Polarity of $\frac{x + 3}{x - 4}$

Question

This problem is giving me loads of confusion. I just need someone to walk through it because I have the answer and I can't get to it to save my life. I have been on it for days. Please help.

$$\frac{x + 3}{x - 4}\le 0$$

Answer 1

Hint:
Let "$+$" be any positive real number, and "$-$" be any negative real number. It should be clear that:

$$\frac{+}{+} > 0;\ \ \frac{-}{-} > 0$$ $$\frac{+}{-} < 0;\ \ \frac{-}{+} < 0$$

If you can figure out for which values of $x$ has $(x+3)$ positive\negative, and which values of $x$ make $(x-4)$ positive\negative, can you use the above rules to figure out where the fraction is negative?

Answer 2

we have only two cases: a) $$x\geq -3$$ and $$x<4$$ or b) $$x\le -3$$ and $$x>4$$ and this is impossible. Thus we have $$-3\le x<4$$