Inequalities with Fractions

When working with inequalities where the unknown appears in a fraction, you must be extra careful. Especially because the denominator must never be 0, and because the fraction can change sign depending on the values of the numerator and denominator.

Example 1

We look at the inequality:

$$ \large \frac{1}{x} > 0 $$

The denominator must not be 0, so $ \large x \neq 0 $.

For $ \large x = -1 $:

$ \large \frac{1}{-1} = -1 \quad$ Not greater than 0.

For $ \large x = 1 $:

$ \large \frac{1}{1} = 1 \quad$ Greater than 0 = the interval applies.

The solution is therefore:

$$ \large x > 0 $$

Example 2

Now we look at:

$$ \large \frac{x}{x-1} < 0 $$

The denominator must not be 0, so $ \large x \neq 1 $.

For $ \large x = -1 $:

$ \large \frac{-1}{-1-1} = \frac{-1}{-2} = \tfrac{1}{2} \quad$ Not less than 0.

For $ \large x = 0 $:

$ \large \frac{0}{0-1} = 0 $ Not less than 0.

For $ \large x = 2 $:

$ \large \frac{2}{2-1} = \frac{2}{1} = 2 \quad$ Not less than 0.

For $ \large x = \tfrac{1}{2} $:

$ \large \frac{\tfrac{1}{2}}{\tfrac{1}{2}-1} = \frac{0.5}{-0.5} = -1 \quad$ Less than 0 = the interval applies.

The solution is therefore:

$$ \large 0 < x < 1 $$

Example 3

A slightly more complex inequality:

$$ \large \frac{x+1}{x-2} \geq 0 $$

The denominator must not be 0, so $ \large x \neq 2 $.

The numerator gives the root $ \large x = -1 $. The denominator gives the boundary $ \large x = 2 $. The number line is divided into three intervals, which we examine:

For $ \large x = -2 $:

$ \large \frac{-2+1}{-2-2} = \frac{-1}{-4} = \tfrac{1}{4} \geq 0 \quad$ True.

For $ \large x = 0 $:

$ \large \frac{0+1}{0-2} = \frac{1}{-2} = -\tfrac{1}{2} \quad$ Not greater than or equal to 0.

For $ \large x = 3 $:

$ \large \frac{3+1}{3-2} = \frac{4}{1} = 4 \geq 0 \quad$ True.

The solution is therefore:

$$ \large x \leq -1 \quad \text{or} \quad x > 2 $$

Summary

For inequalities with fractions, always remember that the denominator must not be 0.
The fraction changes sign when either the numerator or denominator changes sign.
First find roots and boundary points, then test the intervals into which the number line is divided.
The solution can be one or several intervals, depending on the signs.