1/6(z-4)=1/3(z-2)

Solution for 1/6(z-4)=1/3(z-2) equation:

1/6(z-4)=1/3(z-2)

We move all terms to the left:

1/6(z-4)-(1/3(z-2))=0


Domain of the equation: 6(z-4)!=0

z∈R



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

z∈R


We calculate fractions


(3zz/(6(z-4)*3(z-2)))+(-6zz/(6(z-4)*3(z-2)))=0


We calculate terms in parentheses: +(3zz/(6(z-4)*3(z-2))), so:

3zz/(6(z-4)*3(z-2))

We multiply all the terms by the denominator


3zz

Back to the equation:

+(3zz)



We calculate terms in parentheses: +(-6zz/(6(z-4)*3(z-2))), so:

-6zz/(6(z-4)*3(z-2))

We multiply all the terms by the denominator


-6zz

Back to the equation:

+(-6zz)


We get rid of parentheses

3zz-6zz=0