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

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3) Cross Multiplication Method- To understand this method properly we shall take help of a example
`4x+2y-7=0` ......eq1
`5x+3y-8=0` ......eq2
While practicing this method if we want to find value of a particular variable we will take that variable as a numerator and the denominator will be the substraction between the cross multiplication of coefficients of the remaining two variables. Like 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 if you to find value of x then take,


`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`

Video Tutorials

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Shaalaa.com | Pair of Linear Equation in two variable part 14 (Algebraic Cross Multiplication)

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