(4/x-6)+(3/3x-18)=6

Solution for (4/x-6)+(3/3x-18)=6 equation:

(4/x-6)+(3/3x-18)=6

We move all terms to the left:

(4/x-6)+(3/3x-18)-(6)=0


Domain of the equation: x-6)!=0

x∈R



Domain of the equation: 3x-18)!=0

x∈R


We get rid of parentheses

4/x+3/3x-6-18-6=0

We calculate fractions


12x/3x^2+3x/3x^2-6-18-6=0

We add all the numbers together, and all the variables

12x/3x^2+3x/3x^2-30=0

We multiply all the terms by the denominator


12x+3x-30*3x^2=0

We add all the numbers together, and all the variables

15x-30*3x^2=0

Wy multiply elements

-90x^2+15x=0

a = -90; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-90)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-90}=\frac{-30}{-180}
=1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-90}=\frac{0}{-180}
=0
$