2210.32 – Fractional Inequality

If $x$ is a number such that $\dfrac{1}{x}<2$ and $\dfrac{1}{x}>-3$ , then which of the following is true?

$-\dfrac{1}{3} < x < \dfrac{1}{2}$ ,
$-\dfrac{1}{2} < x < 3$ ,
$x > \dfrac{1}{2},$
$x > \dfrac{1}{2}$ or $-\dfrac{1}{3} < x<0$ ,
$x > \dfrac{1}{2}$ or $x < -\dfrac{1}{3}.$

Solution

We have $-3< \dfrac{1}{x}< 2$ . We need to consider two cases, $x$ positive and $x$ negative, because when we multiply both sides of an inequality by a negative number (and we feel in our bones that we’re going to do this), the sense of the inequality is reversed.

If $x$ < 0, then $x < \dfrac{1}{2}$ is always true. From $\dfrac{1}{x}>-3$ we get: $-3x > 1$ so the stronger condition $x < -\dfrac{1}{3}$ holds.

If $x > 0$ , then $-\dfrac{1}{3} < x$ is always true. From $\dfrac{1}{x}<2$ , we get: $1 < 2x$ so the stronger condition $x > \dfrac{1}{2}$ holds.

The two cases, $x$ positive and $x$ negative, lead to the two possibilities listed in (e).