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

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

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

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We calculate fractions


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

We multiply all the terms by the denominator


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

Wy multiply elements

-18x^2+(2x^2-2x)+(3x^2+3x)=0

We get rid of parentheses

-18x^2+2x^2+3x^2-2x+3x=0

We add all the numbers together, and all the variables

-13x^2+x=0

a = -13; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-13)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-13}=\frac{-2}{-26}
=1/13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-13}=\frac{0}{-26}
=0
$