Question

Solve for $x$:

$\frac{\left(x-1\right)}{\left(x-2\right)}+\frac{\left(x-3\right)}{\left(x-4\right)}=\frac{10}{3}$

Answer 1

Step 1: Simplify the expression by taking L.C.M

$\frac{\left(x-1\right)}{\left(x-2\right)}+\frac{\left(x-3\right)}{\left(x-4\right)}=\frac{10}{3}$

$\frac{\left(x-1\right)\left(x-4\right)+\left(x-3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{10}{3}$

$\frac{x^2-4x-x+4+x^2-2x-3x+6}{x^2-4x-2x+8}=\frac{10}{3}$

$\frac{2x^2-10x+10}{x^2-6x+8}=\frac{10}{3}$

$3\left(2x^2-10x+10\right)=10\left(x^2-6x+8\right)$

$6x^2-30x+30=10x^2-60x+80$

$10x^2-60x+80-6x^2+30x-30=0$

$4x^2-30x+50=0$

$2x^2-15x+25=0$ (divide the whole equation by $2$)

Step 2: Factorize the quadratic and find $x$

Using the factorization method,

$2x^2-10x-5x+25=0$

$2x(x-5)-5(x-5)=0$

$\left(2x-5\right)(x-5)=0$

$$\begin{gathered} 2x-5=0orx-5=0 \\ x=\frac{5}{2}orx=5 \end{gathered}$$