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

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



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

We move all terms to the left:

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



Domain of the equation: 6(3z+2)!=0

z∈R





Domain of the equation: 4z)!=0

z!=0/1

z!=0

z∈R



We get rid of parentheses

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

We calculate fractions


4z/(72z^2+48z)+(18z3/(72z^2+48z)=0



We calculate terms in parentheses: +(18z3/(72z^2+48z), so:

18z3/(72z^2+48z

We multiply all the terms by the denominator


18z3

We add all the numbers together, and all the variables

18z^3

We do not support ezpression: z^3

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