Inequation: quadratic difference equations Given:

$$\frac{(x - 3)}{(x-4)} &gt; \frac{(x + 4)}{(x + 3)}$$

Step 1:

$$(x + 3)(x - 3) &gt; (x + 4)(x - 4)$$

Step2 : Solving step 1:

$$x^2 - 3^2 &gt; x^2 - 4^2$$

*Step 3: 

$ 0 &gt; -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?
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 $\frac {a}{a^2}=\frac 1a\gt 0$
The thing is that the sign in the inequality is not an equal sign but a less than sign which requires that the left side be greater than the right side. If you add 9 to -16,  you will get -7. The answer to the inequality is 0>-7 because a negative number is less than 0.