This note is addressed to lilrhino
You asked about understanding the solution to inequalities not TeX.
As I wrote to you, it is all about the factors.
Consider $\displaystyle (x+a)$. That factor is positive for $\displaystyle x>-a$ and is negative for $\displaystyle x<-a$. For $\displaystyle (x+b)$ the factor is positive for $\displaystyle x>-b$ and is negative for $\displaystyle x<-b$.
Now consider $\displaystyle (x+a)(x+b)$, that product is positive if both factors have the same ‘sign’; that product is negative if the factors have different ‘signs’.
This idea can be extended to any number of factors or quotients.
If you want to understand how inequalities work, then you need to understand the fundamental way factors contribute to the ‘signs’ of products and quotients.
In my 2nd question, I came up with 3 intervals to test:This is an inequality that's just one variable (not sure how to proceed):
$\displaystyle \frac{x+1}{2x-3}$ ≥ 0
$\displaystyle (-\infty{, -1})$
$\displaystyle (-1, \frac{3}{2})$
$\displaystyle (\frac{3}{2}, \infty)$
But after testing (-4, 0, 3) I found that the factors were only negative in:
$\displaystyle (-\infty{, -1})$
$\displaystyle (-1, \frac{3}{2})$
So, is this my correct solution set?
Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.
Look at the actual problem:
$\displaystyle \frac{x + 1}{2x - 3} \geq 0$
We wish to test if $\displaystyle \frac{x + 1}{2x - 3}$ is positive, not just the factors.
So test the expression on each interval:
$\displaystyle ( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0$ (Check!)
$\displaystyle \left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0$ (No!)
$\displaystyle \left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0$ (Check!)
So the solution set is $\displaystyle ( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right ) $. (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)
-Dan
Here is another way.
Either the numerator is non-negative and the denominator is postive.
Or,
EIther the denoimator is non-positive and the denoimator is negative.
Thus,
$\displaystyle x+1 \geq 0 \mbox{ and } 2x - 3>0$
Or,
$\displaystyle x+1\leq 0 \mbox{ and }2x- 3<0$
The rest is trivial.