Solve the following system of equations by the method of cross-multiplication:
$ax\ +\ by\ =\ a\ -\ b$
$bx\ –\ ay\ =\ a\ +\ b$

Given:

The given system of equations is:

$ax\ +\ by\ =\ a\ -\ b$

$bx\ –\ ay\ =\ a\ +\ b$

 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,

$ax+by-(a-b)=0$

$bx-ay-(a+b)=0$

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,

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

Therefore,

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

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

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

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

$x=\frac{(-a^2-b^2)\times1}{-a^2-b^2}$ and $-y=\frac{(-a^2-b^2)\times1}{-a^2-b^2}$

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

$x=1$ and $-y=1$

$x=1$ and $y=-1$

The solution of the given system of equations is $x=1$ and $y=-1$.

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Updated on: 10-Oct-2022

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