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?
 A: 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 $\frac {a}{a^2}=\frac 1a\gt 0$
A: $x^2-9>x^2-16$ is obvious to solve. $16>9$ so $x^2-9>x^2-16$.
A: What you did in step 1 amounts to multiply both sides by $(x-4)(x+3)$. Unfortunately, you have to reverse the inequation if this expression is negative, and leave it as is if it is positive. And as you don't know the sign of this product…
You can simplify solving this inequation writing both sides in canonical form:
$$\frac{x-3}{x-4}=1+\frac1{x-4}>\frac{x+4}{x+3}=1+\frac1{x+3}\iff\frac1{x-4}>\frac1{x+3}$$
Multiplying both members by $(x-4)^2(x-3)^2$ (which is positive on the domain of the inequation), we obtain:
$$(x-4)(x+3)^2> (x-4)^2(x+3)\iff7(x-4)(x+3)>0\iff\begin{cases}x<-3\\\text{or}\\x>4\end{cases}$$
A: I think people ought to draw graphs of things as long as these are simple enough to do so. In this case I drew the two hyperbolas indicated in the question. As in Mark Bennet's answer, we see that the red curve is higher between the two vertical asymptotes, that is $-3 < x < 4.$ However, for $x > 4$ and again for $x < -3,$ it appears the green curve is very slightly higher.
 
Also, we see how the picture suggests that the red and green curves are translates of each other. This turns out to be true: if 
$$ f(x) = \frac{x-3}{x-4}, $$ then
$$ f(x + 7) = \frac{x+ 4}{x+3}. $$
A: Where could the inequality reverse? Candidates are the two singularities (i) $x=-3$ and $x=4$ and (ii) the places where $(x+3)(x-3)=(x+4)(x-4)$. By your calculation there are no such places.
Now use three test points, one to the left of $x=-3$, one between $x=-3$ and $x=4$, and one to the right of $x=4$.
For example, take the test point $-4$ to the left of $-3$. It is clear that the left-hand side is greater than the right-hand side, so the inequality holds in the interval $(-\infty,-3)$. Now use two other test points to finish.
A: 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.
