1/3x-12=7/9x+8

Solution for 1/3x-12=7/9x+8 equation:

1/3x-12=7/9x+8

We move all terms to the left:

1/3x-12-(7/9x+8)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x+8)!=0

x∈R


We get rid of parentheses

1/3x-7/9x-8-12=0

We calculate fractions


9x/27x^2+(-21x)/27x^2-8-12=0

We add all the numbers together, and all the variables

9x/27x^2+(-21x)/27x^2-20=0

We multiply all the terms by the denominator


9x+(-21x)-20*27x^2=0

Wy multiply elements

-540x^2+9x+(-21x)=0

We get rid of parentheses

-540x^2+9x-21x=0

We add all the numbers together, and all the variables

-540x^2-12x=0

a = -540; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-540)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-540}=\frac{0}{-1080}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-540}=\frac{24}{-1080}
=-1/45
$