(4/x-4)+(4/2x-8)=6

Solution for (4/x-4)+(4/2x-8)=6 equation:

(4/x-4)+(4/2x-8)=6

We move all terms to the left:

(4/x-4)+(4/2x-8)-(6)=0


Domain of the equation: x-4)!=0

x∈R



Domain of the equation: 2x-8)!=0

x∈R


We get rid of parentheses

4/x+4/2x-4-8-6=0

We calculate fractions


8x/2x^2+4x/2x^2-4-8-6=0

We add all the numbers together, and all the variables

8x/2x^2+4x/2x^2-18=0

We multiply all the terms by the denominator


8x+4x-18*2x^2=0

We add all the numbers together, and all the variables

12x-18*2x^2=0

Wy multiply elements

-36x^2+12x=0

a = -36; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-36)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-36}=\frac{-24}{-72}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-36}=\frac{0}{-72}
=0
$