What is the easiest method to apply cross multiplication method?
We are not sure what exactly you want.
If you are trying to find out how the concept of cross multiplication is to be used for solving a pair of linear equations in 2 variables, here is the method:
Let us take two equations::
2x + 5y = 44
3x + 2y = 22
These equations can be written as
2x + 5y - 44 = 0 .. (1)
3x + 2y - 22 = 0 ...(2)
These equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Now,create a diagram as follows:
Now write down the equation as follows:
xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1
For the given example, a1 = 2, b1 = 5 c1 =-44 & a2 = 3, b2 = 2 and c2 = -22. Substuting this in above we get,
x5(−22)−2(−44)=y(−44)3−(−22)2=12(2)−3(5)
Solving for x, we get x = 2 and solving for y, we get y = 8.
Using the arrow diagram is short cut method. Hope you have understood
We are not sure what exactly you want.
If you are trying to find out how the concept of cross multiplication is to be used for solving a pair of linear equations in 2 variables, here is the method:
Let us take two equations::
2x + 5y = 44
3x + 2y = 22
These equations can be written as
2x + 5y - 44 = 0 .. (1)
3x + 2y - 22 = 0 ...(2)
These equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Now,create a diagram as follows:
Now write down the equation as follows:
xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1
For the given example, a1 = 2, b1 = 5 c1 =-44 & a2 = 3, b2 = 2 and c2 = -22. Substuting this in above we get,
x5(−22)−2(−44)=y(−44)3−(−22)2=12(2)−3(5)
Solving for x, we get x = 2 and solving for y, we get y = 8.
Using the arrow diagram is short cut method. Hope you have understood
