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

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

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

We move all terms to the left:

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


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

We move all terms containing x to the left, all other terms to the right

x!=3

x∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


(x-4)+(x-2)-x*(x-3)+3*(x-3)=0

We multiply parentheses

-x^2+(x-4)+(x-2)+3x+3x-9=0

We get rid of parentheses

-x^2+x+x+3x+3x-4-2-9=0

We add all the numbers together, and all the variables

-1x^2+8x-15=0

a = -1; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·(-1)·(-15)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-1}=\frac{-6}{-2}
=+3
$