Question

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

Answer 1

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

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

$\Rightarrow \frac{x^2-7x+10+x^2-7x+12}{x^2-8x+15}=\frac{10}{3}$

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

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

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

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

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

$\Rightarrow 2x^2-19x+42=0$

$\Rightarrow x=\frac{19\pm \sqrt{{19}^2-4\times 2\times 42}}{2\times 2}$

$\Rightarrow x=\frac{19\pm \sqrt{25}}{4}$

$\Rightarrow x=\frac{19\pm 5}{4}$

$\Rightarrow x=\frac{19+5}{4},\frac{19-5}{4}$

$\therefore x=\frac{24}{4},\frac{14}{4}$

$\therefore x=6,\frac{7}{2}$