Subsection Solving Equations with Fractions Algebraically
If we need a more accurate value for the windspeed, we can solve the equation
\begin{equation*}
\frac{60}{15-x}=9
\end{equation*}
To solve an equation involving an algebraic fraction, we multiply each side of the equation by the denominator of the fraction. This has the effect of clearing the fraction, giving us an equivalent equation without fractions.
Example 7.92.
Solve the equation \(~~\displaystyle{\frac{60}{15-x}=9}\)
Solution.
We multiply both sides of the equation by \(\alert{15 - x}\) to obtain
\begin{align*}
\alert{(15 - x)}\frac{60}{15-x}\amp =9\alert{(15 - x)}\\
60 \amp = 9(15 - x)\amp\amp \blert{\text{Apply the distributive law.}}
\end{align*}
From here we can proceed as usual.
\begin{align*}
60 \amp = 135 - 9x\amp\amp \blert{\text{Subtract 135 from both sides.}}\\
-75 \amp = -9x\amp\amp \blert{\text{Divide by }-9.}\\
8.\overline{3} \amp = x
\end{align*}
The windspeed was \(8.\overline{3}\text{,}\) or \(8\frac{1}{3}\) miles per hour.
Checkpoint 7.93. Practice 1.
Solve \(~\displaystyle{\frac{x^2}{x+4}}=2\)
If the equation contains more than one fraction, we can clear all the denominators at once by multiplying both sides by the LCD of the fractions.
Example 7.94.
Rani times herself as she kayaks 30 miles down the Derwent River with the help of the current. Returning upstream against the current, she manages only 18 miles in the same amount of time. Rani knows that she can kayak at a rate of 12 miles per hour in still water. What is the speed of the current?
Solution.
If we let \(x\) represent the speed of the current, we can use the formula \(\text{time} = \dfrac{\text{distance}}{\text{rate}}\) to fill in the following table.
|
Distance |
Rate |
Time |
| Downstream |
\(30\) |
\(12+x\) |
\(\dfrac{30}{12+x}\) |
| Upstream |
\(18\) |
\(12-x\) |
\(\dfrac{18}{12-x}\) |
Because Rani paddled for equal amounts of time upstream and downstream, we have the equation
\begin{equation*}
\frac{30}{12+x}=\frac{18}{12-x}
\end{equation*}
The LCD for the fractions in this equation is \((12 + x)(12 - x)\text{.}\) We multiply both sides of the equation by the LCD to obtain
\begin{align*}
\alert{(12 + x)(12 - x)}\frac{30}{12+x} \amp =\frac{18}{12-x}\alert{(12 + x)(12 - x)}\\
30 (12 - x) \amp = 18 (12 + x)
\end{align*}
Solving this equation, we find
\begin{align*}
360 - 30x \amp = 216 + 18x\\
144 \amp = 48x\\
3 \amp = x
\end{align*}
The speed of the current is \(3\) miles per hour.
Checkpoint 7.95. Practice 2.
Solve \(\displaystyle{\frac{x}{6-x}=\frac{1}{2}}\)
Checkpoint 7.96. Practice 2.
If the variable in an equation appears in the denominator of a fraction, what should you do to solve the equation?
Technology 7.97. Solving Equations with Fractions Graphically.
We can solve the equation in
Example 7.94 graphically by considering two functions, one for each side of the equation. Graph the two functions
\begin{equation*}
Y_1=\frac{30}{12+x} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{30}{12-x}
\end{equation*}
in the window
\begin{align*}
\text{Xmin} \amp = -9.4 \amp\amp \text{Xmax} = 9.4\\
\text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 10
\end{align*}
to obtain the graph shown below.
The function \(Y_1\) gives the time it takes Rani to paddle 30 miles downstream, and \(Y_2\) gives the time it takes her to paddle \(18\) miles upstream. Both of these times depend on the speed of the current, \(x\text{.}\)
We are looking for a value of
\(x\) that makes
\(Y_1\) and
\(Y_2\) equal. This occurs at the intersection point of the two graphs,
\((3, 2)\text{.}\) Thus, the speed of the current is
\(3\) miles per hour, as we found in
Example 7.94. The
\(y\)-coordinate of the intersection point gives the time Rani paddled on each part of her trip:
\(2\) hours each way.
Subsection Extraneous Solutions
A rational function is undefined for any values of \(x\) that make its denominator equal zero. These values are not in the domain of the function, and they therefore cannot be solutions to equations involving the function. Consider the equation
\begin{equation*}
\frac{x}{x-3}=\frac{3}{x-3}+2
\end{equation*}
When we multiply both sides by the LCD, \(x - 3\text{,}\) we obtain
\begin{equation*}
\alert{(x - 3)}\frac{x}{x-3}=\alert{(x - 3)}\frac{3}{x-3}+\alert{(x - 3)}\cdot 2
\end{equation*}
or
\begin{equation*}
x = 3 + 2x - 6
\end{equation*}
whose solution is
\begin{equation*}
x = 3
\end{equation*}
However, \(x = 3\) is not a solution of the original equation. Both sides of the equation are undefined at \(x = 3\text{.}\) If you graph the two functions
\begin{equation*}
Y_1=\frac{x}{x-3} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{3}{x-3}+2
\end{equation*}
you will find that the graphs never intersect, which means that there is no solution to the original equation.
What went wrong with our method of solution? We multiplied both sides of the equation by \(x - 3\text{,}\) which is zero when \(x = 3\text{,}\) so we really multiplied both sides of the equation by zero. Multiplying by zero does not produce an equivalent equation, and false solutions may be introduced.
An apparent solution that does not satisfy the original equation is called an extraneous solution. Whenever we multiply an equation by an expression containing the variable, we should check that the solution obtained is not excluded from the domain of the rational functions involved.
When solving an equation with fractions algebraically, we must be careful to multiply each term of the equation by the LCD, no matter whether each term involves fractions.
Example 7.98.
Solve the equation \(\displaystyle{\frac{6}{x}+1=\frac{1}{x+2}} \) algebraically.
Solve the same equation graphically.
Solution.
To solve the equation algebraically, we multiply both sides by the LCD, \(x(x + 2)\text{.}\) Notice that we multiply each term on the left side by the LCD, to get
\begin{equation*}
\alert{x(x + 2)}\left(\frac{6}{x}+1\right)=\alert{x(x + 2)}\frac{1}{x+2}
\end{equation*}
or
\begin{equation*}
6(x + 2) + x(x + 2) = x
\end{equation*}
We use the distributive law to remove the parentheses and write the result in standard form:
\begin{align*}
6x + 12 + x^2 + 2x\amp = x\\
x^2 + 7x + 12 \amp = 0
\end{align*}
This is a quadratic equation that we can solve by factoring.
\begin{equation*}
(x + 3)(x + 4) = 0
\end{equation*}
so the solutions are \(x = -3\) and \(x = -4\text{.}\) Neither of these values causes either denominator to equal zero, so they are not extraneous solutions.
-
To solve the equation graphically, graph the two functions
\begin{equation*}
Y_1=\frac{6}{x}+1 \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{1}{x+2}
\end{equation*}
in the window
\begin{align*}
\text{Xmin} \amp = -4.7 \amp\amp \text{Xmax} = 4.7\\
\text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10
\end{align*}
as shown in figure (a).
We see that the first graph has an asymptote at \(x = 0\text{,}\) and the second graph has one at \(x = -2\text{.}\) It appears that the two graphs may intersect in the third quadrant, around \(x = -3\text{.}\) To investigate further, we change the window settings to
\begin{align*}
\text{Xmin} \amp = -4.55 \amp\amp \text{Xmax} = -2.2\\
\text{Ymin} \amp = -1.3 \amp\amp \text{Ymax} = -0.3
\end{align*}
to obtain the close-up view shown in figure (b). In this window, we can see that the graphs intersect in two distinct points, and by using the Trace we find that their \(x\)-coordinates are \(x = -3\) and \(x = -4\text{.}\)
Checkpoint 7.99. QuickCheck 1.
Which of these is the correct first step to solve \(\dfrac{1}{x}+\dfrac{1}{2x+1} = 2\) ?
\(\displaystyle x+2x+1=\dfrac{1}{2}\)
\(\displaystyle 2x+1+x=2\)
\(\displaystyle 2x+1+x=2x(2x+1)\)
\(\displaystyle \dfrac{2x+1}{x(2x+1)}+\dfrac{x}{x(2x+1)} = 2\)
Checkpoint 7.100. Practice 3.
Solve \(~\displaystyle{\frac{9}{x^2+x-2}+\frac{1}{x^2-4}=\frac{4}{x-1}} \)
Checkpoint 7.101. Pause and Reflect.
What are extraneous solutions, and when might they occur?
