112=(4y-2)+(3y-2)+(3y+2)

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Solution for 112=(4y-2)+(3y-2)+(3y+2) equation: 



112=(4y-2)+(3y-2)+(3y+2)

We move all terms to the left:

112-((4y-2)+(3y-2)+(3y+2))=0



We calculate terms in parentheses: -((4y-2)+(3y-2)+(3y+2)), so:

(4y-2)+(3y-2)+(3y+2)

We get rid of parentheses

4y+3y+3y-2-2+2

We add all the numbers together, and all the variables

10y-2

Back to the equation:

-(10y-2)



We get rid of parentheses

-10y+2+112=0

We add all the numbers together, and all the variables

-10y+114=0

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

-10y=-114

y=-114/-10

y=11+2/5

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