11y+0(1-y)=1y+12(1-y)

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Solution for 11y+0(1-y)=1y+12(1-y) equation: 



11y+0(1-y)=1y+12(1-y)

We move all terms to the left:

11y+0(1-y)-(1y+12(1-y))=0

We add all the numbers together, and all the variables

11y+0(-1y+1)-(1y+12(-1y+1))=0



We calculate terms in parentheses: -(1y+12(-1y+1)), so:

1y+12(-1y+1)

We add all the numbers together, and all the variables

y+12(-1y+1)

We multiply parentheses

y-12y+12

We add all the numbers together, and all the variables

-11y+12

Back to the equation:

-(-11y+12)



We get rid of parentheses

11y+0(-1y+1)+11y-12=0

We add all the numbers together, and all the variables

22y+0(-1y+1)-12=0

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

22y+0(-1y+1)=12

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