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{{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Solving Rational Equations

Equations with rational expressions are called rational equations. Like all equations, these can be solved graphically and algebraically. Sometimes, when solving algebraically, extraneous solutions — or solutions that do not satisfy the equation — arise.
Method

## Solving Rational Equations Graphically

A rational function is a function that contains at least one rational expression, such as $y=\dfrac{4x}{x+5}-\dfrac{2}{x}.$ If the dependent variable of the function is exchanged for a constant, say $C,$ the result is a rational equation, $C=\dfrac{4x}{x+5}-\dfrac{2}{x},$

which can be solved graphically. This is done by first graphing the function $y=\frac{4x}{x+5}-\frac{2}{x},$ then finding the $y$-coordinate(s) of the point(s) on the graph that has the $y$-coordinate $C.$ Then, the $x$-coordinate(s) is the solution to the equation.
Exercise

Solve the rational equation graphically. $\dfrac{1}{x+2}+2=3$

Solution

When solving an equation graphically, we first rearrange the equation to get all variable-terms on one side and all constants on the other side. $\dfrac{1}{x+2}+2=3 \quad \Leftrightarrow \quad \dfrac{1}{x+2}=1$ The left-hand side of the equation can now be expressed as a function, $f(x)=\frac{1}{x+2},$ which we graph in a coordinate plane. The solution to the equation is the $x$-value of the point where the function equals $1.$ We'll mark this point on the graph. From the graph, we see that $f(\text{-} 1)=1.$ This means the solution to the equation is $x=\text{-} 1.$ We can verify this by substituting it into the equation.
$\dfrac{1}{x+2}+2=3$
$\dfrac{1}{{\color{#0000FF}{\text{-} 1}}+2}+2\stackrel{?}{=}3$
$\dfrac{1}{1}+2\stackrel{?}{=}3$
$1+2\stackrel{?}{=}3$
$3=3$
Since $x=\text{-} 1$ makes a true statement, it is a solution to the equation.
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Method

## Solving a Rational Equation by Multiplying Denominators

In rational equations, variable terms are in the denominator. To solve a rational equation, it is necessary to move the variables out of the denominator. Because the domain of rational functions is restricted, some solutions might be extraneous. Thus, it is necessary to verify the solutions. Consider the following equation as an example. $\dfrac{1}{x+1} + \dfrac{1}{x-1} = \text{-} \dfrac{1}{2x}$

### 1

Multiply equation by denominators

To begin, it's necessary to get the variable terms out of the denominator(s). Here, $x$ is in the denominator of three different ratios. To get $(x+1)$ out of the denominator of the first ratio, multiply the equation by $(x+1).$ This yields \begin{aligned} {\color{#0000FF}{(x+1)}} \left( \dfrac{1}{x+1} + \dfrac{1}{x-1}\right) &= {\color{#0000FF}{(x+1)}}\left(\text{-} \dfrac{1}{2x}\right) \\ \dfrac{{\color{#0000FF}{x+1}}}{x+1} + \dfrac{{\color{#0000FF}{x+1}}}{x-1} &= \text{-} \dfrac{{\color{#0000FF}{x+1}}}{2x}\\ 1 + \dfrac{x+1}{x-1} &= \text{-} \dfrac{x+1}{2x} \end{aligned} Notice that the variable term is no longer stuck in the denominator of the first ratio. The other denominators $(x-1)$ and $(2x)$ must be handled in the same way. This leads to multiplying the entire equation by $(x+1), (x-1)$ and $(2x),$ which can be done in one step. Consider multiplying the first ratio by all three denominators. $(x+1)(x-1)(2x)\cdot \dfrac{1}{x+1}$ $\dfrac{({\color{#0000FF}{x+1}})(x-1)(2x)}{{\color{#0000FF}{x+1}}}$ $(x-1)(2x)$ As can be seen, the denominator of the ratio is eliminated and what remains is the product of the numerator and the other denominators. This will happen for each ratio in the equation. Now that the work has been explained, solving the equation can begin.

$\dfrac{1}{x+1} + \dfrac{1}{x-1} = \text{-} \dfrac{1}{2x}$
${\color{#0000FF}{(x+1)(x-1)(2x)}} \left(\dfrac{1}{x+1} + \dfrac{1}{x-1}\right) = {\color{#0000FF}{(x+1)(x-1)(2x)}} \left( \text{-} \dfrac{1}{2x} \right)$
$\dfrac{(x+1)(x-1)(2x)}{x+1} + \dfrac{(x+1)(x-1)(2x)}{x-1}=\text{-} \dfrac{(x+1)(x-1)(2x)}{2x}$
$\dfrac{\cancel{(x+1)}(x-1)(2x)}{\cancel{x+1}} + \dfrac{(x+1)\cancel{(x-1)}(2x)}{\cancel{x-1}}=\text{-} \dfrac{(x+1)(x-1)\cancel{(2x)}}{\cancel{2x}}$
$(x-1)(2x) + (x+1)(2x)=\text{-} (x+1)(x-1)$

### 2

Consolidate variable terms
Notice that there are several terms that contain variables. In order to solve the equation, it is necessary to consolidate these terms. To do this, distribute and combine like terms.
$(x-1)(2x) + (x+1)(2x)=\text{-} (x+1)(x-1)$
$2x^2-2x + 2x^2+2x = \text{-} (x^2-x+x-1)$
$4x^2 = \text{-} (x^2-1)$
$4x^2= \text{-} x^2+1$
$5x^2=1$

### 3

Solve the resulting equation

Now that the variable terms have been consolidated, the equation can be solved. Notice here that the resulting equation is quadratic. This will not always be the case. The type of equation determines which method to use. Here, square roots will be used.

$5x^2=1$
$x^2=\dfrac{1}{5}$
$x^2=0.2$
$x=\pm \sqrt{0.2}$
$x =\pm 0.44721 \ldots$
$\begin{array}{l}x_1 = 0.44721 \ldots \\ \\ x_2 =\text{-} 0.44721 \ldots \end{array}$

### 4

Check for extraneous solutions
The solutions found in Step $3$ might be extraneous. Thus, they must be verified in the original equation. First, the solution $x=0.44721 \ldots$ is tested.
$\dfrac{1}{x+1} + \dfrac{1}{x-1} = \text{-} \dfrac{1}{2x}$
$\dfrac{1}{{\color{#0000FF}{0.44721\ldots}}+1} + \dfrac{1}{{\color{#0000FF}{0.44721\ldots}}-1} \stackrel{?}{=} \text{-} \dfrac{1}{2\left({\color{#0000FF}{0.44721\ldots}}\right)}$
$0.69098\ldots + (\text{-} 1.80900\ldots) \stackrel{?}{=} 1.11803\ldots$
$\text{-} 1.11803\ldots = \text{-} 1.11803\ldots$
The first value made a true statement. Thus, it is a solution to the equation. Next $x=\text{-} 0.44721\ldots$ is tested.
$\dfrac{1}{x+1} + \dfrac{1}{x-1} = \text{-} \dfrac{1}{2x}$
$\dfrac{1}{{\color{#0000FF}{\text{-} 0.44721\ldots}}+1} + \dfrac{1}{{\color{#0000FF}{\text{-} 0.44721\ldots}}-1} \stackrel{?}{=} \text{-} \dfrac{1}{2\left({\color{#0000FF}{\text{-} 0.44721\ldots}}\right)}$
$1.80901\ldots + (\text{-} 0.69098\ldots) \stackrel{?}{=} 1.11803\ldots$
$1.11803\ldots = 1.11803\ldots$
This value also makes a true statement. Thus, it is also a solution.
Exercise

Tamara and Jamie are trimming the hedges in their front yard this weekend. Working together, this task takes $7$ hours. If Tamara works alone, it will take her $11$ hours. How long would it take Jamie to trim the hedges on her own?

Solution
To begin, we can think of the time it takes for this task to be completed as a speed or rate expressed as a ratio. Since it takes Tamara $11$ hours to complete $1$ task, we can express her rate as $T=\dfrac{1 \text { task}}{11 \text{ hours}} = \dfrac{1}{11} \text{task/hour}$ It can be seen that in $1$ hour, Tamara will complete $\frac{1}{11}$ of the work. A ratio to express Jamie's rate can be written in the same way. Since the number of hours it takes her to complete the task is unknown, we'll use $x.$ $J=\dfrac{1 \text { task}}{x \text{ hours}} = \dfrac{1}{x} \text{task/hour}$ The time it takes for Tamara and Jamie to complete the task together is $7$ hours. This ratio is $\frac{1 \text{ task}}{7 \text{ hours}}.$ Thus, the following rational equation can be written. $\dfrac{1}{7} = \dfrac{1}{11} + \dfrac{1}{x}.$ Solving this equation will tell us how long it takes Jamie to trim the hedges on her own. To solve, we'll isolate the variable term, then use inverse operations. Remember, before combining fractions, they must have a common denominator.
$\dfrac{1}{7} = \dfrac{1}{11} + \dfrac{1}{x}$
$\dfrac{1}{7}-\dfrac{1}{11}=\dfrac{1}{x}$
$\dfrac{11}{77}-\dfrac{1}{11}=\dfrac{1}{x}$
$\dfrac{11}{77}-\dfrac{7}{77}=\dfrac{1}{x}$
$\dfrac{4}{77}=\dfrac{1}{x}$
$\dfrac{4x}{77}=1$
$4x=77$
$x=\dfrac{77}{4}$
$x=19.25$
Thus, it will take Jamie $19.25$ hours or $19$ hours and $15$ minutes to trim the hedges on her own.
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Concept

## Inverse Functions to Rational Functions

A function is said to be invertible if each $y$-value corresponds to exactly one $x$-value. One way to determine if a function is invertible, is to draw its graph and perform the horizontal line test. The same holds true for rational function.
Exercise

Determine if $f(x)=\frac{3}{x^2-5}$ is invertible. Then, find the inverse of $f(x).$

Solution

A function is invertible if its inverse is a function. One way to determine this is to perform the horizontal line test on the graph of the function. We can move an imaginary horizontal line across the graph. If it intersects the graph at more than one point anywhere, the inverse is not a function. We'll draw an arbitrary horizontal line. Sometimes more than one must be drawn. Since the line intersects the graph twice, the inverse is not a function. Thus, $f$ is not invertible. However, that doesn't stop us from finding the inverse of $f(x).$ To do this, we can replace $f(x)$ with $y$ in the function rule, then switch $x$ and $y.$ $y=\dfrac{3}{x^2-5} \quad \Leftrightarrow \quad x=\dfrac{3}{y^2-5}$ We now solve this equation for $y.$
$x=\dfrac{3}{y^2-5}$
$\left(y^2-5\right)x=3$
$y^2-5=\dfrac{3}{x}$
$y^2=\dfrac{3}{x}+5$
$y=\pm \sqrt{\dfrac{3}{x}+5}$
Thus, the inverse of $f(x)$ is $f^{\text{-} 1}=\pm \sqrt{\dfrac{3}{x}+5}.$
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