# Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method

## Cross Multiplication Method

Cross Multiplication Method- To understand this method properly, we shall take the help of an example
4x+2y-7=0 ......eq1
5x+3y-8=0 ......eq2
While practising this method, if we want to find the value of a particular variable, we will take that variable as a numerator, and the denominator will be the subtraction between the cross multiplication of coefficients of the remaining two variables. In this case, it will be like

x/"(2×-8)-(-7×3)"

y/"(4×-8)-(-7×5)"

1/"(4×3)-(2×5)"

All of them are equal to each other, so to find the value of x,

x/"(2×-8)-(-7×3)" = 1/"(4×3)-(2×5)"

x/"-16+21"=1/"12-10"

x/5= 1/2

x=5/2

To find the value of y,

y/"(4×-8)-(-7×5)"= 1/"(4×3)-(2×5)"

y/"-32+35" = 1/"12-10"

y/3 = 1/2

y= 3/2

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Pair of Linear Equation in two variable part 14 (Algebraic Cross Multiplication) [00:12:42]
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