Question

Solve each of the following systems of equations by the method of cross-multiplication 

${\displaystyle \frac{x+y}{xy}=2}$

${\displaystyle \frac{x-y}{xy}=6}$

Answer 1

The given system of equations is

${\displaystyle \frac{x+y}{xy}=2}$

${\displaystyle \Rightarrow \frac{x}{xy}+\frac{y}{xy}=2}$

${\displaystyle \Rightarrow \frac{1}{y}+\frac{1}{x}=2}$

${\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=2}$ ............(i)

And

${\displaystyle \frac{x-y}{xy}=6}$

${\displaystyle \Rightarrow \frac{x}{xy}-\frac{y}{xy}=6}$

${\displaystyle \Rightarrow \frac{1}{y}-\frac{1}{x}=6}$

${\displaystyle \Rightarrow \frac{1}{x}-\frac{1}{y}=6}$   ......(ii)

Taking ${\displaystyle u=\frac{1}{x}\text{and}v=\frac{1}{y}}$ we get

${\displaystyle u+v=2\Rightarrow u+v-2=0}$ ....(iii)

And u - v = -6 => u - v+ 6 = 0 ........(iv)

Here

${\displaystyle a_1=1,b_1=1,c_1=-2}$

${\displaystyle a_2=1,b_2=-1,c_2=6}$

By cross multiplication

${\displaystyle \Rightarrow \frac{u}{1\times 6-(-2)\times (-1)}=\frac{v}{1\times 6-(-2)\times 1}=\frac{1}{1\times (-1)-1\times 1}}$

${\displaystyle \Rightarrow \frac{u}{6-2}=\frac{-v}{6+2}=\frac{1}{-1-1}}$

${\displaystyle \Rightarrow \frac{u}{4}=\frac{-v}{8}=\frac{1}{-2}}$

Now, ${\displaystyle \frac{u}{4}=\frac{1}{-2}}$

${\displaystyle \Rightarrow u=\frac{4}{-2}=-2}$

And ${\displaystyle \frac{-v}{8}=\frac{1}{-2}}$

${\displaystyle \Rightarrow -v =\frac{8}{-2}=-4}$

${\displaystyle \Rightarrow -v=-4}$

=> v = 4

Now ${\displaystyle x=\frac{1}{u}=\frac{-1}{2},y=\frac{1}{v}=\frac{1}{4}}$

Hence ${\displaystyle x=\frac{-1}{2},y=\frac{1}{4}}$is the solution of the given system of equations.