Algebraic Methods of Solving a Pair of Linear Equations

Algebraic Methods of Solving a Pair of Linear Equations

In the previous section, we discussed how to solve a pair of linear equations graphically. The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like  etc. There is every possibility of making mistakes while reading such coordinates. Is there any alternative method of finding the solution? There are several algebraic methods, which we shall now discuss.

Substitution Method: We shall explain the method of substitution by taking some examples.

Example: Solve the following pair of equations by substitution method:

7x – 15y = 2                                                           (1)
           x + 2y = 3                                                              (2)

Solution:

Step 1: We pick either of the equations and write one variable in terms of the other. Let us consider the Equation (2):

x + 2y = 3

and write it as       x = 3 – 2y                                                    (3)

Step 2: Substitute the value of x in Equation (1). We get
7(3 – 2y) – 15y = 2
i.e.,                 21 – 14y – 15y = 2
i.e.,                           – 29y = –19
Therefore,         
Step 3: Substituting this value of y in Equation (3), we get  

Therefore, the solution is 

Verification: Substituting x = 29 and y = 29, you can verify that both the Equations
(1) and (2) are satisfied.
Remark: We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.

Elimination Method
Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.

Cross - Multiplication Method
So far, you have learnt how to solve a pair of linear equations in two variables by graphical, substitution and elimination methods. Here, we introduce one more algebraic method to solve a pair of linear equations which for many reasons is a very useful method of solving these equations. Before we proceed further, let us consider the following situation.
Let us now see how this method works for any pair of linear equations in two variables of the form

            a₁x + b₁y + c₁ = 0                                                             (1)

and               a₂x + b₂y + c₂ = 0                                                             (2)

To obtain the values of x and y as shown above, we follow the following steps:

Step 1: Multiply Equation (1) by b₂ and Equation (2) by b₁, to get

b₂a₁x + b₂b₁y + b₂c₁ = 0                                                (3)

b₁a₂x + b₁b₂y + b₁c₂ = 0                                                (4)

Step 2: Subtracting Equation (4) from (3), we get:

(b₂a₁ – b₁a₂) x + (b₂b₁ – b₁b₂) y + (b₂c₁ – b₁c₂) = 0

i.e.,    (b₂a₁ – b₁a₂) x = b₁c₂ – b₂c₁

     (5)

 Step 3: Substituting this value of x in (1) or (2), we get

                                                            (6)

Now, two cases arise:
Case 1:  a₁b₂ – a₂b₁ ≠ 0. In this case  Then the pair of linear equations has a unique solution.
Case 2: a₁b₂ – a₂b₁ = 0. If we write

  then a₁ = k a₂, b₁ = k b₂.
Substituting the values of a₁ and b₁ in the Equation (1), we get
                       k(a₂x + b₂y) + c₁ = 0.                                                           (7)
It can be observed that the Equations (7) and (2) can both be satisfied only if

 c₁ = k c₂, i.e.,
If c₁ = k c₂, any solution of Equation (2) will satisfy the Equation (1), and vice versa. So, if   , then there are infinitely many solutions to the pair of linear equations given by (1) and (2).
If c₁ ≠ k c₂, then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.
We can summarise the discussion above for the pair of linear equations given by (1) and (2) as follows:

(i) When  we get a unique solution.

(ii) When   there are infinitely many solutions.

(iii) When    there is no solution.

Note that you can write the solution given by Equations (5) and (6) in the following form:

                                                         (8)

In remembering the above result, the following diagram may be helpful to you:

The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first.
For solving a pair of linear equations by this method, we will follow the following steps:
Step 1: Write the given equations in the form (1) and (2).
Step 2: Taking the help of the diagram above, write Equations as given in (8).
Step 3: Find x and y, provided a₁b₂ – a₁b₁ ≠ 0
Step 2 above gives you an indication of why this method is called the cross-multiplication method.

Equations Reducible to a Pair of Linear Equations in Two Variables
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. We now explain this process through some examples.