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

Given:

The given system of equations is:

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

$ax\ –\ by\ =\ a^2\ -\ b^2$

 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,

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

$ax-by-(a^2-b^2)=0$

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,

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

Therefore,

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

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

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

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

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

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

$x=a$ and $y=b$

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

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Updated on: 10-Oct-2022

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