Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{2x}{(x-4)}+\frac{(2x-5)}{(x-3)}=\frac{25}{3} \\ \Rightarrow \frac{2x(x-3)+(2x-5)(x-4)}{(x-4)(x-3)}=\frac{25}{3} \\ \Rightarrow \frac{2x^2-6x+2x^2-5x-8x+20}{(x^2-7x+12)}=\frac{25}{3} \\ \Rightarrow \frac{4x^2-19x+20}{(x^2-7x+12)}=\frac{25}{3} \\ \Rightarrow 12x^2-57x+60=25x^2-175x+300\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow -13x^2+118x-240=0 \\ \Rightarrow 13x^2-(78+40)x+240=0 \\ \Rightarrow 13x^2-78x-40x+240=0 \\ \Rightarrow 13x(x-6)-40(x-6)=0 \\ \Rightarrow (13x-40)(x-6)=0 \\ \Rightarrow 13x-40=0\text{or}x-6=0 \\ \Rightarrow x=\frac{40}{13}\text{or}x=6 \\ \mathrm{H}\text{ence, the roots of the equation are}\frac{40}{13}\text{and 6.} \end{gathered}$$