Solve $\frac{1}{(x - 2)(x-3)} + \frac{1}{(x -1)(x -2)} = \frac{2}{3}$, $x$ is not equal to $1$, $2$, $3$.

Question

Solve the following equation: $$\frac{1}{(x - 2)(x-3)} + \frac{1}{(x -1)(x -2)} = \frac{2}{3},$$

where $x \ne 1,2,3$

Taking $(x - 2)$ as a common factor I got the quadratic equation $x^2 - 4x = 0$ and got values $x = 0$ and $x = 4$ as solutions.

But if I do not take $(x - 2)$ as a common factor and solve the whole equation then I got the equation $x^4 - 8x^3 + 20x^2 - 16x = 0$ and for this equation I got four values of $x = 0, 2, 2, 4$ as solutions. I just want to ask how could we get $x = 2$ as one solution as it was given in the problem that $x$ is not defined at $x = 2$.

Answer 1

Well if we multiply both sides by $(x-1)(x-2)(x-3)$ we get

$(x-1) +(x-3)=\frac 23 (x-1)(x-2)(x-3)$

$2x -4 = \frac 23 (x-1)(x-2)(x-3)$

$2(x-2) =\frac 23 (x-1)(x-2)(x-3)$. Divide both sides by $x-2$ and we get

$2 = \frac 23(x-1)(x-3)$ which is a basic quadratic equation.

So solve it...

.....

Multiply both side by $\frac 32$ and ...

$3 = (x-1)(x-3) = x^2-4x+3$

$0 = x^2 - 4x = x(x-4)$ so $x=0$ or $x =4$.

Answer 2

$$\frac{1}{(x-2)(x-3)}+\frac{1}{(x-1)(x-2)} = \frac{2}{3}$$ By looking at the equation, we'll quickly conclude that $x$ is not defined at $1,2,3$ $$\frac{1}{x-3}+\frac{1}{x-1} = \frac{2(x-2)}{3}$$ But if we arrange the equation by grouping the $(x-2)$, we'll see that $x$ can be defined at $2$ but not $1,3$ $$\frac{1}{x-3}+\frac{1}{x-1} = \frac{2(x-2)}{3}$$ Solving the equation $$(x-1)+(x-3) = \frac{2}{3}\cdot(x-1)\cdot(x-2)\cdot(x-3)$$ $$2(x-2) = \frac{2}{3}\cdot(x-1)(x-2)(x-3)$$ $$\frac{3}{2}2(x-2) = (x-1)(x-2)(x-3)$$ $$3(x-2) = (x-1)(x-2)(x-3)$$ See $x=2$ is a solution $$3x-6 = x^3-6x^2+11x-6$$ $$x^3-6x^2+8x = 0$$ $$x(x-2)(x-4)$$ So the equation was better written as $$\frac{1}{x-3}+\frac{1}{x-1} = \frac{2(x-2)}{3}$$