<s>[INST] 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....
 [/INST] HINT: use that $$\frac{6x^2-8x-1}{x+5}-\frac{x^2-2}{2}=\frac{(4-x)(x-2)(x-1)}{2(5+x)}$$
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