SSC CHSL Model Paper 3

Given,
$\frac{3}{\mathit{x}+\mathit{y}}+\frac{2}{\mathit{x}-\mathit{y}}=2$ ...(i)
and $\frac{9}{\mathit{x}+\mathit{y}}-\frac{4}{\mathit{x}-\mathit{y}}=1$ ....(ii)
Let $\frac{1}{\mathit{x}+\mathit{y}}=\mathit{a}$ and $\frac{1}{\mathit{x}-\mathit{y}}=\mathit{b}$
∴ $3\mathit{a}+2\mathit{b}=2$ ....(iii)
$9\mathit{a}-4\mathit{b}=1$ ... (iv)
On multiplyin g equation ( iii) by 2 and addition of equation (iv) and new one, then we get
$6\mathit{a}+4\mathit{b}=4$
$9\mathit{a}-4\mathit{b}=1$
$\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}$
$15\mathit{a}=5$
⇒ $\mathit{a}=\frac{5}{15}=\frac{1}{3}$
∴ $\frac{1}{\mathit{x}+\mathit{y}}=\frac{1}{3}$
⇒ $\mathit{x}+\mathit{y}=3$ ...(v)
On putting the value of (a) in equation (iii), we get
$3\times \frac{1}{3}+2\mathit{b}=2$
⇒ $2\mathit{b}=2-1=1$
⇒ $\mathit{b}=\frac{1}{2}\Rightarrow \frac{1}{\mathit{x}-\mathit{y}}=\frac{1}{2}$
⇒ $\mathit{x}-\mathit{y}=2$ ....(vi)
∴ $\mathit{x}+\mathit{y}=3$
$\mathit{x}-\mathit{y}=2$
$\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}\mathit{–}$
$2\mathit{x}=5$
⇒ $\mathit{x}=\frac{5}{2}$
From equation (v),
$\mathit{y}=3-\frac{5}{2}=\frac{1}{2}$
$\therefore \frac{\mathit{x}}{\mathit{y}}=\frac{\frac{5}{2}}{\frac{1}{2}}=5$