Question

Solve for ${}^{\prime }{x}^{\prime }:\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3},x\ne 4,3$

Answer 1

$\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}$

$\Rightarrow$ $\frac{2x(x-3)+(2x-5)(x-4)}{(x-4)(x-3)}=\frac{25}{3}$

$\Rightarrow$ $\frac{2x^2-6x+2x^2-8x-5x+20}{x^2-3x-4x+12}=\frac{25}{3}$

$\Rightarrow$ $\frac{4x^2-19x+20}{x^2-7x+12}=\frac{25}{3}$

$\Rightarrow$ $12x^2-57x+60=25x^2-175x+300$

$\Rightarrow$ $13x^2-118x+240=0$

$\Rightarrow$ $13x^2-78x-40x+240=0$

$\Rightarrow$ $13x(x-6)-40(x-6)=0$

$\Rightarrow$ $(13x-40)(x-6)=0$

$\therefore$ $x=\frac{40}{13}$ and $x=6$