Question

Solve. $\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$.

Answer 1

Solution.

Consider,

$$\begin{gathered} \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7} \\ \Rightarrow \frac{{\left(x-3\right)}^2-{\left(x+3\right)}^2}{\left(x+3\right)\left(x-3\right)}=\frac{48}{7} \\ \Rightarrow \frac{\left(x^2+9-6x\right)-\left(x^2+9+6x\right)}{\left(x^2-9\right)}=\frac{48}{7} \\ \Rightarrow \frac{x^2+9-6x-x^2-9-6x}{\left(x^2-9\right)}=\frac{48}{7} \\ \Rightarrow \frac{-12x}{\left(x^2-9\right)}=\frac{48}{7} \\ \Rightarrow -12x\times 7=48\left(x^2-9\right) \\ \Rightarrow -84x=48x^2-432 \\ \Rightarrow 48x^2+84x-432=0 \\ \Rightarrow 12\left(4x^2+7x-36\right)=0 \\ \Rightarrow 4x^2+7x-36=0 \\ \Rightarrow 4x^2+16x+9x-36=0 \\ \Rightarrow 4x\left(x+4\right)+9\left(x-4\right)=0 \\ \Rightarrow \left(4x+9\right)\left(x+4\right)=0 \\ \Rightarrow \left(4x+9\right)=0\mathrm{or}\left(x+4\right)=0 \\ \therefore x=\frac{-9}{4},x=-4 \end{gathered}$$.

Hence, the solution is $x=-\frac{9}{4},-4$.