Chapter 11.5, Problem 6LC

To determine

A rational equation that has one solution and one extraneous solution.

Answer to Problem 6LC

  $\frac{x-4}{x^2-4}=\frac{-2}{x-2}$

Explanation of Solution

Concept Used:

• Least Common Denominator (LCD) is the smallest number that is divisible by the denominator of all the fractions that are being added or subtracted.
• Solving an equation containing rational expressions
1. Multiply both sides of equations by LCD(least common denominator) of all denominators.
2. Remove any grouping symbols and solve the resulting equation for the variable asked in equation

Calculation:

One such rational equation that has one solution and one extraneous solution is $\frac{x-4}{x^2-4}=\frac{-2}{x-2}$ .

Because the solutions of the equation can be found by cross multiplying as shown below

  $\begin{array}{c}\frac{x-4}{x^2-4}=\frac{-2}{x-2}\\ \left(x-4\right)\cdot \left(x-2\right)=-2\cdot \left(x^2-4\right)\\ x^2-2x-4x+8=-2x^2+8\\ 3x^2-6x=0\\ 3x\left(x-2\right)=0\\ 3x=0\text{or}x-2=0\\ x=0\text{or}x=2\end{array}$

Thus, the solutions of the given equation are $x=0,2$ .

Now to check the solution, substitute $x=0$ in the given equation and check whether left hand side value is equal to the right hand side value or no. So, it gives

  $\begin{array}{c}\frac{0-4}{{\left(0\right)}^2-4}=\frac{-2}{0-2}\\ \frac{-4}{-4}=\frac{-2}{-2}\\ 1=1\text{True Statement}\end{array}$

Again to check the other solution, substitute $x=2$ in the given equation and check whether left hand side value is equal to the right hand side value or no. So, it gives

  $\begin{array}{c}\frac{2-4}{{\left(2\right)}^2-4}=\frac{-2}{2-2}\\ \frac{-2}{4-4}=\frac{-2}{0}\\ \text{Undefined}\end{array}$

So, $x=2$ is an extraneous solution.

Thus, the only solution of the given equation is $x=0$ .