Question

$\frac{(x+3)}{(x-2)}-\frac{(1-x)}{x}=\frac{17}{4},(x=0,2)$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{(x+3)}{(x-2)}-\frac{(1-x)}{x}=\frac{17}{4} \\ \Rightarrow \frac{x(x+3)-(1-x)(x-2)}{(x-2)x}=\frac{17}{4} \\ \Rightarrow \frac{x^2+3x-(x-2-x^2+2x)}{x^2-2x}=\frac{17}{4} \\ \Rightarrow \frac{x^2+3x+x^2-3x+2}{x^2-2x}=\frac{17}{4} \\ \Rightarrow \frac{2x^2+2}{x^2-2x}=\frac{17}{4} \\ \Rightarrow 8x^2+8=17x^2-34x\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow -9x^2+34x+8=0 \\ \Rightarrow 9x^2-34x-8=0 \\ \Rightarrow 9x^2-36x+2x-8=0 \\ \Rightarrow 9x(x-4)+2(x-4)=0 \\ \Rightarrow (x-4)(9x+2)=0 \\ \Rightarrow x-4=0\text{or 9}x+2=0 \\ \Rightarrow x=4\text{or}x=\frac{-2}{9} \\ \mathrm{H}\text{ence, the roots of the equation are 4 and}\frac{-2}{9}\text{.} \end{gathered}$$