Question

1. $\frac{1}{x} - \frac{4}{3x^2} = \frac{1}{3x^2}$

Explanation:

Step1: Eliminate denominators

Multiply each term by \(3x^2\) (the least common denominator) to get rid of the fractions:
\(3x^2 \cdot \frac{1}{x} - 3x^2 \cdot \frac{4}{3x^2} = 3x^2 \cdot \frac{1}{3x^2}\)
Simplify each term:
\(3x - 4 = 1\)

Step2: Solve for \(x\)

Add 4 to both sides:
\(3x - 4 + 4 = 1 + 4\)
\(3x = 5\)

Divide both sides by 3:
\(x = \frac{5}{3}\)

Step3: Check for extraneous solutions

Substitute \(x = \frac{5}{3}\) back into the original equation to verify.
Left - hand side: \(\frac{1}{\frac{5}{3}}-\frac{4}{3(\frac{5}{3})^2}=\frac{3}{5}-\frac{4}{3\times\frac{25}{9}}=\frac{3}{5}-\frac{4}{\frac{25}{3}}=\frac{3}{5}-\frac{12}{25}=\frac{15 - 12}{25}=\frac{3}{25}\)
Right - hand side: \(\frac{1}{3(\frac{5}{3})^2}=\frac{1}{3\times\frac{25}{9}}=\frac{1}{\frac{25}{3}}=\frac{3}{25}\)
Since LHS = RHS, \(x=\frac{5}{3}\) is a valid solution.

Answer:

\(x = \frac{5}{3}\)