2/3x+6=3/9x+23

Solution for 2/3x+6=3/9x+23 equation:

2/3x+6=3/9x+23

We move all terms to the left:

2/3x+6-(3/9x+23)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x+23)!=0

x∈R


We get rid of parentheses

2/3x-3/9x-23+6=0

We calculate fractions


18x/27x^2+(-9x)/27x^2-23+6=0

We add all the numbers together, and all the variables

18x/27x^2+(-9x)/27x^2-17=0

We multiply all the terms by the denominator


18x+(-9x)-17*27x^2=0

Wy multiply elements

-459x^2+18x+(-9x)=0

We get rid of parentheses

-459x^2+18x-9x=0

We add all the numbers together, and all the variables

-459x^2+9x=0

a = -459; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-459)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-459}=\frac{-18}{-918}
=1/51
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-459}=\frac{0}{-918}
=0
$