Solve the following system of equations by the method of cross-multiplication:
$\frac{a^2}{x}-\frac{b^2}{y}=0$
$\frac{a^2b}{x}+\frac{b^2a}{y}=a+b, x, y≠0$

Given:

The given system of equations is:

$\frac{a^2}{x}-\frac{b^2}{y}=0$

$\frac{a^2b}{x}+\frac{b^2a}{y}=a+b, x, y≠0$

 To do: 

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

Solution:  

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

The given system of equations can be written as,

$\frac{a^2}{x}-\frac{b^2}{y}=0$

$a^2u-b^2y=0$........(i)

$\frac{a^2b}{x}+\frac{b^2a}{y}=a+b$

$a^2bu+b^2av-(a+b)=0$.........(ii)

The solution of a linear pair(standard form) of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{-y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

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

$a_1=a^2, b_1=-b^2, c_1=0$ and $a_2=a^2b, b_2=b^2a, c_2=-(a+b)$

Therefore,

$\frac{u}{-b^2\times-(a+b)-(b^2a)\times(0)}=\frac{-v}{a^2\times-(a+b)-a^2b\times(0)}=\frac{1}{a^2\times(b^2a)-a^2b\times (-b^2)}$

$\frac{u}{ab^2+b^3-0}=\frac{-v}{-a^3-a^2b+0}=\frac{1}{a^3b^2+a^2b^3}$

$\frac{u}{b^2(a+b)}=\frac{-v}{-a^2(a+b)}=\frac{1}{a^2b^2(a+b)}$

$u=\frac{b^2(a+b)}{a^2b^2(a+b)}$ and $-v=\frac{-a^2(a+b)}{a^2b^2(a+b)}$

$u=\frac{1}{a^2}$ and $-v=\frac{-1}{b^2}$

$u=\frac{1}{a^2}$ and $v=\frac{1}{b^2}$

This implies,

$x=\frac{1}{\frac{1}{a^2}}$ and $y=\frac{1}{\frac{1}{b^2}}$

$x=a^2$ and $y=b^2$

The solution of the given system of equations is $x=a^2$ and $y=b^2$.

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Updated on: 10-Oct-2022

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