1/2(y-4)=1/3(2-y)

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Solution for 1/2(y-4)=1/3(2-y) equation: 



1/2(y-4)=1/3(2-y)

We move all terms to the left:

1/2(y-4)-(1/3(2-y))=0



Domain of the equation: 2(y-4)!=0

y∈R





Domain of the equation: 3(2-y))!=0

y∈R



We add all the numbers together, and all the variables

1/2(y-4)-(1/3(-1y+2))=0

We calculate fractions


(3y(-)/(2(y-4)*3(-1y+2)))+(-2yy/(2(y-4)*3(-1y+2)))=0



We calculate terms in parentheses: +(3y(-)/(2(y-4)*3(-1y+2))), so:

3y(-)/(2(y-4)*3(-1y+2))

We add all the numbers together, and all the variables

3y0/(2(y-4)*3(-1y+2))

We multiply all the terms by the denominator


3y0

We add all the numbers together, and all the variables

3y

Back to the equation:

+(3y)





We calculate terms in parentheses: +(-2yy/(2(y-4)*3(-1y+2))), so:

-2yy/(2(y-4)*3(-1y+2))

We multiply all the terms by the denominator


-2yy

Back to the equation:

+(-2yy)



We get rid of parentheses

3y-2yy=0

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