Chapter 12.5, Problem 47SR

To determine

To solve the equation and state any extraneous solution.

Answer to Problem 47SR

The solution of the equation is $x\text{=}-\frac{3}{2}$

Extraneous solution of the equation is $x=-1,x=1$

Explanation of Solution

Given information:

The given expression is $\frac{x}{x+1}=\frac{x-6}{x-1}$

Calculation:

The solution of the expression

  $\begin{array}{l}\frac{x}{x+1}=\frac{x-6}{x-1}\\ \frac{x}{x+1}\left(\frac{x-1}{x-1}\right)=\frac{x-6}{x-1}\left(\frac{x+1}{x+1}\right)\left[\text{write }\frac{x}{x+1}\text{ and }\frac{x-6}{x-1}\text{ using LCD,}\left(x+1\right)\left(x-1\right)\right]\\ \frac{x^2-x}{x^2-1}=\frac{x^2-5x-6}{x^2-1}\left[\text{simplify}\right]\\ x^2-x=x^2-5x-6\left[\text{multiply both sides by}\left(x^2-1\right)\right]\\ -x=-5x-6\left[\text{subtract }{x}^2\text{ from both sides}\right]\\ \text{4}x\text{=}-6\left[\text{ add 5}x\text{ to both sides}\right]\\ x=-\frac{3}{2}\left[\text{divide both sides by 4}\right]\end{array}$

Therefore,

Solution of the equation is $x\text{=}-\frac{3}{2}$

For extraneous solution put denominator of the equation $=0$

  $\begin{array}{l}x+1=0\\ x=-1\left[\text{subtract 1 from both sides}\right]\\ x-1=0\\ x=1\left[\text{add 1 to both sides}\right]\end{array}$

Therefore,

Extraneous solution of the equation is $x=-1,x=1$