Solve each of the following systems of equations by the method of cross-multiplication :
$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$
where $x \neq 0$ and $y \neq 0$
GIVEN:
$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
Rewriting the equation again
$\frac{2}{x}+\frac{3}{y}-13=0$
$\frac{5}{x}-\frac{4}{y}+2=0$
Taking $u=\frac{1}{x}$ and $v=\frac{1}{y}$
$2 u+3 v-13=0$$\ldots \ldots(1)$
$5 u-4 v+2=0$...$.(2)$
By cross multiplication method we get from eq. (1) and eq. (2)
$\frac{u}{(6)-(52)}=\frac{-v}{(4)-(-65)}=\frac{1}{(-8)-(15)}$
$\frac{u}{-46}=\frac{-v}{69}=\frac{1}{-23}$
$\Rightarrow \frac{u}{-46}=\frac{1}{-23}$
$\Rightarrow u=2$
And
$\frac{-v}{69}=\frac{1}{-23}$
$-v=\frac{69}{-23}$
$\Rightarrow v=3$
We know that
Taking $u=\frac{1}{x}$ and $v=\frac{1}{y}$
$2=\frac{1}{x} \Rightarrow x=\frac{1}{2}$
and
$3=\frac{1}{y} \Rightarrow y=\frac{1}{3}$
Hence we get the value of $x=\frac{1}{2}$ and $\mathrm{y}=\frac{1}{3}$