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

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