2(9y-4)(y+1)-7), so:

2(9y-4)(y+1)-7

We multiply parentheses ..

2(+9y^2+9y-4y-4)-7

We multiply parentheses

18y^2+18y-8y-8-7

We add all the numbers together, and all the variables

18y^2+10y-15

Back to the equation:

-(18y^2+10y-15)



We multiply parentheses

18y^2+6y-27y-(18y^2+10y-15)-9=0

We get rid of parentheses

18y^2-18y^2+6y-27y-10y+15-9=0

We add all the numbers together, and all the variables

-31y+6=0

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

-31y=-6

y=-6/-31

y=6/31

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