Question

If $\frac{2}{x}+\frac{3}{y}=6\mathrm{and}\frac{1}{x}+\frac{1}{2y}=2$ then
(a) $x=1,y=\frac{2}{3}$ (b) $x=\frac{2}{3},y=1$ (c) $x=1,y=\frac{3}{2}$ (d) $x=\frac{3}{2},y=1$

Answer 1

The given equations are
$$\begin{gathered} \frac{2}{x}+\frac{3}{y}=6.....\left(i\right) \\ \frac{1}{x}+\frac{1}{2y}=2.....\left(ii\right) \end{gathered}$$
Multiplying (ii) by 2 and subtracting it from (ii), we get
$$\begin{gathered} \frac{3}{y}-\frac{1}{y}=6-4 \\ \Rightarrow \frac{2}{y}=2 \\ \Rightarrow y=1 \end{gathered}$$
Substituting y = 1 in (ii), we get
$$\begin{gathered} \frac{1}{x}+\frac{1}{2}=2 \\ \Rightarrow \frac{1}{x}=2-\frac{1}{2}=\frac{3}{2} \\ \Rightarrow x=\frac{2}{3} \end{gathered}$$
Hence, the correct answer is option (b).