Question

or extraneous solutions.

1. \\(\frac{1}{3n} = \frac{1}{3} + \frac{1}{n}\\)

Explanation:

Step1: Find common denominator

The common denominator of \(3n\), \(3\), and \(n\) is \(3n\). Multiply each term by \(3n\) to eliminate denominators:
\(3n\times\frac{1}{3n}=3n\times\frac{1}{3}+3n\times\frac{1}{n}\)
Simplify each term: \(1 = n + 3\)

Step2: Solve for \(n\)

Subtract \(n\) and \(3\) (or directly subtract 3 from both sides):
\(1 - 3 = n\)
So, \(n=-2\)

Step3: Check for extraneous solutions

Substitute \(n = - 2\) back into the original equation:
Left - hand side: \(\frac{1}{3\times(-2)}=-\frac{1}{6}\)
Right - hand side: \(\frac{1}{3}+\frac{1}{-2}=\frac{2 - 3}{6}=-\frac{1}{6}\)
Since LHS = RHS, \(n=-2\) is a valid solution. Also, we need to check the domain: \(n
eq0\) (because denominators \(3n\) and \(n\) cannot be zero), and \(-2
eq0\), so it is not extraneous.

Answer:

\(n = - 2\)