Chapter 12.5, Problem 48SR

To determine

To solve the equation and state any extraneous solution.

Answer to Problem 48SR

The solution of the equation is $\text{n=}-3$

No extraneous solution is possible.

Explanation of Solution

Given information:

The given expression is $\frac{2n}{3}+\frac{1}{2}=\frac{2n-3}{6}$

Calculation:

The solution of the expression

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

Therefore,

Solution of the equation is $\text{n=}-3$

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

The denominators of the equation is integers,

Therefore,

No extraneous solution is possible