1/9x+6=1/3x

Solution for 1/9x+6=1/3x equation:

1/9x+6=1/3x

We move all terms to the left:

1/9x+6-(1/3x)=0


Domain of the equation: 9x!=0

x!=0/9

x!=0

x∈R



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1/9x-(+1/3x)+6=0

We get rid of parentheses

1/9x-1/3x+6=0

We calculate fractions


3x/27x^2+(-9x)/27x^2+6=0

We multiply all the terms by the denominator


3x+(-9x)+6*27x^2=0

Wy multiply elements

162x^2+3x+(-9x)=0

We get rid of parentheses

162x^2+3x-9x=0

We add all the numbers together, and all the variables

162x^2-6x=0

a = 162; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·162·0
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*162}=\frac{0}{324}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*162}=\frac{12}{324}
=1/27
$