2/3(z-9)=1/6(z+18)

Solution for 2/3(z-9)=1/6(z+18) equation:

2/3(z-9)=1/6(z+18)

We move all terms to the left:

2/3(z-9)-(1/6(z+18))=0


Domain of the equation: 3(z-9)!=0

z∈R



Domain of the equation: 6(z+18))!=0

z∈R


We calculate fractions


(12zz/(3(z-9)*6(z+18)))+(-3zz/(3(z-9)*6(z+18)))=0


We calculate terms in parentheses: +(12zz/(3(z-9)*6(z+18))), so:

12zz/(3(z-9)*6(z+18))

We multiply all the terms by the denominator


12zz

Back to the equation:

+(12zz)



We calculate terms in parentheses: +(-3zz/(3(z-9)*6(z+18))), so:

-3zz/(3(z-9)*6(z+18))

We multiply all the terms by the denominator


-3zz

Back to the equation:

+(-3zz)


We get rid of parentheses

12zz-3zz=0