Chapter 11.8, Problem 2CYU

To determine

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

Answer to Problem 2CYU

The value is $12$ . Since, $t=12$ results in a zero in the denominator of the original equation, it is an extraneous solution.

Explanation of Solution

Given:

  $\frac{t+3}{5}=\frac{2t+3}{9}$

Calculation:

The equation is solved below.

  $\begin{array}{l}\frac{t+3}{5}=\frac{2t+3}{9}\\ 9\left(t+3\right)=5\left(2t+3\right)\\ 9t+27=10t+15\\ 12=t\end{array}$

Conclusion:

Therefore, the value is $12$ . Since, $t=12$ results in a zero in the denominator of the original equation, it is an extraneous solution.