Solve the following system of equations by the method of cross-multiplication:
$\frac{5}{(x\ +\ y)}\ –\ \frac{2}{(x\ -\ y)}\ =\ -1$
$\frac{15}{(x\ +\ y)}\ +\ \frac{7}{(x\ –\ y)}\ =\ 10$

Given:

The given system of equations is:

$\frac{5}{(x\ +\ y)}\ –\ \frac{2}{(x\ -\ y)}\ =\ -1$

$\frac{15}{(x\ +\ y)}\ +\ \frac{7}{(x\ –\ y)}\ =\ 10$

 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,

$5u-2v=-1$ and $15u+7v=10$

$5u-2v+1=0$ and $15u+7v-10=0$

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,

$a_1=5, b_1=-2, c_1=1$ and $a_2=15, b_2=7, c_2=-10$

Therefore,

$\frac{u}{(-2)\times(-10)-7\times1}=\frac{-v}{5\times(-10)-15\times1}=\frac{1}{5\times7-(-2)\times15}$

$\frac{u}{20-7}=\frac{-v}{-50-15}=\frac{1}{35+30}$

$\frac{u}{13}=\frac{-v}{-65}=\frac{1}{65}$

$\frac{u}{13}=\frac{1}{65}$ and $\frac{-v}{-65}=\frac{1}{65}$

$u=\frac{13\times1}{65}$ and $-v=\frac{-65\times1}{65}$

$u=\frac{13}{65}$ and $-v=\frac{-65}{65}$

$u=\frac{1}{5}$ and $-v=-1$

$u=\frac{1}{5}$ and $v=1$

This implies,

$x+y=\frac{1}{u}=\frac{1}{\frac{1}{5}}=5$---(i)

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

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

$x+y+x-y=5+1$

$2x=6$

$x=\frac{6}{2}=3$

Substituting $x=3$ in equation (i), we get,

$3+y=5$

$y=5-3=2$

The solution of the given system of equations is $x=3$ and $y=2$.

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Updated on: 10-Oct-2022

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