Proof of Cross multiplication method formula

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.

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

Eliminate a variable to find remaining variable

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 $b_1$.

Step – 1

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

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

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

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

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

Step – 2

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

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

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

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

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

Step – 3

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.

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

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

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

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

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

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

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

Eliminate the known variable to find the unknown

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 $a_1$.

Step – 1

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

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

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

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

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

Step – 2

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

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

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

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

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

Step – 3

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.

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

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

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

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

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

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

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

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

Cross multiplication formula

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.