Inequalities involving fractions


calculus

algebra
It’s easy to solve inequalities like the following:
$\begin{split} & 2x + 5 \leq 7 \newline & \implies 2x \leq 7  5 \newline & \implies 2x \leq 2 \newline & \implies x \geq 1 \newline \end{split}$We reverse the sign because $x$ coefficient was negative. If you substituted $0$ for $x$ into the equation, it holds. This is standard inequality exercise all of us can do. But while solving some examples, I was met with a sample below:
$\frac{x  7}{x + 3} \gt 2$If you went straight to treat this just like you would with equalities, you’ll end up with the solution $13 \gt x$. Which is wrong. If you substituted $14$, for example, into the equation, you get $1.9$ which is not greater than 2.
How do we solve this then? First, we need to understand that the denominator $(x + 3)$ cannot be $0$. From that understanding, it means our $x$ solution can be on either sides of $3$ on the number line.
$3$ because when $x + 3 = 0$, then $x = 3$. I hope that was clear.
This means we have to solve twice; once for when $x + 3 > 0$ (ie, $x > 3$) and when $x + 3 < 0$ (ie, $x < 3$).
When $(x + 3) > 0$
$\begin{split} & \frac{x  7}{x + 3} \gt 2\newline \implies & x  7 \gt 2(x + 3)\newline \implies & x  7 \gt 2x + 6\newline \implies & 13 \gt x \end{split}$But we already established that this is false. And also $x < 13$ contradicts $x > 3$.
Now let’s solve for when $(x + 3) < 0$. Because $(x + 3)$ is negative (ie, less than), we will flip the inequality. You can think of this as multiplying through by $1$ (or $\frac{1}{1}$).
$\begin{split} & \frac{x  7}{x + 3} \lt 2\newline \implies & x  7 \lt 2(x + 3)\newline \implies & x  7 \lt 2x + 6\newline \implies & 13 \lt x \end{split}$When we test with $12$, the inequality should now hold. And $x > 13$ satisfies $x < 3$. That is if you listed numbers greater than $13$, you’ll find some that fall below $3$. This means $13 < x < 3$. If you test values in that range, you’ll find that they satisfy the inequality.
If you’re lost, you may need to take this slowly with a number of repetitions.
Note that, you wouldn’t always have your answers on only one partition on the number line. They could be on both.
Some patterns I noticed while solving some examples are:

When the solution contradicts the range defined by a direction, discard it; as seen with solving for when $x + 3 > 0$.

If it satifies the range defined by the direction, keep it.

The solution or the range may overlap each other. You should keep the only one that covers both. For example, say in the direction of $x < 3$, you got $x < 10$. Notice here that all values below $10$ also fall under $3$. So we keep just $x < 3$ as our solution.
Try this:
