Solve Rational Equations

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

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.

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.

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. 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.

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 $x\ne c$ next to the equation.

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

If $x=0$ , then $\tfrac{1}{x}$ is undefined, so write $x\ne0$ . The LCD of $\tfrac{1}{x}$ , $\tfrac{1}{3}$ , and $\tfrac{5}{6}$ is $6x$ . Multiply both sides by $6x$ , distribute, and solve:

\begin{array}{lrcl} \text{Clear the fractions.} & 6x\left(\tfrac{1}{x}+\tfrac{1}{3}\right) &=& 6x\left(\tfrac{5}{6}\right) \\[10pt] \text{Distribute and simplify.} & 6+2x &=& 5x \\[4pt] \text{Solve.} & 6 &=& 3x \\[4pt] &&& x=2 \end{array}

We did not get $0$ as an algebraic solution. Check $x=2$ in the original equation:

\begin{array}{rcl} \tfrac{1}{2}+\tfrac{1}{3} &\overset{?}{=}& \tfrac{5}{6} \\[10pt] \tfrac{3}{6}+\tfrac{2}{6} &\overset{?}{=}& \tfrac{5}{6} \\[10pt] \tfrac{5}{6} &=& \tfrac{5}{6}\ \checkmark \end{array}

The solution is $x=2$ .

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

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

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.

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

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

Note $y\ne0$ . The LCD is $y^2$ . Clear the fractions and solve the resulting quadratic equation using the Zero Product Property:

\begin{array}{lrcl} \text{Clear the fractions.} & y^2\left(1-\tfrac{5}{y}\right) &=& y^2\left(-\tfrac{6}{y^2}\right) \\[10pt] \text{Distribute and multiply.} & y^2-5y &=& -6 \\[4pt] \text{Write 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=0 &\text{or}& y-3=0 \\[4pt] \text{Solve.} & y=2 &\text{or}& y=3 \end{array}

Neither solution is excluded. Check both values in the original equation.

For $y=2$ :

\begin{array}{rcl} 1-\tfrac{5}{2} &\overset{?}{=}& -\tfrac{6}{2^2} \\[10pt] \tfrac{2}{2}-\tfrac{5}{2} &\overset{?}{=}& -\tfrac{6}{4} \\[10pt] -\tfrac{3}{2} &=& -\tfrac{3}{2}\ \checkmark \end{array}

For $y=3$ :

\begin{array}{rcl} 1-\tfrac{5}{3} &\overset{?}{=}& -\tfrac{6}{3^2} \\[10pt] \tfrac{3}{3}-\tfrac{5}{3} &\overset{?}{=}& -\tfrac{6}{9} \\[10pt] -\tfrac{2}{3} &=& -\tfrac{2}{3}\ \checkmark \end{array}

The solutions are $y=2$ and $y=3$ .

Solve $1-\tfrac{2}{x}=\tfrac{15}{x^2}$ . Enter both solutions, separated by commas.

Solve $1-\tfrac{4}{y}=\tfrac{12}{y^2}$ . Enter both solutions, separated by commas.

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

Example. Solve $\tfrac{2}{x+2}+\tfrac{4}{x-2}=\tfrac{x-1}{x^2-4}$ .

The denominators show that $x\ne-2$ and $x\ne2$ . Since $x^2-4=(x+2)(x-2)$ , the LCD is $(x+2)(x-2)$ .

\begin{array}{lrcl} \text{Multiply both sides by the LCD.} & (x+2)(x-2)\left(\tfrac{2}{x+2}+\tfrac{4}{x-2}\right) &=& (x+2)(x-2)\left(\tfrac{x-1}{x^2-4}\right) \\[10pt] \text{Distribute.} & (x+2)(x-2)\tfrac{2}{x+2}+(x+2)(x-2)\tfrac{4}{x-2} &=& (x+2)(x-2)\tfrac{x-1}{(x+2)(x-2)} \\[10pt] \text{Remove common factors and simplify.} & 2(x-2)+4(x+2) &=& x-1 \\[4pt] \text{Distribute.}\quad & 2x-4+4x+8 &=& x-1 \\[4pt] \text{Solve.} & 6x+4 &=& x-1 \\[4pt] & 5x &=& -5 \\[4pt] &&& x=-1 \end{array}

The value $-1$ is not excluded. In the original equation it gives $\tfrac{2}{1}+\tfrac{4}{-3}=\tfrac{-2}{-3}$ , or $\tfrac{2}{3}=\tfrac{2}{3}$ . The solution is $x=-1$ .

Solve $\tfrac{2}{x+1}+\tfrac{1}{x-1}=\tfrac{1}{x^2-1}$ .

Solve $\tfrac{5}{y+3}+\tfrac{2}{y-3}=\tfrac{5}{y^2-9}$ .

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

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

Factor $m^2-5m+4=(m-4)(m-1)$ , so $m\ne4$ and $m\ne1$ . The LCD is $(m-4)(m-1)$ .

Multiply both sides by the LCD:

\begin{aligned} &(m-4)(m-1)\left(\tfrac{m+11}{(m-4)(m-1)}\right) \\ &\quad=(m-4)(m-1)\left(\tfrac{5}{m-4}-\tfrac{3}{m-1}\right). \end{aligned}

Distribute, then remove the common factors:

\begin{aligned} &(m-4)(m-1)\tfrac{m+11}{(m-4)(m-1)} \\ &\quad=(m-4)(m-1)\tfrac{5}{m-4} -(m-4)(m-1)\tfrac{3}{m-1} \\ &m+11=5(m-1)-3(m-4). \end{aligned}

Now solve the resulting equation:

\begin{array}{lrcl} \text{Solve the resulting equation.} & m+11 &=& 5m-5-3m+12 \\[4pt] &&& 4=m \end{array}

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

Solve $\tfrac{x+13}{x^2-7x+10}=\tfrac{6}{x-5}-\tfrac{4}{x-2}$ .

Solve $\tfrac{y-6}{y^2+3y-4}=\tfrac{2}{y+4}+\tfrac{7}{y-1}$ .

The equation 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. Solve $\tfrac{y}{y+6}=\tfrac{72}{y^2-36}+4$ .

Factor $y^2-36=(y-6)(y+6)$ , so $y\ne6$ and $y\ne-6$ . The LCD is $(y-6)(y+6)$ .

\begin{array}{lrcl} \text{Clear the fractions.} & (y-6)y &=& 72+4(y-6)(y+6) \\[4pt] \text{Simplify.} & y^2-6y &=& 72+4y^2-144 \\[4pt] \text{Solve the resulting equation.} & 0 &=& 3y^2+6y-72 \\[4pt] &&& 0=3(y^2+2y-24) \\[4pt] &&& 0=3(y+6)(y-4) \\[4pt] &&& y=-6,\ y=4 \end{array}

$y=-6$ is extraneous. Checking $y=4$ gives $\tfrac{4}{10}=\tfrac{72}{-20}+4=\tfrac{4}{10}$ . The solution is $y=4$ .

Solve $\tfrac{x}{x+4}=\tfrac{32}{x^2-16}+5$ .

Solve $\tfrac{y}{y+8}=\tfrac{128}{y^2-64}+9$ .

In some cases, all the algebraic solutions are extraneous.

Example. Solve $\tfrac{x}{2x-2}-\tfrac{2}{3x+3}=\tfrac{5x^2-2x+9}{12x^2-12}$ .

Factoring the denominators gives $2(x-1)$ , $3(x+1)$ , and $12(x-1)(x+1)$ , so $x\ne1$ and $x\ne-1$ . The LCD is $12(x-1)(x+1)$ .

\begin{array}{lrcl} \text{Clear the fractions and simplify.} & 6x(x+1)-4(x-1)\cdot2 &=& 5x^2-2x+9 \\[4pt] \text{Distribute.} & 6x^2+6x-8x+8 &=& 5x^2-2x+9 \\[4pt] \text{Solve.} & x^2-1 &=& 0 \\[4pt] &&& (x-1)(x+1)=0 \\[4pt] &&& x=1\text{ or }x=-1 \end{array}

Both $x=1$ and $x=-1$ are extraneous solutions. The equation has no solution.

Solve $\tfrac{y}{5y-10}-\tfrac{5}{3y+6}=\tfrac{2y^2-19y+54}{15y^2-60}$ .

Solve $\tfrac{z}{2z+8}-\tfrac{3}{4z-8}=\tfrac{3z^2-16z-16}{8z^2+16z-64}$ .

Example. Solve $\tfrac{4}{3x^2-10x+3}+\tfrac{3}{3x^2+2x-1}=\tfrac{2}{x^2-2x-3}$ .

Factor all denominators:

\frac{4}{(3x-1)(x-3)}+\frac{3}{(3x-1)(x+1)} =\frac{2}{(x-3)(x+1)}, \qquad x\ne-1,\ x\ne\tfrac{1}{3},\ x\ne3.

The LCD is $(3x-1)(x+1)(x-3)$ . Clearing fractions gives

4(x+1)+3(x-3)=2(3x-1).

Distribute and solve:

\begin{array}{rcl} 4x+4+3x-9 &=& 6x-2 \\[4pt] 7x-5 &=& 6x-2 \\[4pt] x &=& 3 \end{array}

The only algebraic solution, $x=3$ , would make a denominator zero. The algebraic solution is extraneous. There is no solution to this equation.

Solve $\tfrac{15}{x^2+x-6}-\tfrac{3}{x-2}=\tfrac{2}{x+3}$ .

Solve $\tfrac{5}{x^2+2x-3}-\tfrac{3}{x^2+x-2}=\tfrac{1}{x^2+5x+6}$ .

Use rational functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

Example. For the rational function $f(x)=\tfrac{2x-6}{x^2-8x+15}$ , (a) find the domain of the function, (b) solve $f(x)=1$ , and (c) find the points on the graph at this function value.

The domain of a rational function is all real numbers except those that make the rational expression undefined. Set the denominator equal to zero:

\begin{array}{rcl} x^2-8x+15 &=& 0 \\[4pt] (x-3)(x-5) &=& 0 \\[4pt] x &=& 3,\ 5 \end{array}

So the domain is all real numbers except $x\ne3$ and $x\ne5$ .

To solve $f(x)=1$ , substitute the rational expression and factor the denominator:

\begin{array}{rcl} \tfrac{2x-6}{(x-3)(x-5)} &=& 1 \\[10pt] 2x-6 &=& x^2-8x+15 \\[4pt] 0 &=& x^2-10x+21 \\[4pt] 0 &=& (x-7)(x-3) \\[4pt] x-7=0 &\text{or}& x-3=0 \\[4pt] x=7 &\text{or}& x=3 \end{array}

However, $x=3$ is outside the domain, so discard that root as extraneous. The value of the function is $1$ when $x=7$ , so the point on the graph is $(7,1)$ .

For $f(x)=\tfrac{8-x}{x^2-7x+12}$ , find the values excluded from the domain, separated by commas.

For $f(x)=\tfrac{8-x}{x^2-7x+12}$ , solve $f(x)=3$ . Enter both solutions, separated by commas.

For $f(x)=\tfrac{8-x}{x^2-7x+12}$ , find the points on the graph where $f(x)=3$ , separated by commas.

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.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD:

\begin{array}{lrcl} &m&=&\tfrac{y-y_1}{x-x_1} \\[10pt] \text{Multiply both sides by }x-x_1. & m(x-x_1)&=&\left(\tfrac{y-y_1}{x-x_1}\right)(x-x_1) \\[10pt] \text{Simplify.} & m(x-x_1)&=&y-y_1 \\[4pt] \text{Rewrite with the }y\text{ terms on the left.} & y-y_1&=&m(x-x_1) \end{array}

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through a point in Chapter 3. We will add one more step to solve for $y$ .

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

Note $x\ne3$ . Clear the fractions by multiplying both sides by $x-3$ , then isolate the term with $y$ :

\begin{array}{rcl} (x-3)m &=& y-2 \\[4pt] xm-3m+2 &=& y \end{array}

Thus $y=mx-3m+2$ .

Solve $m=\tfrac{y-5}{x-4}$ for $y$ .

Solve $m=\tfrac{y-1}{x+5}$ for $y$ .

Remember to multiply both sides by the LCD in the next example.

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

Note $c\ne0$ and $m\ne0$ . The LCD is $cm$ .

\begin{array}{lrcl} \text{Clear the fractions.} & cm\left(\tfrac{1}{c}+\tfrac{1}{m}\right) &=& cm(1) \\[10pt] \text{Distribute and simplify.} & m+c &=& cm \\[4pt] \text{Collect the terms with }c\text{ to the right.} & m &=& cm-c \\[4pt] \text{Factor the expression on the right.} & m &=& c(m-1) \\[4pt] \text{Divide by }m-1. & \tfrac{m}{m-1} &=& c \end{array}

Even though we excluded $c=0$ and $m=0$ from the original equation, we must also now state that $m\ne1$ . Thus $c=\tfrac{m}{m-1}$ .

Solve $\tfrac{1}{a}+\tfrac{1}{b}=c$ for $a$ .

Solve $\tfrac{2}{x}+\tfrac{1}{3}=\tfrac{1}{y}$ for $y$ .

Key terms. A rational equation is an equation that contains a rational expression. An extraneous solution to a rational equation is an algebraic solution that would cause an expression in the original equation to be undefined.

Adapted from OpenStax Intermediate Algebra 2e, Section 7.4, by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at OpenStax. Changes: adapted the section for interactive web delivery and converted Try It exercises to immediate-feedback questions.

Simplify Complex Rational Expressions Solve Applications with Rational Equations