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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-5(-1y)-(-4(+2y)+5+12y)+3=0

We multiply parentheses

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



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

-4(+2y)+5+12y

determiningTheFunctionDomain
-4(+2y)+12y+5

We add all the numbers together, and all the variables

12y-4(+2y)+5

We multiply parentheses

12y-8y+5

We add all the numbers together, and all the variables

4y+5

Back to the equation:

-(4y+5)



We get rid of parentheses

5y-4y-5+3=0

We add all the numbers together, and all the variables

y-2=0

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

y=2

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