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

GIVEN:

$\frac{57}{x+y}+\frac{6}{x-y}=5$

$\frac{38}{x+y}+\frac{21}{x-y}=9$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

$\frac{57}{x+y}+\frac{6}{x-y}-5=0$

$\frac{38}{x+y}+\frac{21}{x-y}-9=0$

let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$

Now rewriting the given equation as

$57 u+6 v-5=0$..(1)

$38 u+21 v-9=0$..$.(2)$

By cross multiplication method we get

$\frac{u}{(-9 \times 6)-(-5 \times 21)}=\frac{-v}{(-9 \times 57)-(-5 \times 38)}=\frac{1}{(21 \times 57)-(38 \times 6)}$

$\frac{u}{(-54)-(-105)}=\frac{-v}{(-513)-(-190)}=\frac{1}{(1197)-(228)}$

$\frac{u}{51}=\frac{-v}{-323}=\frac{1}{969}$

$\frac{u}{51}=\frac{v}{323}=\frac{1}{969}$

Consider the following for u

$\frac{u}{51}=\frac{1}{969}$

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

Consider the following for v

$\frac{\mathrm{V}}{323}=\frac{1}{969}$

$v=\frac{1}{3}$

We know that

$\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$

$\frac{1}{x+y}=\frac{1}{19}$

$x+y=19$ ..(3)

$\frac{1}{x-y}=\frac{1}{3}$

$x-y=3$ ...(4)

Now adding eq. (3) and (4) we get $x=11$

And after substituting the value of $\mathrm{x}$ in eq. (4) we get $y=8$

Hence we get the value of $x=11$ and $y=8$