Solve the following system of equations by the method of cross-multiplication:
$x\ +\ ay\ =\ b$
$ax\ -\ by\ =\ c$

Given:

The given system of equations is:

$x\ +\ ay\ =\ b$

$ax\ -\ by\ =\ c$

 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,

$x+ay-b=0$

$ax-by-c=0$

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,

$a_1=1, b_1=a, c_1=-b$ and $a_2=a, b_2=-b, c_2=-c$

Therefore,

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

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

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

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

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

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

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

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Updated on: 10-Oct-2022

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