# Cross multiplication Method

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In this section we will discuss cross multiplication method. It is also known as Cramer's rule.
Let the two equations be ,
$a_{1}x + b_{1}y + c_{1}$ = 0
$a_{2}x + b_{2}y + c_{2}$ = 0

be a system of linear equations in two variables x and y such that
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$

i.e $a_{1}. b_{2} - a_{2}.b_{1} \neq$ 0
Then the system has a unique solution given by

$x = \frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$ and $y = \frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

## Examples on Cross multiplication method

Example 1 :
Find the solution of the given equations.
2x + 3y = 17 and 3x – 2y = 6
$a_{1}$= 2 , $b_{1}$ = 3 , and $c_{1}$= -17
$a_{2}$= 3 , $b_{1}$ = -2 and $c_{2}$ = -6

$x= \frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$ and y = $\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$

$x= \frac{3 \times (-6) - (-2) \times (-17)}{2 \times (-2) - 3 \times 3}$ and y = $\frac{(-17) \times 3 - (-6) \times 2}{2 \times (-2) - 3 \times 3}$

$x =\frac{-18 -34}{-4-9}$ and $y = \frac{-51 +12 }{-4-9}$

$x= \frac{-52}{-13}$ and $y =\frac{-39}{-13}$
x = 4 and y = 3

Linear equation in two variables

Solving linear equation by graphical method.
Substitution method.
Solving system of equation by elimination method
Cross multiplication method or Cramer’s rule

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