2/3z-3/9z=9

Solution for 2/3z-3/9z=9 equation:

2/3z-3/9z=9

We move all terms to the left:

2/3z-3/9z-(9)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 9z!=0

z!=0/9

z!=0

z∈R


We calculate fractions


18z/27z^2+(-9z)/27z^2-9=0

We multiply all the terms by the denominator


18z+(-9z)-9*27z^2=0

Wy multiply elements

-243z^2+18z+(-9z)=0

We get rid of parentheses

-243z^2+18z-9z=0

We add all the numbers together, and all the variables

-243z^2+9z=0

a = -243; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-243)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-243}=\frac{-18}{-486}
=1/27
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-243}=\frac{0}{-486}
=0
$