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Q.
Solution of $\frac{x-7}{x+3} > 2$ is
Linear Inequalities
Solution:
We have, $\frac{x-7}{x+3} > 2$
or $\quad\frac{x-7}{x+3} - 2 > 0$ or $\frac{x+13}{x+3} < 0$
$\Rightarrow \quad$ {$x + 13 > 0$ and $x + 3 < 0$} or {$x+ 13 < 0$ and $x + 3 > 0$}
$\Rightarrow \quad$ {$ x >-13$ and $x < - 3$} or {$x < - 13$ and $x > - 3$}
it is not possible
$\Rightarrow \quad x \in \left(-13,\,-3\right)$