If \(\frac{2}{x}+\frac{3}{y}=13\) and \(\frac{3}{x}-\frac{2}{y}=4\), what is the value of \(\frac{1}{x}\)?

Correct answer: \(\frac{38}{13}\)

Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). The equations become \(2u+3v=13\) and \(3u-2v=4\). Multiply the first equation by 2 and the second by 3 to get \(4u+6v=26\) and \(9u-6v=12\). Adding them gives \(13u=38\), so \(u=\frac{38}{13}\). Hence, \(\frac{1}{x}=\frac{38}{13}\). The value \(\frac{31}{13}\) is for \(v=\frac{1}{y}\), not for \(\frac{1}{x}\). Exam tip: For reciprocal terms, substitute \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) first to form linear equations.