1/6x+1/3=1/9x+2

Solution for 1/6x+1/3=1/9x+2 equation:

1/6x+1/3=1/9x+2

We move all terms to the left:

1/6x+1/3-(1/9x+2)=0


Domain of the equation: 6x!=0

x!=0/6

x!=0

x∈R



Domain of the equation: 9x+2)!=0

x∈R


We get rid of parentheses

1/6x-1/9x-2+1/3=0

We calculate fractions


486x^2/486x^2+81x/486x^2+(-54x)/486x^2-2=0

Fractions to decimals

81x/486x^2+(-54x)/486x^2-2+1=0

We multiply all the terms by the denominator


81x+(-54x)-2*486x^2+1*486x^2=0

Wy multiply elements

-972x^2+486x^2+81x+(-54x)=0

We get rid of parentheses

-972x^2+486x^2+81x-54x=0

We add all the numbers together, and all the variables

-486x^2+27x=0

a = -486; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·(-486)·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*-486}=\frac{-54}{-972}
=1/18
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*-486}=\frac{0}{-972}
=0
$