2/3x-3=5/9x+13

Solution for 2/3x-3=5/9x+13 equation:

2/3x-3=5/9x+13

We move all terms to the left:

2/3x-3-(5/9x+13)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x+13)!=0

x∈R


We get rid of parentheses

2/3x-5/9x-13-3=0

We calculate fractions


18x/27x^2+(-15x)/27x^2-13-3=0

We add all the numbers together, and all the variables

18x/27x^2+(-15x)/27x^2-16=0

We multiply all the terms by the denominator


18x+(-15x)-16*27x^2=0

Wy multiply elements

-432x^2+18x+(-15x)=0

We get rid of parentheses

-432x^2+18x-15x=0

We add all the numbers together, and all the variables

-432x^2+3x=0

a = -432; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-432)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-432}=\frac{-6}{-864}
=1/144
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-432}=\frac{0}{-864}
=0
$