solving inequalities with fractions on both sides Solve this inequality: $\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5}$.
My solution: 
Multiply $(x+5)^2$ on both sides: 
$\frac{(x^2 -2)(x+5)^2}{2} = \frac{x^4 + 10x^3 + 23x^2 -20x -50}{2}$
$(6x^2 -8x - 1)(x+5) = 6x^3 + 22x^2 -41x - 5 $
$\frac{x^4 + 10x^3 + 23x^2 -20x -50}{2} < 6x^3 + 22x^2 -41x - 5$
$0.5x^4 - x^3 - 10.5x^2 + 31x - 20 < 0 $
I don't think doing this way will lead to solving the question....

This is very similar to Meshal's approach, but you can avoid having to factor a quartic. It gives the general idea for a slightly different approach as well.
Move all terms to the same side:
$$\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5} \iff \frac{x^2 -2}{2} - \frac{6x^2 -8x - 1}{x+5} < 0$$
Now put everything on a common denominator and factor the cubic in the numerator:
$$\frac{x^3-7 x^2+14 x-8}{2(x+5)} < 0 \overset{\;\color{green}{(*)}}{\iff}  \frac{(x-1) (x-2) (x-4)}{2(x+5)} < 0$$
You can make a sign table of this fraction which leads immediately to the solution set and/or follow a scheme like Meshal's.
You can also notice that all roots (three in the numerator and one in the denominator) are simple roots (i.e. their multiplicity is 1) and thus sign changes occur in each root.
These roots are in increasing order $-5$, $1$, $2$ and $4$ and clearly the inequality is not satisfied for $x<-5$ (check with a value or reason for $x \to -\infty$); so it is satisfied for $x \in (-5,1) \cup (2,4)$.

To factor in step $\color{green}{(*)}$, you can follow general algorithms or group as follows:
$$\begin{align}
\color{blue}{x^3}\color{red}{-7 x^2+14 x}\color{blue}{-8} 
& = \color{blue}{(x-2)(x^2+2x+4)}\color{red}{-7x(x-2)} \\
& = (x-2)(x^2-5x+4) \\
& = (x-2)(x-1)(x-4) 
\end{align}$$
