Question

$\frac{(x-3)}{(x+3)}-\frac{(x+3)}{(x-3)}=6\frac{6}{7},(x\ne -3,3)$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{(x-3)}{(x+3)}-\frac{(x+3)}{(x-3)}=6\frac{6}{7} \\ \Rightarrow \frac{(x-3)(x-3)-(x+3)(x+3)}{(x+3)(x-3)}=\frac{48}{7} \\ \Rightarrow \frac{(x^2-6x+9)-(x^2+6x+9)}{(x^2-9)}=\frac{48}{7} \\ \Rightarrow \frac{-12x}{(x^2-9)}=\frac{48}{7} \\ \Rightarrow \frac{-x}{(x^2-9)}=\frac{4}{7}\left[\mathrm{On}\text{dividing both sides by 12}\right] \\ \Rightarrow (4x^2-36)=-7x\left[\mathrm{On}\text{cross multiplying}\right] \\ \Rightarrow 4x^2+7x-36=0 \\ \Rightarrow 4x^2+16x-9x-36=0 \\ \Rightarrow 4x(x+4)-9(x+4)=0 \\ \Rightarrow (4x-9)(x+4)=0 \\ \Rightarrow 4x-9=0\text{or}x+4=0 \\ \Rightarrow x=\frac{9}{4}\text{or}x=-4 \\ \mathrm{H}\text{ence, the roots of the equation are}\frac{9}{4}\text{and}-\text{4.} \end{gathered}$$