Example 4 : 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, y = 19/29

Step 3 : Substituting this value of y in Equation (3), we get
x = 3 – 2(19/29)
i.e., x = (49/29)

Therefore, the solution is x = 49/29, y = 19/29.

Verification : Substituting x = 49/29 and y = 19/29, you can verify that both the Equations (1) and (2) are satisfied.

Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.

Step 2 : Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and 10 below, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.

Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

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.

Example 5 : Solve the following question—Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”

Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively. Then, the pair of linear equations that represent the situation is
s – 7 = 7(t – 7), i.e., s – 7t + 42 = 0 (1)
and s + 3 = 3(t + 3), i.e., s – 3t = 6 (2)

Using Equation (2), we get s = 3t + 6. Putting this value of s in Equation (1), we get
(3t + 6) – 7t + 42 = 0,
i.e., 4t = 48, which gives t = 12.
Putting this value of t in Equation (2), we get s = 3(12) + 6 = 42. So, Aftab and his daughter are 42 and 12 years old, respectively.

Example 6 : In a shop the cost of 2 pencils and 3 erasers is 9 and the cost of 4 pencils and 6 erasers is18. Find the cost of each pencil and each eraser.

Solution : The pair of linear equations formed were:
2x + 3y = 9 (1)
4x + 6y = 18 (2)

We first express the value of x in terms of y from the equation 2x + 3y = 9, to get x = (9−3y)/2 (3)
Now we substitute this value of x in Equation (2), to get
4((9−3y)/2) + 6y = 18
i.e., 18 – 6y + 6y = 18
i.e., 18 = 18.
This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same. Therefore, Equations (1) and (2) have infinitely many solutions.

Example 7 : Two rails are represented by the equations
x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other?

Solution : The pair of linear equations formed were:
x + 2y – 4 = 0 (1)
2x + 4y – 12 = 0 (2)

We express x in terms of y from Equation (1) to get x = 4 – 2y.
Now, we substitute this value of x in Equation (2) to get
2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0
i.e., – 4 = 0
which is a false statement.
Therefore, the equations do not have a common solution. So, the two rails will not cross each other.