Chapter 5, Problem 22T

Solution

To calculate: The solution of $\frac{x-2}{x-1}=\frac{x+2}{x+4}$ and check for extraneous solutions.

The required solution is $x=6$ and it is not extraneous.

Given information:

The equation $\frac{x-2}{x-1}=\frac{x+2}{x+4}$.

Calculation:

Consider the equation $\frac{x-2}{x-1}=\frac{x+2}{x+4}$.

Cross multiply.

  $\left(x-2\right)\left(x+4\right)=\left(x+2\right)\left(x-1\right)$

Use distributive property.

  $x^2+4x-2x-8=x^2-x+2x-2$

Simplify.

  $\begin{array}{c}{x}^2+2x-8=x^2+x-2\\ 2x-8=x-2\\ x=6\end{array}$

So, the solution is $x=6$.

To check, substitute $x=6$ in $\frac{x-2}{x-1}=\frac{x+2}{x+4}$.

  $\begin{array}{c}\frac{6-2}{6-1}=\frac{6+2}{6+4}\\ \frac{4}{5}=\frac{8}{10}\\ \frac{4}{5}=\frac{4}{5}\end{array}$

Here, the left hand side of the equation is equal to the right hand side and thus $x=6$ is not extraneous solution.

Therefore, the required solution is $x=6$ and it is not extraneous.