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Cross Multiplication - Pair Of Linear Equations In Two Variables

Cross Multiplication- Pair Of Linear Equations In Two Variables

In Maths, cross multiplication method is used to solve linear equation in two variables. This is the simplest method and gives the accurate value of the variables. Cross multiplication is only applicable when we have a pair of linear equations in two variables. 

Let us suppose that a1x + b1y + c1 = 0 and a2x + b2x + c2 = 0 are the two equations which has to be solved. By using cross multiplication, we will get the values x and y such as:

�=�1 �2 − �2 �1�2 �1 − �1 �2

�=�1 �2 − �2 �1�2 �1 − �1 �2

Where

�2 �1 − �1 �2 ≠ 0

And the final solution is:

��1 �2 − �2 �1=��1 �2 − �2 �1=1�2 �1 − �1 �2

The solution of the simultaneous linear equation can be divided into two broad categories, Graphical Method and Algebraic method. The algebraic method can be sub-divided into three categories: Substitution method, Elimination method and cross-multiplication method.

Derivation of Cross multiplication method

In general, a pair of linear equations in two variables can be represented as:

�1 � + �1 � + �1=0 ��� �2 � + �2 � + �2=0

For solving pair of linear equations in two variables following steps are followed:

Given a pair of linear equations in two variables;

�1 � + �1 � + �1=0…(1)
�2 � + �2 � + �2=0…(2)

Step i) Equation (1) is multiplied by b2 and Equation (2) by b1 .

�2 �1 � + �2 �1 � + �2 �1=0…(3)
�1 �2 � + �1 �2 � + �1 �2=0…(4)

Step ii) Subtracting Equation (4) from (3):

(�2 �1 − �1 �2)� + (�2 �1 − �1 �2)� + (�2 �1 − �1 �2)=0
⇒(�2 �1 − �1 �2)�=�1 �2 − �2 �1
⇒ �=�1 �2 − �2 �1�2 �1 − �1 �2, ����� �2 �1 − �1 �2 ≠ 0

Step iii) The value of “x” obtained as such is substituted either in Equation (1) or Equation (2). Hence, the value of “y” obtained is:

�=�1 �2 − �2 �1�2 �1 − �1 �2

The solution of the equations is given as:

��1 �2 − �2 �1=��1 �2 − �2 �1=1�2 �1 − �1 �2…(5)

The above method is called is called ‘Cross-Multiplication Method’, as the following cross-multiplication technique can be used to simplify the solution and hence will help in memorizing it. To memorize the method of cross multiplication, for solving linear equations in two variables the following diagram is helpful:

Cross Multiplication for pair of linear equations

The arrows indicate the multiplication of the values connected through the arrow and then the second product is subtracted from the first. The obtained result is then substituted as the denominator of the variables and 1, as mentioned above the arrow and then the entire values obtained are equated to form equation (5).

��1 �2 − �2 �1=��1 �2 − �2 �1=1�2 �1 − �1 �2

From here, x and y are evaluated provided that;

�1 �2 − �2 �1≠0

Therefore, this method is called cross-multiplication as depicted.

In such a method, the condition for consistency of pair of linear equation in two variables must be checked, which are as follows:

  1.  If a1a2 ≠ b1b2 , then we get a unique solution and the pair of linear equations in the two variables are consistent.
  2.  If a1a2 = b1b2 = c1c2, then there exists infinitely many solutions and the pair of lines are coincident and therefore, dependent and consistent.
  3.  If a1a2 = b1b2 ≠ c1c2, then there exists no solution and the pair of linear equations in two variables is said to be inconsistent.

Solved Examples

Consider the following example:

Example 1: Solve the following linear equations using cross multiplication method.

3� − 4�=2
� − 2�=7

Solution: The above equation can be rewritten as:

3� − 4�=2
−2� + �=7

By method of cross multiplication,

��1 �2 − �2 �1=��1 �2 − �2 �1=1�2 �1 − �1 �2

Substituting the value in the above equation;

�28 + 2=�4 + 21=13 − 8
⇒ �30=�25=−15
⇒ �=−6 , �=−5

Example 2: Find the value of variables which satisfies the following equation:

2x + 5y = 20 and 3x+6y =12.

Solution: Given,

2x+5y = 20 

3x + 6y = 12

By cross multiplication method we know;

��1 �2 − �2 �1=��1 �2 − �2 �1=1�2 �1 − �1 �2

Hence, by substituting the values in the above equation we get;

x/[(5).(12)-(6).(20)] = y/[(20).(3)-(12).(2)] = 1/[(5).(3)-(6).(2)]

x/(60-120) = y/(60 – 24) = 1/(15-12)

x/(-60) = y/36 = ⅓

Now,

x/-60 = ⅓

x= -20 

And

y/36 = ⅓

y = 12

Therefore, x=-20 and y =12 is the point where the given equations intersect.


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