(7/x-4)+(8/3x-12)=6

Solution for (7/x-4)+(8/3x-12)=6 equation:

(7/x-4)+(8/3x-12)=6

We move all terms to the left:

(7/x-4)+(8/3x-12)-(6)=0


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

x∈R



Domain of the equation: 3x-12)!=0

x∈R


We get rid of parentheses

7/x+8/3x-4-12-6=0

We calculate fractions


21x/3x^2+8x/3x^2-4-12-6=0

We add all the numbers together, and all the variables

21x/3x^2+8x/3x^2-22=0

We multiply all the terms by the denominator


21x+8x-22*3x^2=0

We add all the numbers together, and all the variables

29x-22*3x^2=0

Wy multiply elements

-66x^2+29x=0

a = -66; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·(-66)·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*-66}=\frac{-58}{-132}
=29/66
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*-66}=\frac{0}{-132}
=0
$