Question

Solving for x and y by substitution method :

$\frac{m}{x}+\frac{n^2}{y}=1+n\frac{m^2}{x}-\frac{n^2}{y}=m-n,(x\ne 0,y\ne 0)$

x and y are :

• A.
  $m+1,n$
• B.
  $m,n$
• C.
  $m,n-1$
• D.
  $m+1,n+1$

Answer 1

The correct option is B

$m,n$
Let $\frac{1}{x}=a,\frac{1}{y}=b\frac{m}{x}+\frac{n^2}{y}=1+n\Rightarrow am+n^{2}b=1+n...\left(1\right)\frac{m^2}{x}-\frac{n^2}{y}=m-n\Rightarrow a{m}^2-n^{2}b=m-n......\left(2\right)Fromequation\left(1\right),weget:am+n^{2}b=1+n\Rightarrow n^{2}b=1+n-am....\left(3\right)Substituting{n}^{2}b=1+n-amin\left(2\right),weget:a{m}^2-n^{2}b=m-n\Rightarrow a{m}^2+am=m-n+1+n\Rightarrow am(m+1)=(m+1)\Rightarrow am=1\therefore a=\frac{1}{m}Substitutingthevalueofa=\frac{1}{m}inequation\left(1\right),Weget:am+n^{2}b=1+n\Rightarrow m\left(\frac{1}{m}\right)+n^{2}b=1+n\Rightarrow 1+n^{2}b=1+n\Rightarrow n^{2}b=n\Rightarrow b=\frac{n}{n^2}\Rightarrow b=\frac{1}{n}\because a=\frac{1}{m}\Rightarrow \frac{1}{x}=\frac{1}{m}\Rightarrow x=mb=\frac{1}{n}\Rightarrow \frac{1}{y}=\frac{1}{n}\Rightarrow y=n$