7/3x+6=1/9x

Solution for 7/3x+6=1/9x equation:

7/3x+6=1/9x

We move all terms to the left:

7/3x+6-(1/9x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 9x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7/3x-(+1/9x)+6=0

We get rid of parentheses

7/3x-1/9x+6=0

We calculate fractions


63x/27x^2+(-3x)/27x^2+6=0

We multiply all the terms by the denominator


63x+(-3x)+6*27x^2=0

Wy multiply elements

162x^2+63x+(-3x)=0

We get rid of parentheses

162x^2+63x-3x=0

We add all the numbers together, and all the variables

162x^2+60x=0

a = 162; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·162·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*162}=\frac{-120}{324}
=-10/27
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*162}=\frac{0}{324}
=0
$