x+1/3x-4=3x-2/3x+4

Solution for x+1/3x-4=3x-2/3x+4 equation:

x+1/3x-4=3x-2/3x+4

We move all terms to the left:

x+1/3x-4-(3x-2/3x+4)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 3x+4)!=0

x∈R


We get rid of parentheses

x+1/3x-3x+2/3x-4-4=0

We multiply all the terms by the denominator


x*3x-3x*3x-4*3x-4*3x+1+2=0

We add all the numbers together, and all the variables

x*3x-3x*3x-4*3x-4*3x+3=0

Wy multiply elements

3x^2-9x^2-12x-12x+3=0

We add all the numbers together, and all the variables

-6x^2-24x+3=0

a = -6; b = -24; c = +3;
Δ = b2-4ac
Δ = -242-4·(-6)·3
Δ = 648
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-18\sqrt{2}}{2*-6}=\frac{24-18\sqrt{2}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+18\sqrt{2}}{2*-6}=\frac{24+18\sqrt{2}}{-12}
$