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Mathematics LibreTexts

4.3: Solve Applications with Rational Equations

  • Page ID
    29060
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    Learning Objectives
    • Solve rational inequalities
    • Solve an inequality with rational functions
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Find the value of \(x-5\) when ⓐ \(x=6\) ⓑ \(x=-3\) ⓒ \(x=5\)
      If you missed this problem, review Example 1.2.16.
    2. Solve: \(8-2 x<12\)
      If you missed this problem, review Example 2.6.13.
    3. Write in interval notation: \(-3 \leq x<5 \)
      If you missed this problem, review Example 2.6.4.

    Solve Rational Inequalities

    We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

    Having just learned to solve rational equations we are now ready to solve rational inequalities. A rational inequality is an inequality that contains a rational expression.

    Rational Inequality

    A rational inequality is an inequality that contains a rational expression.

    Inequalities such as\(\quad \dfrac{3}{2 x}>1, \quad \dfrac{2 x}{x-3}<4, \quad \dfrac{2 x-3}{x-6} \geq x,\quad\) and \(\quad \dfrac{1}{4}-\dfrac{2}{x^{2}} \leq \dfrac{3}{x}\quad \) are rational inequalities as they each contain a rational expression.

    When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

    Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

    When we solve an equation and the result is \(x=3\), we know there is one solution, which is 3.

    When we solve an inequality and the result is \(x>3\), we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a critical point and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

    clipboard_e564fdf3437198939cd8bf44061cfa74e.png

    To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

    Next we determine the critical points to use to divide the number line into intervals. A critical point is a number which make the rational expression zero or undefined.

    We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

    We write the solution in interval notation being careful to determine whether the endpoints are included.

    Example \(\PageIndex{1}\)

    Solve and write the solution in interval notation: \(\dfrac{x-1}{x+3} \geq 0\)

    Solution

    Step 1. Write the inequality as one quotient on the left and zero on the right.

    Our inequality is in this form.\[\dfrac{x-1}{x+3} \geq 0 \nonumber \]

    Step 2. Determine the critical points—the points where the rational expression will be zero or undefined.

    The rational expression will be zero when the numerator is zero. Since \(x-1=0\) when \(x=1\), then 1 is a critical point.

    The rational expression will be undefined when the denominator is zero. Since \(x+3=0\) when \(x=-3\), then -3 is a critical point.

    The critical points are 1 and -3.

    Step 3. Use the critical points to divide the number line into intervals.

    clipboard_e431a0043d1c5bb7e4b4d0de15dc59cbc.png

    The number line is divided into three intervals:

    \[(-\infty,-3) \quad (-3,1) \quad (1,\infty) \nonumber \]

    Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

    To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

    \[\text { Interval }(-\infty,-3) \nonumber \]

    The number -4 is in the interval \((-\infty,-3)\). Test \(x=-4\) in the expression in the numerator and the denominator.

    The numerator:

    \[\begin{array}{l} {x-1} \\ {-4-1} \\ {-5} \\ {\text {Negative}} \end{array} \nonumber \]

    The denominator:

    \[\begin{array}{l} {x+3} \\ {-4+3} \\ {-1} \\ {\text {Negative}} \end{array} \nonumber \]

    Above the number line, mark the factor \(x-1\) negative and mark the factor \(x+3\) negative.

    Since a negative divided by a negative is positive, mark the quotient positive in the interval \((-\infty,-3)\)

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