(12-3y)/6=(y-4)/2

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Solution for (12-3y)/6=(y-4)/2 equation: 



(12-3y)/6=(y-4)/2

We move all terms to the left:

(12-3y)/6-((y-4)/2)=0

We add all the numbers together, and all the variables

(-3y+12)/6-((y-4)/2)=0

We calculate fractions


(-3y)/()+(-((y-4)*6)/()=0



We calculate terms in parentheses: +(-((y-4)*6)/(), so:

-((y-4)*6)/(

We multiply all the terms by the denominator


-((y-4)*6)



We calculate terms in parentheses: -((y-4)*6), so:

(y-4)*6

We multiply parentheses

6y-24

Back to the equation:

-(6y-24)



We get rid of parentheses

-6y+24

Back to the equation:

+(-6y+24)



We get rid of parentheses

(-3y)/()-6y+24=0

We multiply all the terms by the denominator


(-3y)-6y*()+24*()=0

We add all the numbers together, and all the variables

(-3y)-6y*()=0

We get rid of parentheses

-3y-6y*()=0

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