The formula of cross multiplication method is proved in algebraic form by taking a system of linear equations in two variables $a_{1}x+b_{1}y+c_{1} = 0$ and $a_{2}x+b_{2}y+c_{2} = 0$. It is actually derived in algebra by the elimination method. The simultaneous linear equations in two variables $x$ and $y$ can be written as follows.

- $a_{1}x+b_{1}y = -c_{1}$
- $a_{2}x+b_{2}y = -c_{2}$

The variable $y$ can be eliminated from the system of linear equations by multiplying the linear equation $a_{1}x+b_{1}y = -c_{1}$ by $b_2$ and $a_{2}x+b_{2}y = -c_{2}$ by $b_1$.

Firstly, multiply both sides of the linear equation in two variables $a_{1}x+b_{1}y = -c_{1}$ by $b_{2}$.

$\implies$ $b_{2}(a_{1}x+b_{1}y)$ $=$ $b_{2} \times (-c_{1})$

$\implies$ $b_{2} \times a_{1}x+ b_{2} \times b_{1}y$ $=$ $b_{2} \times (-c_{1})$

$\implies$ $b_{2}a_{1}x+ b_{2}b_{1}y$ $=$ $-b_{2}c_{1}$

$\,\,\, \therefore \,\,\,\,\,\,$ $a_{1}b_{2}x+ b_{1}b_{2}y$ $=$ $-b_{2}c_{1}$

Similarly, multiply both sides of the linear equation in two variables $a_{2}x+b_{2}y = -c_{2}$ by $b_{1}$.

$\implies$ $b_{1}(a_{2}x+b_{2}y)$ $=$ $b_{1} \times (-c_{2})$

$\implies$ $b_{1} \times a_{2}x+ b_{1} \times b_{2}y$ $=$ $b_{1} \times (-c_{2})$

$\implies$ $b_{1}a_{2}x+ b_{1}b_{2}y$ $=$ $-b_{1}c_{2}$

$\,\,\, \therefore \,\,\,\,\,\,$ $a_{2}b_{1}x+ b_{1}b_{2}y$ $=$ $-b_{1}c_{2}$

Now, let’s eliminate the $y$ from the simultaneous linear equations in two variables.

- $a_{1}x+b_{1}y = -c_{1}$ became $a_{1}b_{2}x+b_{1}b_{2}y$ $=$ $-b_{2}c_{1}$
- $a_{2}x+b_{2}y = -c_{2}$ became $a_{2}b_{1}x+b_{1}b_{2}y$ $=$ $-b_{1}c_{2}$

$b_{1}b_{2}y$ is a same term in the both linear equations and it can be eliminated by subtracting them. So, subtract the second linear equation from the first linear equation.

$\implies$ $(a_{1}b_{2}x+b_{1}b_{2}y)$ $-$ $(a_{2}b_{1}x+b_{1}b_{2}y)$ $=$ $(-b_{2}c_{1})$ $-$ $(-b_{1}c_{2})$

$\implies$ $a_{1}b_{2}x+b_{1}b_{2}y$ $-a_{2}b_{1}x-b_{1}b_{2}y$ $=$ $-b_{2}c_{1}+b_{1}c_{2}$

$\implies$ $a_{1}b_{2}x-a_{2}b_{1}x$ $+b_{1}b_{2}y-b_{1}b_{2}y$ $=$ $b_{1}c_{2}-b_{2}c_{1}$

$\implies$ $a_{1}b_{2}x-a_{2}b_{1}x$ $\require{cancel} +\cancel{b_{1}b_{2}y}-\cancel{b_{1}b_{2}y}$ $=$ $b_{1}c_{2}-b_{2}c_{1}$

$\implies$ $a_{1}b_{2}x-a_{2}b_{1}x$ $=$ $b_{1}c_{2}-b_{2}c_{1}$

$\implies$ $(a_{1}b_{2}-a_{2}b_{1})x$ $=$ $b_{1}c_{2}-b_{2}c_{1}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = \dfrac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

In earlier step, the value of $x$ is evaluated by eliminating $y$ from the simultaneous linear equations. In the same way, the value of $y$ can also be derived mathematically by eliminating $x$ from the system of linear equations. So, multiply the linear equation $a_{1}x+b_{1}y = -c_{1}$ by $a_2$ and $a_{2}x+b_{2}y = -c_{2}$ by $a_1$.

Multiply both sides of the linear equation in two variables $a_{1}x+b_{1}y = -c_{1}$ by $a_{2}$.

$\implies$ $a_{2}(a_{1}x+b_{1}y)$ $=$ $a_{2} \times (-c_{1})$

$\implies$ $a_{2} \times a_{1}x+ a_{2} \times b_{1}y$ $=$ $a_{2} \times (-c_{1})$

$\implies$ $a_{2}a_{1}x+ a_{2}b_{1}y$ $=$ $-a_{2}c_{1}$

$\,\,\, \therefore \,\,\,\,\,\,$ $a_{1}a_{2}x+ a_{2}b_{1}y$ $=$ $-c_{1}a_{2}$

In the same way, multiply both sides of the linear equation in two variables $a_{2}x+b_{2}y = -c_{2}$ by $a_{1}$.

$\implies$ $a_{1}(a_{2}x+b_{2}y)$ $=$ $a_{1} \times (-c_{2})$

$\implies$ $a_{1} \times a_{2}x+ a_{1} \times b_{2}y$ $=$ $a_{1} \times (-c_{2})$

$\implies$ $a_{1}a_{2}x+ a_{1}b_{2}y$ $=$ $-a_{1}c_{2}$

$\,\,\, \therefore \,\,\,\,\,\,$ $a_{1}a_{2}x+ a_{1}b_{2}y$ $=$ $-c_{2}a_{1}$

Now, let us eliminate the $x$ from a pair of simultaneous linear equations in two variables.

- $a_{1}x+b_{1}y = -c_{1}$ became $a_{1}a_{2}x+ a_{2}b_{1}y$ $=$ $-c_{1}a_{2}$
- $a_{2}x+b_{2}y = -c_{2}$ became $a_{1}a_{2}x+ a_{1}b_{2}y$ $=$ $-c_{2}a_{1}$

The two linear equations have a same term $a_{1}a_{2}x$ and the variable $x$ can be eliminated from them by subtraction. So, subtract the second linear equation from the first linear equation.

$\implies$ $(a_{1}a_{2}x+a_{2}b_{1}y)$ $-$ $(a_{1}a_{2}x+ a_{1}b_{2}y)$ $=$ $(-c_{1}a_{2})$ $-$ $(-c_{2}a_{1})$

$\implies$ $a_{1}a_{2}x+a_{2}b_{1}y$ $-a_{1}a_{2}x-a_{1}b_{2}y$ $=$ $-c_{1}a_{2}+c_{2}a_{1}$

$\implies$ $a_{1}a_{2}x-a_{1}a_{2}x$ $+a_{2}b_{1}y-a_{1}b_{2}y$ $=$ $c_{2}a_{1}-c_{1}a_{2}$

$\implies$ $\require{cancel} \cancel{a_{1}a_{2}x}-\cancel{a_{1}a_{2}x}$ $+a_{2}b_{1}y-a_{1}b_{2}y$ $=$ $c_{2}a_{1}-c_{1}a_{2}$

$\implies$ $a_{2}b_{1}y-a_{1}b_{2}y$ $=$ $c_{2}a_{1}-c_{1}a_{2}$

$\implies$ $(a_{2}b_{1}-a_{1}b_{2})y$ $=$ $c_{2}a_{1}-c_{1}a_{2}$

$\implies$ $y = \dfrac{c_{2}a_{1}-c_{1}a_{2}}{a_{2}b_{1}-a_{1}b_{2}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $y = \dfrac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

Therefore, the values of $x$ and $y$ are derived mathematically for a system of linear equations in two variables $a_{1}x+b_{1}y+c_{1} = 0$ and $a_{2}x+b_{2}y+c_{2} = 0$.

$x \,=\, \dfrac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

$y \,=\, \dfrac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

$a_{1}b_{2}-a_{2}b_{1}$ is a common expression in the denominator of the both equations. So, move the expression to left hand side of the equations.

$\implies$ $\dfrac{x}{b_{1}c_{2}-b_{2}c_{1}} \,=\, \dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}$

$\implies$ $\dfrac{y}{c_{1}a_{2}-c_{2}a_{1}} \,=\, \dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{x}{b_{1}c_{2}-b_{2}c_{1}}$ $\,=\,$ $\dfrac{y}{c_{1}a_{2}-c_{2}a_{1}}$ $\,=\,$ $\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}$

This is called a formula for the cross multiplication method and used to find the variables of a system of linear equations in two variables.

Latest Math Topics

Apr 18, 2022

Apr 14, 2022

Apr 05, 2022

Mar 18, 2022

Mar 05, 2022

Latest Math Problems

Apr 06, 2022

Mar 22, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved