Inequalities involving fractions

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:

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

2. 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:

   $$\frac{x + 4}{x - 3} < 2$$