8(y+2)+3=(3y-4)-4

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Solution for 8(y+2)+3=(3y-4)-4 equation: 



8(y+2)+3=(3y-4)-4

We move all terms to the left:

8(y+2)+3-((3y-4)-4)=0

We multiply parentheses

8y-((3y-4)-4)+16+3=0



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

(3y-4)-4

We get rid of parentheses

3y-4-4

We add all the numbers together, and all the variables

3y-8

Back to the equation:

-(3y-8)



We add all the numbers together, and all the variables

8y-(3y-8)+19=0

We get rid of parentheses

8y-3y+8+19=0

We add all the numbers together, and all the variables

5y+27=0

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

5y=-27

y=-27/5

y=-5+2/5

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