Solve the following system of equations by the method of cross-multiplication:
$a^2x+b^2y=c^2$
$b^2x+a^2y=d^2$

Given:

The given system of equations is:

$a^2x+b^2y=c^2$

$b^2x+a^2y=d^2$

 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,

$a^2x+b^2y-c^2=0$

$b^2x+a^2y-d^2=0$

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,

$a_1=a^2, b_1=b^2, c_1=-c^2$ and $a_2=b^2, b_2=a^2, c_2=-d^2$

Therefore,

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

$\frac{x}{-b^2d^2+a^2c^2}=\frac{-y}{-a^2d^2+b^2c^2}=\frac{1}{a^4-b^4}$

$x=\frac{-b^2d^2+a^2c^2}{a^4-b^4}$ and $-y=\frac{-a^2d^2+b^2c^2}{a^4-b^4}$

$x=\frac{a^2c^2-b^2d^2}{a^4-b^4}$ and $y=\frac{a^2d^2-b^2c^2}{a^4-b^4}$

The solution of the given system of equations is $x=\frac{a^2c^2-b^2d^2}{a^4-b^4}$ and $y=\frac{a^2d^2-b^2c^2}{a^4-b^4}$.

Tutorialspoint

Updated on: 10-Oct-2022

26 Views

Kickstart Your Career

Get certified by completing the course

Get Started