8y+4(6+y)=3(y-4)+10y

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Solution for 8y+4(6+y)=3(y-4)+10y equation:



8y+4(6+y)=3(y-4)+10y
We move all terms to the left:
8y+4(6+y)-(3(y-4)+10y)=0
We add all the numbers together, and all the variables
8y+4(y+6)-(3(y-4)+10y)=0
We multiply parentheses
8y+4y-(3(y-4)+10y)+24=0
We calculate terms in parentheses: -(3(y-4)+10y), so:
3(y-4)+10y
We add all the numbers together, and all the variables
10y+3(y-4)
We multiply parentheses
10y+3y-12
We add all the numbers together, and all the variables
13y-12
Back to the equation:
-(13y-12)
We add all the numbers together, and all the variables
12y-(13y-12)+24=0
We get rid of parentheses
12y-13y+12+24=0
We add all the numbers together, and all the variables
-1y+36=0
We move all terms containing y to the left, all other terms to the right
-y=-36
y=-36/-1
y=+36

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