1/3x-3=1/9x+13

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Solution for 1/3x-3=1/9x+13 equation:



1/3x-3=1/9x+13
We move all terms to the left:
1/3x-3-(1/9x+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
1/3x-1/9x-13-3=0
We calculate fractions
9x/27x^2+(-3x)/27x^2-13-3=0
We add all the numbers together, and all the variables
9x/27x^2+(-3x)/27x^2-16=0
We multiply all the terms by the denominator
9x+(-3x)-16*27x^2=0
Wy multiply elements
-432x^2+9x+(-3x)=0
We get rid of parentheses
-432x^2+9x-3x=0
We add all the numbers together, and all the variables
-432x^2+6x=0
a = -432; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-432)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-432}=\frac{-12}{-864} =1/72 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-432}=\frac{0}{-864} =0 $

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