Question

$\frac{1}{(x-2)}+\frac{1}{x}=\frac{8}{(2x+5)},x\ne 0,\frac{-5}{2}$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{(x-2)}+\frac{1}{x}=\frac{8}{(2x+5)} \\ \Rightarrow \frac{x+(x-2)}{x(x-2)}=\frac{8}{(2x+5)} \\ \Rightarrow \frac{(2x-2)}{x(x-2)}=\frac{8}{(2x+5)} \\ \Rightarrow (2x-2)(2x+5)=8x(x-2)\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow 4x^2+6x-10=8x^2-16x \\ \Rightarrow 4x^2-22x+10=0 \\ \Rightarrow 2(2x^2-11x+5)=0 \\ \Rightarrow 2x^2-11x+5=0 \\ \Rightarrow 2x^2-(10+1)x+5=0 \\ \Rightarrow 2x^2-10x-x+5=0 \\ \Rightarrow 2x(x-5)-1(x-5)=0 \\ \Rightarrow (x-5)(2x-1)=0 \\ \Rightarrow x-5=0\text{or 2}x-\text{1 = 0} \\ \Rightarrow x=5\text{or}x=\frac{1}{2} \\ \text{Hence, the roots of the equation are 5 and}\frac{1}{2}\text{.} \end{gathered}$$