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2.3: Rational Inequalities

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    Solving rational inequalities involves finding the zeroes of the numerator and denominator, then using these values to investigate solution set regions on the number line.

    Example 2.3.1

    Solve the inequalities and write the solution sets in interval notation:

    1. \(\dfrac{x − 1}{x + 1} ≥ 0\)
    2. \(\dfrac{2x − 3}{x + 1} ≤ 0\)
    3. \(\dfrac{x + 2}{x − 2} ≥ 0\)
    Solution
    1. \(\begin{array} &&\dfrac{x − 1}{x + 1} ≥ 0 &\text{Example problem} \\ &\dfrac{x − 1}{x + 1} ≥ 0 &\text{The quotient must be greater than or equal to \(0\).} \\ &x − 1 = 0,\; x = 1 &\text{Find the zeroes of the numerator} \\ &x + 1 = 0, x = \;−1 &\text{Find the zeroes of the denominator} \end{array}\)
    clipboard_ec11760d50b919ddd83bc64d27535e4b7.png
    The zeroes divide the number line into \(3\) regions, \(x < −1\), \(−1 < x < 1\), \(x > 1\)

    \(\begin{array} &&\text{For } x < −1, \text{ choose } x = −2. \;\;\dfrac{−2 − 1}{−2 + 1} = \dfrac{−3}{−1} = 3 ≥ 0 \\ &\text{Replacing \(-2\) for \(x\) results in the answer \(3\), which is greater than or equal to \(0\). This region \(x < −1\) is included in the solution set.} \\[0.25in] &\text{For } −1 < x < 1, \text{ choose } x = 0.\;\; \dfrac{0 − 1}{0 + 1} = \dfrac{−1}{1} = −1 < 0 \\ &\text{Replacing \(0\) for \(x\) results in the answer \(-1\), which is less than \(0\), not fulfilling the given inequality in the problem.} \\ &\text{This region \(−1 < x < 1\) is excluded from the solution set.} \\[0.25in] &\text{For } x > 1, \text{ choose } x = 2.\;\; \dfrac{2 − 1}{2 + 1} = \dfrac{1}{3} ≥ 0 \\ &\text{Replacing \(2\) for \(x\) results in the answer \(\dfrac{1}{3}\), which is greater than or equal to \(0\). This region \(x > 1\) is included in the solution set.} \\[0.25in] &(−∞, −1) ∪ (1, ∞) \\ &\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)

    1. \(\begin{array} &&\dfrac{2x − 3}{x + 1} ≤ 0 &\text{Example problem} \\ &\dfrac{2x − 3}{x + 1} ≤ 0 &\text{The quotient must be less than or equal to \(0\).} \\ &2x − 3 = 0,\; x = 1.5 &\text{Find the zeroes of the numerator} \\ &x + 1 = 0,\; x = −1 &\text{Find the zeroes of the denominator} \end{array}\)
    clipboard_eb1a3fd2706ba7b03d6acdeb3ad51ec07.png
    The zeroes divide the number line into \(3\) regions, \(x < −1\), \(−1 < x < 1.5\), \(x > 1.5\)

    \(\begin{array} &&\text{For } -1< x, \text{ choose } x = -2. \;\;\dfrac{2(-2) - 3}{-2 + 1} = \dfrac{−7}{-1} = 7 ≥ 0 \\ &\text{Replacing \(-2\) for \(x\) results in the answer \(7\), which is greater than or equal to \(0\). This region \(-1 < x\) is excluded in the solution set.} \end{array}\)

    \(\begin{array} &&\text{For } -1< x < 1.5, \text{ choose } x = 0. \;\;\dfrac{2(0) - 3}{0 + 1} = \dfrac{−3}{1} = -3 ≤ 0 \\ &\text{Replacing \(0\) for \(x\) results in the answer \(-3\), which is less than or equal to \(0\). This region \(-1 < x < 1.5\) is included in the solution set.} \end{array}\)

    \(\begin{array} &&\text{For } x > 1.5, \text{ choose } x =2. \;\;\dfrac{2(2) - 3}{2 + 1} = \dfrac{1}{3} ≥ 0 \\ &\text{Replacing \(2\) for \(x\) results in the answer \(\dfrac{1}{3}\), which is greater than or equal to \(0\). This region \(x > 1.5 \) is excluded in the solution set.} \end{array}\)

    \(\begin{array} &&\text{(-1, 1.5)} \end{array}\)

    \(\begin{array} &&\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)

     

    1. \(\begin{array} &&\dfrac{x + 2}{x − 2} ≥ 0 &\text{Example problem} \\ &\dfrac{x + 2}{x − 2} ≥ 0 &\text{The quotient must be greater than or equal to \(0\).} \\ &x + 2 = 0, \;\;x = −2 &\text{Find the zeroes of the numerator} \\ &x − 2 = 0,\;\; x = 2 &\text{Find the zeroes of the denominator} \end{array}\)
    clipboard_e3374fd2eab55feb8c3dfe2d5ebd8d557.png
    The zeroes divide the number line into \(3\) regions, \(x < −2\), \(−2 < x < 2\), \(x > 2\)

    \(\begin{array} &&\text{For } x < −2, \text{ choose } x = −3. \dfrac{−3 + 2}{−3 − 2} = \dfrac{−1}{−5} = \dfrac{1}{5} ≥ 0 \\ &\text{Replacing \(-3\) for \(x\) results in the answer \(\dfrac{1}{5}\), which is greater than or equal to \(0\). This region \(x < −2\) is included in the solution set.} \\[0.25in] &\text{For } −2 < x < 2, \text{ choose } x = 0. \;\; \dfrac{0 + 2}{0 − 2} = \dfrac{2}{−2} = −1 < 0 \\ &\text{Replacing \(0\) for \(x\) results in the answer \(-1\), which is less than \(0\), not fulfilling the given inequality in the problem.} \\ &\text{This region \(−2 < x < 2\) is not included in the solution set.} \\[0.25in] &\text{For } x > 2, \text{ choose } x = 3. \;\; \dfrac{3 + 2}{3 − 2} = \dfrac{5}{1} = 5 ≥ 0 \\ &\text{Replacing \(3\) for \(x\) results in the answer \(5\), which is greater than or equal to \(0\). This region \(x > 2\) is included in the solution set.} \\[0.25in] &(−∞, −2) ∪ (2, ∞) \\ &\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)

    Exercise 2.3.1
    1. \(\dfrac{x + 3}{x − 2} ≥ 0\)
    2. \(\dfrac{x − 2}{x − 1} ≤ 0\)
    3. \(\dfrac{2x − 1}{x + 2} ≤ 0\)
    4. \(\dfrac{2x − 3}{x + 1} ≥ 0\)

    This page titled 2.3: Rational Inequalities is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Victoria Dominguez, Cristian Martinez, & Sanaa Saykali (ASCCC Open Educational Resources Initiative) .