Question

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

Answer 1

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}$
Given $x\ne 3,5$
By taking L.C.M., we get
$\frac{(x-5)(x-2)+(x-3)(x-4)}{(x-3)(x-5)}=\frac{10}{3}$
On multiplying the brackets, we get
$\therefore \frac{2x^2-14x+22}{x^2-8x+15}=\frac{10}{3}$
On simplifying, we have
$2x^2-19x+42=0$
By factorizing,
$(x-6)(2x-7)=0$
$\therefore x=6,x=\frac{7}{2}$