Solve the following system of equations by the method of cross-multiplication:

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

Given:

The given system of equations is:

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

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

 To do: 

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

Solution:  

The given system of equations can be written as,

$\frac{1}{a}x+\frac{1}{b}y-(a+b)=0$

$\frac{1}{a^2}x+\frac{1}{b^2}y-2=0$

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,

$a_1=\frac{1}{a}, b_1=\frac{1}{b}, c_1=-(a+b)$ and $a_2=\frac{1}{a^2}, b_2=\frac{1}{b^2}, c_2=-2$

Therefore,

$\frac{x}{\frac{1}{b}\times(-2)-\frac{1}{b^2}\times-(a+b)}=\frac{-y}{\frac{1}{a}\times(-2)-\frac{1}{a^2}\times-(a+b)}=\frac{1}{\frac{1}{a}\times\frac{1}{b^2}-\frac{1}{a^2}\times \frac{1}{b}}$

$\frac{x}{\frac{-2}{b} +\frac{a}{b^{2}} +\frac{1}{b}} =\frac{-y}{\frac{-2}{a} +\frac{1}{a} +\frac{b}{a^{2}}} =\frac{1}{\frac{1}{ab^{2}} -\frac{1}{a^{2} b}}$

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

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

$x=\frac{\frac{a-b}{b^{2}}}{\frac{a-b}{a^{2} b^{2}}}$ and $-y=\frac{\frac{-(a-b)}{a^{2}}}{\frac{a-b}{a^{2} b^{2}}}$

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

$x=a^2$ 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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