Chapter 11.8, Problem 3CYU

To determine

To solve: the each equation and state the extraneous solution.

Answer to Problem 3CYU

The value is $0,-\frac{4}{5}$ . Since, $a=-\frac{4}{5}$ results in a zero in the denominator of the original equation, it is an extraneous solution.

Explanation of Solution

Given:

  $\frac{a+3}{a}-\frac{6}{5a}=\frac{1}{a}$

Calculation:

The equation is solved below.

  $\begin{array}{l}\frac{a+3}{a}-\frac{6}{5a}=\frac{1}{a}\\ \frac{5a+15-6}{5a}=\frac{1}{a}\\ 5a^2+9a=5a\\ 5a^2+4a=0\\ a\left(5a+4\right)=0\\ a=0,-\frac{4}{5}\end{array}$

Conclusion:

Therefore, the value is $0,-\frac{4}{5}$ . Since, $a=-\frac{4}{5}$ results in a zero in the denominator of the original equation, it is an extraneous solution.