2(40-5y)=10y+(1-y)

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Solution for 2(40-5y)=10y+(1-y) equation:



2(40-5y)=10y+(1-y)
We move all terms to the left:
2(40-5y)-(10y+(1-y))=0
We add all the numbers together, and all the variables
2(-5y+40)-(10y+(-1y+1))=0
We multiply parentheses
-10y-(10y+(-1y+1))+80=0
We calculate terms in parentheses: -(10y+(-1y+1)), so:
10y+(-1y+1)
We get rid of parentheses
10y-1y+1
We add all the numbers together, and all the variables
9y+1
Back to the equation:
-(9y+1)
We get rid of parentheses
-10y-9y-1+80=0
We add all the numbers together, and all the variables
-19y+79=0
We move all terms containing y to the left, all other terms to the right
-19y=-79
y=-79/-19
y=4+3/19

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