Y3m=2(m-5)+12

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Solution for Y3m=2(m-5)+12 equation: 



3Y=2(Y-5)+12

We move all terms to the left:

3Y-(2(Y-5)+12)=0



We calculate terms in parentheses: -(2(Y-5)+12), so:

2(Y-5)+12

We multiply parentheses

2Y-10+12

We add all the numbers together, and all the variables

2Y+2

Back to the equation:

-(2Y+2)



We get rid of parentheses

3Y-2Y-2=0

We add all the numbers together, and all the variables

Y-2=0

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

Y=2

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