Solve the following system of equations by the method of cross-multiplication:
$\frac{57}{(x\ +\ y)}\ +\ \frac{6}{(x\ –\ y)}\ =\ 5$
$\frac{38}{(x\ +\ y)}\ +\ \frac{21}{(x\ –\ y)}\ =\ 9$

Given:

The given system of equations is:

$\frac{57}{(x\ +\ y)}\ +\ \frac{6}{(x\ –\ y)}\ =\ 5$

$\frac{38}{(x\ +\ y)}\ +\ \frac{21}{(x\ –\ y)}\ =\ 9$

 To do: 

Here, we have to solve the given system of equations by the method of cross-multiplication.

Solution:  

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

The given system of equations can be written as,

$57u+6v=5$ and $38u+21v=9$

$57u+6v-5=0$ and $38u+21v-9=0$

The solution of a linear pair(standard form) of equations $a_1u+b_1v+c_1=0$ and $a_2u+b_2v+c_2=0$ is given by,

$\frac{u}{b_1c_2-b_2c_1}=\frac{-v}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

Comparing the above equations with the standard form of the equations, we get,

$a_1=57, b_1=6, c_1=-5$ and $a_2=38, b_2=21, c_2=-9$

Therefore,

$\frac{u}{6\times(-9)-21\times(-5)}=\frac{-v}{57\times(-9)-38\times(-5)}=\frac{1}{57\times21-38\times6}$

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

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

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

$u=\frac{51\times1}{969}$ and $-v=\frac{-323\times1}{969}$

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

$u=\frac{1}{19}$ and $-v=\frac{-1}{3}$

$u=\frac{1}{19}$ and $v=\frac{1}{3}$

This implies,

$x+y=\frac{1}{u}=\frac{1}{\frac{1}{19}}=19$---(i)

$x-y=\frac{1}{v}=\frac{1}{\frac{1}{3}}=3$----(ii)

Adding equations (i) and (ii), we get,

$x+y+x-y=19+3$

$2x=22$

$x=\frac{22}{2}=11$

Substituting $x=11$ in equation (i), we get,

$11+y=19$

$y=19-11=8$

The solution of the given system of equations is $x=11$ and $y=8$.