Algebraic Methods of Solving a Pair of Linear Equations - 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`