(1)/(3)(z+1)=(1)/(6)(3z-5)

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Solution for (1)/(3)(z+1)=(1)/(6)(3z-5) equation: 



(1)/(3)(z+1)=(1)/(6)(3z-5)

We move all terms to the left:

(1)/(3)(z+1)-((1)/(6)(3z-5))=0



Domain of the equation: 3(z+1)!=0

z∈R





Domain of the equation: 6(3z-5))!=0

z∈R



We calculate fractions


(6z3/(3(z+1)*6(3z-5)))+(-3zz/(3(z+1)*6(3z-5)))=0



We calculate terms in parentheses: +(6z3/(3(z+1)*6(3z-5))), so:

6z3/(3(z+1)*6(3z-5))

We multiply all the terms by the denominator


6z3

We add all the numbers together, and all the variables

6z^3

We do not support ezpression: z^3

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