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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