Question

On solving $\frac{2}{x}+\frac{3}{y}=3,\frac{3}{x}+\frac{6}{y}=5,$ we get:
(a) x = 2, y = 3
(b) x = $\frac{1}{2}$, y = $\frac{1}{3}$
(c) x = 1, y = 3
(d) x = 2, y = 6

Answer 1

The correct option is (c).
The given equations are as follows:
$\frac{2}{x}+\frac{3}{y}=3$
$\frac{3}{x}+\frac{6}{y}=5$
Putting $\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$ in the given equations, we get:
2u + 3v = 3 ...(i)
3u + 6v = 5 ...(ii)
On multiplying (i) by 2, we get:
4u + 6v = 6 ...(iii)
On subtracting (ii) from (iii), we get:
u = 1 ⇒ $\frac{1}{x}=1\Rightarrow x=1$
Putting u = 1 in (i), we get:
2 + 3v = 3 ⇒ 3v = 1 ⇒ v = $\frac{1}{3}$⇒ $\frac{1}{y}=\frac{1}{3}\Rightarrow y=3$
∴​ x = 1 and y = 3