The techniques for solving rational equations are extensions of techniques you already know. Recall that when there are fractions in an equation you can multiply through by the denominator to clear the fraction. The same technique helps turn rational expressions into polynomials that you already know how to solve. When you multiply by a constant there is no problem, but when you multiply through by a value that varies and could possibly be zero interesting things happen.

Since every equation is trivially true when both sides are multiplied by zero, how do you account for this when solving rational equations?

### Finding Solutions to Rational Equations

The first step in solving rational equations is to transform the equation into a polynomial equation. This is accomplished by clearing the fraction which means multiplying the entire equation by the common denominator of all the rational expressions. Then you should solve using what you already know. The last thing to check once you have the solutions is that they do not make the denominators of any part of the equation equal to zero when substituted back into the original equation. If so, that solution is called **extraneous** and is a “fake” solution that was introduced when both sides of the equation were multiplied by a number that happened to be zero.

Take the following rational equation:

\begin{align*}x-\frac{5}{x+3}=12\end{align*}

To find the solutions of the equation, first multiply all parts of the equation by \begin{align*}(x+3)\end{align*}

\begin{align*}x(x+3)-5 &= 12(x+3)\\
x^2+3x-5-12x-36 &= 0\\
x^2-9x-41 &= 0\\
x &= \frac{-(-9) \pm \sqrt{(-9)^2-4 \cdot 1 \cdot (-41)}}{2 \cdot 1}\\
x &= \frac{9 \pm 7 \sqrt{5}}{2}\end{align*}

The only potential extraneous solution would have been -3 since that is the number that makes the denominator of the original equation zero. Therefore, both answers are possible.

### Examples

#### Example 1

Earlier, you were asked to account for the extra solutions introduced when both sides of an equation are multiplied by a variable. In order to deal with the possible extra solutions, you must check each solution to see if it makes the denominator of any fraction in the original equation zero. If it does, it is called an extraneous solution.

#### Example 2

Solve the following rational equation

\begin{align*}\frac{3x}{x+4}-\frac{1}{x+2}=-\frac{2}{x^2+6x+8}\end{align*}

Multiply each part of the equation by the common denominator of \begin{align*}x^2+6x+8=(x+2)(x+4)\end{align*}

\begin{align*}(x+2)(x+4) \left[\frac{3x}{x+4}-\frac{1}{x+2}\right] &= \left[\frac{-2}{(x+2)(x+4)}\right](x+2)(x+4)\\
3x(x+2)-(x+4) &= -2\\
3x^2+6x-x-2 &= 0\\
3x^2+5x-2 &= 0\\
(3x-1)(x+2) &= 0\\
x &= \frac{1}{3}, -2\end{align*}

Note that -2 is an extraneous solution. The only actual solution is \begin{align*}x=\frac{1}{3}\end{align*}

#### Example 3

Solve the following rational equation for \begin{align*}y\end{align*}

\begin{align*}x=2+\frac{1}{2+\frac{1}{y+1}}\end{align*}

This question can be done multiple ways. You can use the clearing fractions technique twice.

\begin{align*}\left(2+\frac{1}{y+1}\right)x &= \left[2+\frac{1}{2+\frac{1}{y+1}}\right] \left(2+\frac{1}{y+1}\right)\\
2x+\frac{x}{y+1} &= 2\left(2+\frac{1}{y+1}\right)+1\\
2x+\frac{x}{y+1} &= 4+\frac{2}{y+1}+1\\
(y+1)\left[2x+\frac{x}{y+1}\right] &= \left[5+\frac{2}{y+1}\right](y+1)\\
2x(y+1)+x &= 5(y+1)+2\\
2xy+2x+x &= 5y+5+2\end{align*}

Now just get the \begin{align*}y\end{align*}

\begin{align*}2xy-5y &= -3x+7\\
y(2x-5) &= -3x+7\\
y &= \frac{-3x+7}{2x-5}\end{align*}

#### Example 4

Solve the following rational equation.

\begin{align*}\frac{3x}{x-5}+4=x\end{align*}

\begin{align*}\frac{3x}{x-5}+4=x\end{align*}

\begin{align*}3x+4x-20 &= x^2-5x\\
0 &= x^2-12x+20\\
0 &= (x-2)(x-10)\\
x &= 2, 10\end{align*}

Neither solution is extraneous.

#### Example 5

In electrical circuits, resistance can be solved for using rational expressions. This is an electric circuit diagram with three resistors. The first resistor \begin{align*}R_1\end{align*}

The equation of value is:

\begin{align*}R=R_1+\frac{R_2R_3}{R_2+R_3}\end{align*}

\begin{align*}R = R_1+\frac{R_2R_3}{R_2+R_3}\end{align*}

\begin{align*}100 &= 22+\frac{x \cdot 22}{x+22}\\
78(x+22) &= 22x\\
78x+1716 &= 22x\\
56x &= -1716\\
x &= -30.65\end{align*}

A follow up question would be to ask whether or not ohms can be negative which is beyond the scope of this text.

### Review

Solve the following rational equations. Identify any extraneous solutions.

- \begin{align*}\frac{2x-4}{x}=\frac{16}{x}\end{align*}
- \begin{align*}\frac{4}{x+1}-\frac{x}{x+1}=2\end{align*}
- \begin{align*}\frac{5}{x+3}+\frac{2}{x-3}=1\end{align*}
- \begin{align*}\frac{3}{x-4}-\frac{5}{x+4}=6\end{align*}
- \begin{align*}\frac{x}{x+1}-\frac{6}{x+2}=4\end{align*}
- \begin{align*}\frac{x}{x-4}-\frac{4}{x-4}=8\end{align*}
- \begin{align*}\frac{4x}{x-2}+3=1\end{align*}
- \begin{align*}\frac{-2x}{x+1}+6=-x\end{align*}
- \begin{align*}\frac{1}{x+2}+1=-2x\end{align*}
- \begin{align*}\frac{-6x-3}{x+1}-3=-4x\end{align*}
- \begin{align*}\frac{x+3}{x}-\frac{3}{x+3}=\frac{6}{x^2+3x}\end{align*}
- \begin{align*}\frac{x-4}{x}-\frac{2}{x-4}=\frac{8}{x^2-4x}\end{align*}
- \begin{align*}\frac{x+6}{x}-\frac{2}{x+6}=\frac{12}{x^2+6x}\end{align*}
- \begin{align*}\frac{x+5}{x}-\frac{3}{x+5}=\frac{15}{x^2+5x}\end{align*}
- Explain what it means for a solution to be extraneous.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.6.