<s>[INST] Inequation: quadratic difference equations Given:
$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$
Step 1:
$$(x + 3)(x - 3) > (x + 4)(x - 4)$$
Step2 : Solving step 1:
$$x^2 - 3^2 > x^2 - 4^2$$
*Step 3: 
$ 0 > -16 + 9$ ???
As you see, I can delete the $x^2$, but there is no point in doing that.
What should be the next step?
 [/INST] One way to do this more carefully is to take $$\frac {x-3}{x-4}-\frac {x+4}{x+3}=\frac 7{(x-4)(x+3)}\gt 0$$ and this is clearly true iff $(x-4)(x+3)\gt 0$
To do this more formally, note that from $a\gt 0$ we have $a^2\gt 0$ and we can deduce $\frac {a}{a^2}=\frac 1a\gt 0$
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