Question

Solve for $x$: $\frac{(x-2)}{(x-3)}+\frac{(x-4)}{\left(x-5\right)}=\frac{10}{3},x\ne 3,5$

Answer 1

$\frac{(x-2)}{(x-3)}+\frac{(x-4)}{\left(x-5\right)}=\frac{10}{3}$

$\frac{(x-2)\left(x-5\right)+\left(x-4\right)\left(x-3\right)}{(x-3)\left(x-5\right)}=\frac{10}{3}$

$\frac{x^2-5x-2x+10+x^2-3x-4x+12}{x^2-5x-3x+15}=\frac{10}{3}$

$\frac{2x^2-14x+22}{x^2-8x+15}=\frac{10}{3}$

$3\left(2x^2-14x+22\right)=10\left(x^2-8x+15\right)$

$6x^2-42x+66=10x^2-80x+150$

$10x^2-80x+150-6x^2+42x-66=0$

$4x^2-38x+84=0$

$2x^2-19x+42=0$(dividing the whole equation by $2$)

Using the factorization method,

$2x^2-12x-7x+42=0$

$2x(x-6)-7(x-6)=0$

$\left(2x-7\right)(x-6)=0$

$$\begin{gathered} \left(2x-7\right)=0or\left(x-6\right)=0 \\ x=\frac{7}{2}orx=6 \end{gathered}$$

Hence, the required value is $x=\frac{7}{2}orx=6$