--11(y+2)+5(3y-5)=3(y-4)+10

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Solution for --11(y+2)+5(3y-5)=3(y-4)+10 equation: 



--11(y+2)+5(3y-5)=3(y-4)+10

We move all terms to the left:

--11(y+2)+5(3y-5)-(3(y-4)+10)=0

We add all the numbers together, and all the variables

-11(y+2)+5(3y-5)-(3(y-4)+10)=0

We multiply parentheses

-11y+15y-(3(y-4)+10)-22-25=0



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

3(y-4)+10

We multiply parentheses

3y-12+10

We add all the numbers together, and all the variables

3y-2

Back to the equation:

-(3y-2)



We add all the numbers together, and all the variables

4y-(3y-2)-47=0

We get rid of parentheses

4y-3y+2-47=0

We add all the numbers together, and all the variables

y-45=0

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

y=45

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