In the previous target, you learned how to perform operations on rational expressions. We will now solve equations that involve these expressions. You will extend that knowledge in future classes when you graph rational functions. Being able to solve an equation with rational expressions will naturally transfer to finding the domain and zeros of those functions.
Steps for solving equations with rational expressions:
- Find the LCD (least common denominator) of the expressions in the equation.
- Identify any potential extraneous solutions (values that would cause the denominator to be \(0\)).
- Multiply the entire equation by the LCD.
- Simplify and solve.
- Check to make sure you don’t have any extraneous solutions.
In terms of potential extraneous solutions, we need to eliminate any solutions that would make the denominator \(0\) in the original equation. The most straightforward way to do this is to identify which values \(x\) cannot be by finding all of the values that would make the LCD to be \( 0\). If we get any of those values while solving, we will eliminate those solutions.
Example 1: Solve \(\Large{\frac{x-1}{2x+5}=\frac{x+3}{x-2}}\).
Step 1: The LCD of the expressions in the equation is \(\left(2x+5\right)\left(x-2\right)\). If you need any help on finding the LCD, refer to the previous target.
Step 2: If we set the LCD equal to \(0\), we get \(\left(2x+5\right)\left(x-2\right)=0\). This means that the potential extraneous solutions are \(-\frac{5}{2}\) and \(2\) because those would be the values of x that would cause the original equation to have expressions that are undefined. If we get either \(-\frac{5}{2}\) or \(2\) when we solve, we will eliminate those solutions.
Step 3: We will multiply the entire equation by the LCD.
\(\left(2x+5\right)\left(x-2\right)\left(\large\frac{x-1}{2x+5}\right)\normalsize=\left(2x+5\right)\left(x-2\right)\left(\large\frac{x+3}{x-2}\right)\)
Step 4: Notice that for each of the rational expressions in the original equation, there will be factors that divide out when multiplying by the LCD. For example, in the first fraction, \(\left(2x+5\right)\) will divide out, and in the second fraction, \(\left(x-2\right)\) will divide out. This leaves us with the following equation which we can simplify and solve.
\(\begin{align}\left(x-2\right)\left(x-1\right)&=\left(2x+5\right)\left(x+3\right)\\
x^2-3x+2&=2x^2+11x+15\\
x^2+14x+13&=0\\
\left(x+13\right)\left(x+1\right)&=0\\
x&=-13,-1\end{align}\)
Step 5: When we check our solutions above with the potential extraneous solutions we identified in Step 1, we notice that none of these need to be excluded so the solutions to the equation are \(x=-13,-1\).
Example 2: Solve \(\Large{\frac{2x-3}{x+5}+\frac{x+12}{x^2+3x-10}=\frac{-x+4}{x-2}}\).
Example 3: Solve \(\large{\frac{2x-5}{x-4}-\normalsize{2}=\large\frac{6}{x^2-16}}\).
Step 1: By factoring the denominator on the right side of the equation, we get:
\(\large{\frac{2x-5}{x-4}-\normalsize{2}=\large\frac{6}{\left(x+4\right)\left(x-4\right)}}\)
The LCD of these expressions is \(\left(x+4\right)\left(x-4\right)\).
Step 2: If we set the LCD equal to \(0\) and solve, we get potential extraneous solutions of \(4\) and \(-4\).
Step 3: We will multiply the entire equation by the LCD (including the constant!).
\(\left(x+4\right)\left(x-4\right)\left(\large\frac{2x-5}{x-4}\right)-2\left(x+4\right)\left(x-4\right)=\left(\large\frac{6}{\left(x+4\right)\left(x-4\right)}\right)\left(x+4\right)\left(x-4\right)\)
Step 4: Just like in the other examples, we have some factors that divide out and we are left with the following equation to solve.
\(\begin{align}\left(x+4\right)\left(2x-5\right)-2\left(x+4\right)\left(x-4\right)&=6\\
2x^2+3x-20-2\left(x^2-16\right)&=6\\
2x^2+3x-20-2x^2+32&=6\\
3x+12&=6\\
3x&=-6\\
x&=-2\end{align}\)
Step 5: The solution we obtained above is not one of the potential extraneous solutions we identified in Step 2 so the solution to this equation is \(x=-2\).
Quick Check
Solve each of the following equations.
1) \(\Large\frac{x-5}{x^2-8x}=\frac{2}{x}\)
2) \(\Large\frac{6}{x^2-x-12}+\frac{x-1}{x+3}=-4\)
3) \(\Large\frac{x}{x-1}+\frac{5}{x+3}=\frac{4}{x+3}\)
Quick Check Solutions
Solve each of the following equations.
1) \(\Large\frac{x-5}{x^2-8x}=\frac{2}{x}\)
2) \(\Large\frac{6}{x^2-x-12}+\frac{x-1}{x+3}=-4\)
3) \(\Large\frac{x}{x-1}+\frac{5}{x+3}=\frac{4}{x+3}\)
Quick Check Solutions