This is one of the methods we use to solve system of linear equations.
Let us consider the following system of linear equations.
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
We have to write the coefficients of the equations and do cross multiplication as shown below.
We write the coefficient of y and constant term and two more columns by repeating the coefficients of x and y as follows.
The result is given by
The solution is
Example 1 :
Solve the following system of equations using cross multiplication method :
2x + 7y - 5 = 0
-3x + 8y = -11
Solution :
First we have to change the given linear equations in the form a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0.
2x + 7y - 5 = 0
-3x + 8y + 11 = 0
x/(77 + 40) = y/(15 - 22) = 1/[16 + 21]
x/117 = y/(-7) = 1/37
x/117 = 1/37 x = 117/37 |
y/(-7) = 1/37 y = -7/37 |
Therefore the solution is (117/37, -7/37).
Example 2 :
Solve the following system of equations using cross multiplication method :
3x + 4y = 24
20x - 11y = 47
Solution :
Write the given equations in the form of
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
3x + 4y - 24 = 0 ----(1)
20x - 11y - 47 = 0 ----(2)
x/(-188 - 264) = y/(-480 - (-141)) = 1/(-33 - 80)
x/(-452) = y/(-480+141)) = 1/(-33-80)
x/(-452) = y/(-339) = 1/(-113)
x/(-452) = 1/(-113) x = (-452)/(-113) x = 4 |
y/(-339) = 1/(-113) y = (-339)/(-113) y = 3 |
Therefore solution is (4, 3).
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