CR.19: Solving Rational Equations

Learning Outcomes

Solve rational equations by clearing denominators
Identify extraneous solutions in a rational equation

Equations that contain rational expressions are called rational equations. For example, $\frac{2x+1}{4}=\frac{x}{3}$ is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships.

One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator and then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:

Solve $\frac{1}{2}x-3=2-\frac{3}{4}x$ by clearing the fractions in the equation first.

Multiply both sides of the equation by $4$ , the common denominator of the fractional coefficients.

$\begin{array}{c}\frac{1}{2}x-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\\\,\,\,\,4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2x-12=8-3x\\\underline{+3x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+3x}\\5x-12=8\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\underline{+12}\,\,\,\,\underline{+12} \\5x=20\\x=4\end{array}$

We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a term that has a polynomial in the numerator.

Example

Solve the equation $\frac{x+5}{8}=\frac{7}{4}$ .

Show Solution

Find the least common denominator of $4$ and $8$ . Remember, to find the LCD, identify the greatest number of times each factor appears in each factorization. Here, $2$ appears $3$ times, so $2\cdot2\cdot2$ , or $8$ , will be the LCD.

Multiply both sides of the equation by the common denominator, $8$ , to keep the equation balanced and to eliminate the denominators.

$\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}$

Simplify and solve for x.

$\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}$

Check the solution by substituting $9$ for x in the original equation.

$\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}$

Therefore, $x=9$ .

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.

Example

Solve the equation $\frac{8}{x+1}=\frac{4}{3}$ .

Show Solution

Clear the denominators by multiplying each side by the common denominator. The common denominator is $3\left(x+1\right)$ since $3\text{ and }x+1$ do not have any common factors.

$\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}$

Simplify common factors.

$\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}$

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

$\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}$

Check the solution in the original equation.

$\begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\\\frac{4}{3}=\frac{4}{3}\end{array}$

Therefore, $x=5$ .

Try It

Solve the equation $\frac{2x+1}{x}=\frac{3}{2}$ .

Show Solution

$x=-2$

Example

Solve the equation $\frac{x}{3}+1=\frac{4}{3}$ .

Show Solution

Both fractions in the equation have a denominator of $3$ . Multiply every term on both sides of the equation (not just the fractions!) by $3$ .

$3\left( \frac{x}{3}+1 \right)=3\left( \frac{4}{3} \right)$

Apply the distributive property and multiply $3$ by each term within the parentheses. Then simplify and solve for x.

$\begin{array}{r}3\left( \frac{x}{3} \right)+3\left( 1 \right)=3\left( \frac{4}{3} \right)\\\\\cancel{3}\left( \frac{x}{\cancel{3}} \right)+3\left( 1 \right)=\cancel{3}\left( \frac{4}{\cancel{3}} \right)\\\\ x+3=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3}\,\,\,\,\,\underline{-3}\\\\x=1\end{array}$

Therefore, $x=1$ .

Try It

Solve the equation $\frac{8}{y}+\frac{4}{y}=4$ .

Show Solution

$y=3$

In the video that follows, we present two ways to solve rational equations with both integer and variable denominators.

Excluded Values and Extraneous Solutions

As you have seen, some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by $0$ is undefined, you must exclude values of the variable that would result in a denominator of $0$ . These values are called excluded values. Let us look at an example.

Example

Solve the equation $\frac{2x-5}{x-5}=\frac{15}{x-5}$ .

Show Solution

Determine any values for x that would make the denominator $0$ .

$\frac{2x-5}{x-5}=\frac{15}{x-5}$

$5$ is an excluded value because it makes the denominator $x-5$ equal to $0$ .

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.

$\begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}$

Check the solution in the original equation.

$\begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}$

Therefore, $x=10$ .

In the following video, we present an example of solving a rational equation with variables in the denominator.

You have seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that do not work in the original form of the equation. These types of answers are called extraneous solutions. That is why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions.

Example

Solve the equation $\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}$ .

Show Solution

Determine any values for m that would make the denominator $0$ . The value $−4$ is an excluded value because it makes $m+4$ equal to $0$ .

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

$\begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left( m-4 \right)\end{array}$

$\begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}$

Check the solutions in the original equation.

Since $m=−4$ leads to division by $0$ , it is an extraneous solution.

Notice, however, that latex]m=4[/latex] is a solution that results in a true statement.

$\begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}$

$-4$ is excluded because it leads to division by $0$ .

$\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}$

Therefore, $m=4$ .

Example

Solve the equation $\frac{6}{x+9}+\frac{5}{x+3}=\frac{4x}{x^2+12x+27}$ .

Show Solution

It is best to factor this one first: $\dfrac{6}{x+9}+\dfrac{5}{x+3}=\dfrac{4x}{(x+9)(x+3)}$

Determine any values for x that would make the denominator $0$ . The values $−9$ and $-3$ are excluded values because it makes at least one denominator equal to $0$ .

We want to make all the denominators the same, so we need to multiply each fraction by what it is missing in order to get common denominators:

$\dfrac{6}{x+9} \left(\dfrac{x+3}{x+3}\right)+\dfrac{5}{x+3} \left(\dfrac{x+9}{x+9}\right)=\dfrac{4x}{(x+9)(x+3)}$

Now we multiply across the numerators:

$\dfrac{6x+18}{(x+9)(x+3)}+\dfrac{5x+45}{(x+9)(x+3)}=\dfrac{4x}{(x+9)(x+3)}$

Now we can add the two fractions on the left side of the equation:

$\dfrac{11x+63}{(x+9)(x+3)}=\dfrac{4x}{(x+9)(x+3)}$

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.

$\begin{array}{l}11x+63=4x\\\,\,\,63=-7x\\\,\,\,x=-9\end{array}$

Check the solutions in the original equation.

Since $x=−9$ leads to division by $0$ , it is an extraneous solution.

Therefore, there are no solutions.

Example

Solve the equation $\frac{x+5}{x^2-5x+6}-\frac{8}{x^2-7x+12}=\frac{x+5}{x^2-6x+8}$ .

Show Solution

It is best to start by factoring this one.

$\dfrac{x+5}{(x-2)(x-3)}-\dfrac{8}{(x-4)(x-3)}=\dfrac{x+5}{(x-4)(x-2)}$

Determine any values for x that would make the denominator $0$ . The values $2$ , $3$ , and $4$ are excluded values because it makes at least one denominator equal to $0$ .

In the last problem, we got all the denominators the same. Now we will show a different method, which involves multiplying both sides of the equation by the LCD.

The LCD is $(x-2)(x-3)(x-4)$ .

Now we will multiply both sides by the LCD:

$(x-2)(x-3)(x-4)\left(\dfrac{x+5}{(x-2)(x-3)}-\dfrac{8}{(x-4)(x-3)}\right)=(x-2)(x-3)(x-4)\left(\dfrac{x+5}{(x-4)(x-2)}\right)$

Use the distributive property to distribute the LCD.

$\dfrac{(x-2)(x-3)(x-4)(x+5)}{(x-2)(x-3)}-\dfrac{8(x-2)(x-3)(x-4)}{(x-4)(x-3)}=\dfrac{(x-2)(x-3)(x-4)(x+5)}{(x-4)(x-2)}$

Now we can cancel out common factors:

$(x+5)(x-4)-8(x-2)=(x-5)(x-3)$

Next, use the distributive property to expand each side:

$x^2+x-20-8x+16=x^2-8x+15$

Combine all like terms on the left side of the equation.

$x^2-7x-4=x^2-8x+15$

Subtract the $x^2$ from both sides and solve for x.

$-7x-4=-8x+15$
$x-4=15$
$x=19$

Check the solutions in the original equation.

Since $x=19$ does not lead to division by $0$ , it is a solution.

Therefore, $x=19$ .

Try It

Solve the equation $\frac{x+4}{x^2-3x+2}-\frac{5}{x^2-4x+3}=\frac{x-4}{x^2-5x+6}$ .

Show Solution

$x=6$

Summary

You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common denominator of all fractions so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that do not satisfy the original form of the equation because they produce untrue statements or are excluded values that cause a denominator to equal $0$ .

Chapter CR: Corequisite Review

Learning Outcomes

Example

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Try It

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Try It

Excluded Values and Extraneous Solutions

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Try It

Summary