Question

If $\frac{1}{x}+\frac{2}{y}=4\mathrm{and}\frac{3}{y}-\frac{1}{x}=11$ then
(a) x = 2, y = 3 (b) x = − 2, y = 3 (c) $x=-\frac{1}{2},y=3$ (d) $x=-\frac{1}{2},y=\frac{1}{3}$

Answer 1

The given system of equations is
$$\begin{gathered} \frac{1}{x}+\frac{2}{y}=4.....\left(i\right) \\ \frac{3}{y}-\frac{1}{x}=11.....\left(ii\right) \end{gathered}$$
Adding (i) and (ii), we get
$$\begin{gathered} \frac{2}{y}+\frac{3}{y}=15 \\ \Rightarrow \frac{5}{y}=15\Rightarrow y=\frac{5}{15}=\frac{1}{3} \end{gathered}$$
Now, putting $y=\frac{1}{3}$ in (i), we have
$\frac{1}{x}+2\times 3=4\Rightarrow \frac{1}{x}=4-6\Rightarrow x=-\frac{1}{2}$
Thus, $x=-\frac{1}{2},y=\frac{1}{3}$.
Hence, the correct answer is option (d).