5.6: Solve Rational Equations (2024)

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    Learning Objectives
    • Solve rational equations
    • Use rational functions
    • Solve a rational equation for a specific variable
    Warm up
    1. Solve: \(\dfrac{1}{6} x+\dfrac{1}{2}=\dfrac{1}{3}\)
    2. Solve: \(n^2-5n-36=0\)
    3. Solve \(5x+2x=10\)
    Answer
    1. -1
    2. \(n=9, n=-4\)
    3. \(x=\dfrac{10}{7}\)

    After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve a rational equation.

    Definition: RATIONAL EQUATION

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

    You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

    \[\text {Rational Expression }\quad \quad \text{ Rational Equation} \nonumber \]

    \[\dfrac{1}{8} x+\dfrac{1}{2} \quad \quad \dfrac{1}{8} x+\dfrac{1}{2}=\dfrac{1}{4} \nonumber \]

    \[\dfrac{y+6}{y^{2}-36} \quad \quad \quad \dfrac{y+6}{y^{2}-36}=y+1 \nonumber \]

    \[\dfrac{1}{n-3}+\dfrac{1}{n+4} \quad \quad \quad \quad \dfrac{1}{n-3}+\dfrac{1}{n+4}=\dfrac{15}{n^{2}+n-12} \nonumber \]

    We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

    We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

    So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

    An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation.

    Definition: EXTRANEOUS Solution TO A RATIONAL EQUATION

    An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

    We note any possible extraneous solutions, \(c\), by writing \(x\neq c\) next to the equation.

    Example \(\PageIndex{1}\)

    Solve: \[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \]

    Solution

    Step 1. Note any value of the variable that would make any denominator zero.

    If \(x=0\), then \(\dfrac{1}{x}\) is undefined. So we'll write \(x \neq 0\) next to the equation.

    \[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}, x \neq 0 \nonumber \]

    Step 2. Find the least common denominator of all denominators in the equation.

    Find the LCD of \(\dfrac{1}{x}\), \(\dfrac{1}{3}\), and \(\dfrac{5}{6}\)

    The LCD is \(6x\).

    Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.

    Multiply both sides of the equation by the LCD, \(6x\).

    \[{\color{red}6 x} \cdot\left(\dfrac{1}{x}+\dfrac{1}{3}\right)={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \]

    Use the Distributive Property.

    \[{\color{red}6 x} \cdot \dfrac{1}{x}+{\color{red}6 x} \cdot \dfrac{1}{3}={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \]

    Simplify - and notice, no more fractions!

    \[6+2 x=5 x \nonumber \]

    Step 4. Solve the resulting equation.

    Simplify.

    \[\begin{aligned} &6=3 x\\ &2=x \end{aligned} \nonumber \]

    Step 5. Check.

    If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.

    We did not get 0 as an algebraic solution.

    \[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \]

    We substitute \(x=2\) into the original equation.

    \[\begin{aligned} \frac{1}{2}+\frac{1}{3}&\overset{?}{=}\frac{5}{6} \\ \frac{3}{6}+\frac{2}{6}&\overset{?}{=}\frac{5}{6} \\ \frac{5}{6}&=\frac{5}{6} \surd \end{aligned} \nonumber \]

    The solution is \(x=2\)

    Exercise \(\PageIndex{2}\)

    Solve: \[\dfrac{1}{y}+\dfrac{2}{3}=\dfrac{1}{5} \nonumber \]

    Answer

    \(y=-\dfrac{15}{7}\)

    Exercise \(\PageIndex{3}\)

    Solve: \[\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x} \nonumber \]

    Answer

    \(x=\dfrac{15}{3}\)

    The steps of this method are shown.

    how to Solve equations with rational expressions.
    • Step 1. Note any value of the variable that would make any denominator zero.
    • Step 2. Find the least common denominator of all denominators in the equation.
    • Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
    • Step 4. Solve the resulting equation.
    • Step 5. Check:
      • If any values found in Step 1 are algebraic solutions, discard them.
      • Check any remaining solutions in the original equation.

    We always start by noting the values that would cause any denominators to be zero.

    Example \(\PageIndex{4}\)

    Solve: \[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \]

    Solution

    Note any value of the variable that would make any denominator zero.

    \[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}, y \neq 0 \nonumber \]

    Find the least common denominator of all denominators in the equation. The LCD is \(y^2\).

    Clear the fractions by multiplying both sides of the equation by the LCD.

    \[y^{2}\left(1-\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber \]

    Distribute.

    \[y^{2} \cdot 1-y^{2}\left(\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber \]

    Multiply.

    \[y^{2}-5 y=-6 \nonumber \]

    Solve the resulting equation. First write the quadratic equation in standard form.

    \[y^{2}-5 y+6=0 \nonumber \]

    Factor.

    \[(y-2)(y-3)=0 \nonumber \]

    Use the Zero Product Property.

    \[y-2=0 \text { or } y-3=0 \nonumber \]

    Solve.

    \[y=2 \text { or } y=3 \nonumber \]

    Check. We did not get \(0\) as an algebraic solution.

    Check \(y=2\) and \(y=3\)in the original equation.

    \[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \quad \quad \quad 1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \]

    \[1-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{2^{2}} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{3^{2}} \nonumber \]

    \[1-\dfrac{5}{2} \overset{?}{=}-\dfrac{6}{4} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]

    \[\dfrac{2}{2}-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad \dfrac{3}{3}-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]

    \[-\dfrac{3}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad -\dfrac{2}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]

    \[-\dfrac{3}{2}=-\dfrac{3}{2} \surd \quad \quad \quad -\dfrac{2}{3}=-\dfrac{2}{3} \surd \nonumber \]

    The solution is \(y=2,y=3\)

    Exercise \(\PageIndex{5}\)

    Solve: \[1-\dfrac{2}{x}=\dfrac{15}{x^{2}} \nonumber \]

    Answer

    \(x=-3, x=5\)

    Exercise \(\PageIndex{6}\)

    Solve: \[1-\dfrac{4}{y}=\dfrac{12}{y^{2}} \nonumber \]

    Answer

    \(y=-2, y=6\)

    In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

    Example \(\PageIndex{7}\)

    Solve: \[\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4} \nonumber \]

    Solution

    Note any value of the variable that would make any denominator zero.

    \[\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{(x+2)(x-2)}, x \neq-2, x \neq 2 \nonumber \]

    Find the least common denominator of all denominators in the equation. The LCD is \((x+2)(x-2)\).

    Clear the fractions by multiplying both sides of the equation by the LCD.

    \[(x+2)(x-2)\left(\dfrac{2}{x+2}+\dfrac{4}{x-2}\right)=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \]

    Distribute.

    \[(x+2)(x-2) \dfrac{2}{x+2}+(x+2)(x-2) \dfrac{4}{x-2}=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \]

    Remove common factors.

    \[\cancel {(x+2)}(x-2) \dfrac{2}{\cancel {x+2}}+(x+2){\cancel {(x-2)}} \dfrac{4}{\cancel {x-2}}=\cancel {(x+2)(x-2)}\left(\dfrac{x-1}{\cancel {x^{2}-4}}\right) \nonumber \]

    Simplify.

    \[2(x-2)+4(x+2)=x-1 \nonumber \]

    Distribute.

    \[2 x-4+4 x+8=x-1 \nonumber \]

    Solve.

    \[\begin{aligned} 6 x+4&=x-1\\ 5 x&=-5 \\ x&=-1 \end{aligned}\]

    Check: We did not get 2 or −2 as algebraic solutions.

    Check \(x=-1\) in the original equation.

    \[\begin{aligned} \dfrac{2}{x+2}+\dfrac{4}{x-2} &=\dfrac{x-1}{x^{2}-4} \\ \dfrac{2}{(-1)+2}+\dfrac{4}{(-1)-2} &\overset{?}{=} \dfrac{(-1)-1}{(-1)^{2}-4} \\ \dfrac{2}{1}+\dfrac{4}{-3} &\overset{?}{=} \dfrac{-2}{-3} \\ \dfrac{6}{3}-\dfrac{4}{3} &\overset{?}{=} \dfrac{2}{3} \\ \dfrac{2}{3} &=\dfrac{2}{3} \surd \end{aligned} \nonumber \]

    The solution is \(x=-1\).

    Exercise \(\PageIndex{8}\)

    Solve: \[\dfrac{2}{x+1}+\dfrac{1}{x-1}=\dfrac{1}{x^{2}-1} \nonumber \]

    Answer

    \(x=\dfrac{2}{3}\)

    Exercise \(\PageIndex{9}\)

    Solve: \[\dfrac{5}{y+3}+\dfrac{2}{y-3}=\dfrac{5}{y^{2}-9} \nonumber \]

    Answer

    \(y=2\)

    In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.

    Example \(\PageIndex{10}\)

    Solve: \[\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1} \nonumber \]

    Solution

    Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator.

    \[\dfrac{m+11}{(m-4)(m-1)}=\dfrac{5}{m-4}-\dfrac{3}{m-1}, m \neq 4, m \neq 1 \nonumber \]

    Find the least common denominator of all denominators in the equation. The LCD is \((m-4)(m-1)\)

    Clear the fractions by multiplying both sides of the equation by the LCD.

    \[(m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1)\left(\dfrac{5}{m-4}-\dfrac{3}{m-1}\right) \nonumber \]

    Distribute.

    \[(m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1) \dfrac{5}{m-4}-(m-4)(m-1) \dfrac{3}{m-1} \nonumber \]

    Remove common factors.

    \[\cancel {(m-4)(m-1)}\left(\dfrac{m+11}{\cancel {(m-4)(m-1)}}\right)=\cancel {(m-4)}(m-1) \dfrac{5}{\cancel{m-4}}-(m-4)\cancel {(m-1)} \dfrac{3}{\cancel {m-1}} \nonumber \]

    Simplify.

    \[m+11=5(m-1)-3(m-4) \nonumber \]

    Solve the resulting equation.

    \[\begin{aligned} m+11&=5 m-5-3 m+12 \\ 4&=m \end{aligned} \nonumber \]

    Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution.

    There is no solution to this equation.

    Exercise \(\PageIndex{11}\)

    Solve: \[\dfrac{x+13}{x^{2}-7 x+10}=\dfrac{6}{x-5}-\dfrac{4}{x-2} \nonumber \]

    Answer

    There is no solution.

    Exercise \(\PageIndex{13}\)

    Solve: \[\dfrac{y-6}{y^{2}+3 y-4}=\dfrac{2}{y+4}+\dfrac{7}{y-1} \nonumber \]

    Answer

    There is no solution.

    The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

    Example \(\PageIndex{14}\)

    Solve: \[\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4 \nonumber \]

    Solution

    Factor all the denominators, so we can note any value of the variable that would make any denominator zero.

    \[\dfrac{y}{y+6}=\dfrac{72}{(y-6)(y+6)}+4, y \neq 6, y \neq-6 \nonumber \]

    Find the least common denominator. The LCD is \((y-6)(y+6)\)

    Clear the fractions.

    \[(y-6)(y+6)\left(\dfrac{y}{y+6}\right)=(y-6)(y+6)\left(\dfrac{72}{(y-6)(y+6)}+4\right) \nonumber \]

    Simplify.

    \[(y-6) \cdot y=72+(y-6)(y+6) \cdot 4 \nonumber \]

    Simplify.

    \[y(y-6)=72+4\left(y^{2}-36\right) \nonumber \]

    Solve the resulting equation.

    \[\begin{aligned} y^{2}-6 y&=72+4 y^{2}-144\\ 0&=3 y^{2}+6 y-72 \\ 0&=3\left(y^{2}+2 y-24\right) \\ 0&=3(y+6)(y-4) \\ y&=-6, y=4 \end{aligned} \nonumber \]

    Check.

    \(y=-6\) is an extraneous solution. Check \(y=4\) in the original equation.

    \[\begin{aligned} \dfrac{y}{y+6} &=\dfrac{72}{y^{2}-36}+4 \\ \dfrac{4}{4+6} &\overset{?}{=}\dfrac{72}{4^{2}-36}+4 \\ \dfrac{4}{10} &\overset{?}{=} \dfrac{72}{-20}+4 \\ \dfrac{4}{10} &\overset{?}{=} -\dfrac{36}{10}+\dfrac{40}{10} \\ \dfrac{4}{10} &=\dfrac{4}{10} \surd \end{aligned} \nonumber \]

    The solution is \(y=4\).

    Exercise \(\PageIndex{15}\)

    Solve: \[\dfrac{x}{x+4}=\dfrac{32}{x^{2}-16}+5 \nonumber \]

    Answer

    \(x=3\)

    Exercise \(\PageIndex{16}\)

    Solve: \[\dfrac{y}{y+8}=\dfrac{128}{y^{2}-64}+9 \nonumber \]

    Answer

    \(y=7\)

    5.6: Solve Rational Equations (2024)

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