Question

The solution of the equations $\frac{m}{x}+\frac{n}{y}=a$ and $\frac{n}{x}+\frac{m}{y}=b$ is given by

• A. $x=\frac{n^2+m^2}{am-bn},y=\frac{m^2-n^2}{bm-an}$
• B. $x=\frac{m^2-n^2}{am-bn},y=\frac{n^2-m^2}{bm-an}$
• C. $x=\frac{m^2-n^2}{am-bn},y=\frac{m^2-n^2}{bm-an}$
• D. $x=\frac{n^2-m^2}{am-bn},y=\frac{n^2-m^2}{bm-an}$

Answer 1

Answer: (C)

$x=\frac{m^2-n^2}{am-bn},y=\frac{m^2-n^2}{bm-an}$

$\frac{m}{x}+\frac{n}{y}=a,\frac{n}{x}+\frac{m}{y}=b$

Let, $\frac{1}{x}=j$ and $\frac{1}{y}=k$
$\frac{m}{x}+\frac{n}{y}=a,\frac{n}{x}+\frac{m}{y}=b$

$\therefore mj+nk=a$ ...(1)

and, $nj+mk=b$ ...(2)
Multiply eq(1) by n and eq(2) by m and subtract,
$mnj+n^{2}k-mnj-m^{2}k=an-bm$
$\therefore k(n^2-m^2)=an-bm$

$k=\frac{an-bm}{n^2-m^2}$
Multiply eq(1) by m and eq(2) by n and subtract,
$m^{2}j+nmk-n^{2}j-mnk=am-bn$
$j(m^2-n^2)=am-bn$
$j=\frac{am-bn}{m^2-n^2}$
Thus, $x=\frac{1}{j}=\frac{m^2-n^2}{am-bn}$

and $y=\frac{1}{k}=\frac{n^2-m^2}{an-bm}$