Question

Solve for x :
$\frac{x-3}{x-4}+\frac{x-5}{x-6}=\frac{10}{3};x\ne 4,6$

Answer 1

By cross multiplying,
$\frac{(x-3)(x-6)+(x-5)(x-4)}{(x-4)(x-6)}=\frac{10}{3}$

$\Rightarrow \frac{x^2-9x+18x+x^2-9x+20}{x^2-10x+24}=\frac{10}{3}$

$\Rightarrow \frac{2x^2-18x+38}{x^2-10x+24}=\frac{10}{3}$

$\Rightarrow 2(x^2-9x+19)=\frac{10}{3}[x^2-10x+24]$

$\Rightarrow x^2-9x+19=\frac{5}{3}[x^2-10x+24]$

$\Rightarrow 3x^2-27x+57=5x^2-50x+120$

$\Rightarrow 2x^2-23x+63=0$

$\Rightarrow (x-7)(2x-9)=0$

$\Rightarrow x=7or\frac{9}{2}$