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

Question:

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 $y \neq 0$

Solution:

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 $\mathrm{y}=\frac{1}{3}$