Solve Rational Equations

By the end of this section, you will be able to: solve rational equations, and solve a rational equation for a specific variable.

We have simplified many rational expressions so far in this chapter. Now we will solve rational equations. The definition of a rational equation is similar to the definition of equation we used earlier.

Rational equation. A rational equation is two rational expressions connected by an equal sign.

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

Rational expression	Rational equation
$\tfrac{1}{8}x + \tfrac{1}{2}$	$\tfrac{1}{8}x + \tfrac{1}{2} = \tfrac{1}{4}$
$\tfrac{y+6}{y^2-36}$	$\tfrac{y+6}{y^2-36} = y + 1$
$\tfrac{1}{n-3} + \tfrac{1}{n+4}$	$\tfrac{1}{n-3} + \tfrac{1}{n+4} = \tfrac{15}{n^2+n-12}$

Solve Rational Equations

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: 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 the 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.

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. Solve: $\tfrac{1}{x} + \tfrac{1}{3} = \tfrac{5}{6}$ .

\begin{array}{lrcl} & \tfrac{1}{x} + \tfrac{1}{3} &=& \tfrac{5}{6} \\[4pt] \text{Note any value that makes a denominator zero: } x \neq 0. & \tfrac{1}{x} + \tfrac{1}{3} &=& \tfrac{5}{6}, \; x \neq 0 \\[4pt] \text{The LCD of all denominators is } 6x. \text{ Multiply both sides by it.} & 6x\left(\tfrac{1}{x} + \tfrac{1}{3}\right) &=& 6x\left(\tfrac{5}{6}\right) \\[4pt] \text{Distribute.} & 6x \cdot \tfrac{1}{x} + 6x \cdot \tfrac{1}{3} &=& 6x \cdot \tfrac{5}{6} \\[4pt] \text{Simplify — no more fractions!} & 6 + 2x &=& 5x \\[4pt] \text{Solve the resulting equation.} & 6 &=& 3x \\[4pt] & 2 &=& x \end{array}

Check: we did not get $0$ as an algebraic solution, so substitute $x = 2$ into the original equation.

\begin{array}{rcl} \tfrac{1}{x} + \tfrac{1}{3} &=& \tfrac{5}{6} \\[2pt] \tfrac{1}{2} + \tfrac{1}{3} &\overset{?}{=}& \tfrac{5}{6} \\[2pt] \tfrac{3}{6} + \tfrac{2}{6} &\overset{?}{=}& \tfrac{5}{6} \\[2pt] \tfrac{5}{6} &=& \tfrac{5}{6} \; \checkmark \end{array}

The solution is $x = 2$ .

Solve equations with rational expressions.

Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation.
Clear the fractions by multiplying both sides of the equation by the LCD.
Solve the resulting equation.
Check.
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.

Solve: $\tfrac{1}{y} + \tfrac{2}{3} = \tfrac{1}{5}$ .

We always start by noting the values that would cause any denominators to be zero. When one of the denominators is a quadratic, remember to factor it first to find the LCD.

Example. Solve: $1 - \tfrac{5}{y} = -\tfrac{6}{y^2}$ .

The denominators are $y$ and $y^2$ , so $y \neq 0$ . The LCD is $y^2$ . Multiply both sides by $y^2$ and simplify.

\begin{array}{lrcl} & 1 - \tfrac{5}{y} &=& -\tfrac{6}{y^2}, \; y \neq 0 \\[4pt] \text{Multiply both sides by } y^2 \text{ and distribute.} & y^2 - 5y &=& -6 \\[4pt] \text{Write the quadratic equation in standard form.} & y^2 - 5y + 6 &=& 0 \\[4pt] \text{Factor.} & (y-2)(y-3) &=& 0 \\[4pt] \text{Use the Zero Product Property.} & y = 2 &\text{or}& y = 3 \end{array}

We did not get $0$ as an algebraic solution, and checking $y = 2$ and $y = 3$ in the original equation makes both sides equal. The solutions are $y = 2$ and $y = 3$ .

Example. Solve: $\tfrac{5}{3u-2} = \tfrac{3}{2u}$ .

The denominators are $3u - 2$ and $2u$ , so $u \neq \tfrac{2}{3}$ and $u \neq 0$ . The LCD is $2u(3u-2)$ .

\begin{array}{lrcl} & \tfrac{5}{3u-2} &=& \tfrac{3}{2u}, \; u \neq \tfrac{2}{3},\ u \neq 0 \\[4pt] \text{Multiply both sides by } 2u(3u-2) \text{ and remove common factors.} & 2u(5) &=& (3u-2)(3) \\[4pt] \text{Multiply.} & 10u &=& 9u - 6 \\[4pt] \text{Solve the resulting equation.} & u &=& -6 \end{array}

We did not get $0$ or $\tfrac{2}{3}$ as an algebraic solution, so $u = -6$ is a valid solution. Checking it gives $-\tfrac{1}{4} = -\tfrac{1}{4}$ .

Solve: $\tfrac{1}{x-1} = \tfrac{2}{3x}$ . (Enter the value of $x$ .)

Extraneous solutions

Sometimes an algebraic solution turns out to be a value we already excluded — one that makes a denominator zero. That value is an extraneous solution and must be discarded. If it was the only algebraic solution, the equation has no solution.

Example. Solve: $\tfrac{m+11}{m^2-5m+4} = \tfrac{5}{m-4} - \tfrac{3}{m-1}$ .

Factor the first denominator: $m^2 - 5m + 4 = (m-4)(m-1)$ . So $m \neq 4$ and $m \neq 1$ , and the LCD is $(m-4)(m-1)$ .

\begin{array}{lrcl} & \tfrac{m+11}{(m-4)(m-1)} &=& \tfrac{5}{m-4} - \tfrac{3}{m-1}, \; m \neq 4,\ m \neq 1 \\[4pt] \text{Clear the fractions and remove common factors.} & m + 11 &=& 5(m-1) - 3(m-4) \\[4pt] \text{Solve the resulting equation.} & m + 11 &=& 5m - 5 - 3m + 12 \\[4pt] & m + 11 &=& 2m + 7 \\[4pt] & 4 &=& m \end{array}

The only algebraic solution is $4$ , but we noted that $4$ would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.

For a no-solution case like this one, the check step is what catches the trap: the excluded value and the algebraic solution are the same number.

Solve: $\tfrac{x}{2x-2} - \tfrac{2}{3x+3} = \tfrac{5x^2-2x+9}{12x^2-12}$ . After clearing fractions, the equation simplifies to $x^2 - 1 = 0$ , giving $x = 1$ or $x = -1$ . What is the solution set?

Example (extraneous root discarded). Solve: $\tfrac{n}{12} + \tfrac{n+3}{3n} = \tfrac{1}{n}$ .

Here $n \neq 0$ , and the LCD is $12n$ . Clearing fractions gives $n \cdot n + 4(n+3) = 12 \cdot 1$ , which simplifies to $n^2 + 4n = 0$ , so $n(n+4) = 0$ and $n = 0$ or $n = -4$ . Since $n = 0$ is an excluded value, it is an extraneous solution and we discard it. Checking $n = -4$ gives $-\tfrac{1}{4} = -\tfrac{1}{4}$ , so the only solution is $n = -4$ .

In each of these problems, identifying the excluded values before solving is what tells us which algebraic solutions to keep.

For the equation $\tfrac{4}{q-4} - \tfrac{3}{q-3} = 1$ , before solving you note the values that make a denominator zero. One of them is $q = 4$ . What is the other excluded value of $q$ ?

Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable, using the same steps: note the excluded values, clear the fractions with the LCD, and then isolate the variable we want.

Example. Solve $\tfrac{D}{T} = R$ for $T$ .

\begin{array}{lrcl} & \tfrac{D}{T} &=& R, \; T \neq 0 \\[4pt] \text{Multiply both sides by the LCD, } T. & T\left(\tfrac{D}{T}\right) &=& T(R) \\[4pt] \text{Simplify.} & D &=& T \cdot R \\[4pt] \text{Divide both sides by } R \text{ to isolate } T. & \tfrac{D}{R} &=& T \end{array}

Solve $\tfrac{A}{L} = W$ for $L$ . Enter the expression that $L$ equals.

Example. Solve $m = \tfrac{x-2}{y-3}$ for $y$ .

Note $y \neq 3$ . Multiply both sides by the LCD, $y - 3$ .

\begin{array}{lrcl} & m &=& \tfrac{x-2}{y-3}, \; y \neq 3 \\[4pt] \text{Multiply both sides by } y - 3 \text{ and simplify.} & (y-3)m &=& x - 2 \\[4pt] & ym - 3m &=& x - 2 \\[4pt] \text{Isolate the term with } y. & ym &=& x - 2 + 3m \\[4pt] \text{Divide both sides by } m \text{ to isolate } y. & y &=& \tfrac{x - 2 + 3m}{m} \end{array}

Some formulas look simple but cannot be solved instantly for either denominator — be sure to follow all the steps.

Example. Solve $\tfrac{1}{c} + \tfrac{1}{m} = 1$ for $c$ .

Here $c \neq 0$ and $m \neq 0$ . The LCD is $cm$ .

\begin{array}{lrcl} & \tfrac{1}{c} + \tfrac{1}{m} &=& 1, \; c \neq 0,\ m \neq 0 \\[4pt] \text{Multiply both sides by } cm \text{ and distribute.} & m + c &=& cm \\[4pt] \text{Collect the terms with } c \text{ on the right.} & m &=& cm - c \\[4pt] \text{Factor the expression on the right.} & m &=& c(m-1) \\[4pt] \text{Divide both sides by } m - 1 \text{ to isolate } c. & \tfrac{m}{m-1} &=& c \end{array}

Notice that even though we excluded $c = 0$ and $m = 0$ from the original equation, we must also now state that $m \neq 1$ .

Solve $\tfrac{y-2}{x+1} = \tfrac{2}{3}$ for $x$ . Enter the expression that $x$ equals.

Key terms

rational equation — two rational expressions connected by an equal sign. extraneous solution to a rational equation — an algebraic solution that would cause any of the expressions in the original equation to be undefined; it must be discarded. excluded value — a value of the variable that makes a denominator equal to zero, so the rational expression is undefined there.

This section is adapted from Elementary Algebra 2e, Section 8.6: Solve Rational Equations by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: condensed the worked examples into aligned step tables and prose, recast the “How To” procedures as callouts, folded the extraneous-solution cases (no-solution and discarded-root) into a single subsection; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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