Question

Question 9 (vii)
Solve the following pairs of equation
$\frac{2xy}{x+y}=\frac{3}{2},\frac{xy}{2x-y}=\frac{-3}{10},x+y\ne 0,2x-y\ne 0$

Answer 1

Given pair of equations is
$\frac{2xy}{x+y}=\frac{3}{2},wherex+y\ne 0\Rightarrow \frac{x+y}{2xy}=\frac{2}{3}\frac{x}{xy}+\frac{y}{xy}=\frac{4}{3}\frac{1}{y}+\frac{1}{x}=\frac{4}{3}and\frac{xy}{2x-y}=\frac{-3}{10},where2x-y\ne 0\Rightarrow \frac{2x-y}{xy}=\frac{-10}{3}$
$\Rightarrow \frac{2x}{xy}-\frac{y}{xy}=\frac{-10}{3}\Rightarrow \frac{2}{y}-\frac{1}{x}=\frac{-10}{3}$
Now, piu $\frac{1}{x}=uand\frac{1}{y}=v$, then the pair of equations becomes
$v+u=\frac{4}{3}and2v-u=\frac{-10}{3}$
on adding both equations, we get
$3v=\frac{4}{3}-\frac{10}{3}=\frac{-6}{3}\Rightarrow 3v=-2\Rightarrow v=\frac{-2}{3}$
Now, put the value of v in Eq. (iii), we get
$\frac{-2}{3}+u=\frac{4}{3}\Rightarrow u=\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2x=\frac{1}{u}=\frac{1}{2}and\frac{1}{v}=\frac{1}{\left(\frac{-2}{3}\right)}=\frac{-3}{2}$
Hence, the required values of x and y are $\frac{1}{2}and\frac{-3}{2}$ respectively