Class 10

Topic Covered

♦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 ( sqrt 3 , 2 sqrt 7 ) , (-1.75 , 3.3 ) , ( 4/13 , 1/19 ) , etc.

Substitution Method :

=> We shall explain the method of substitution by taking some examples.
Q 3159378214

Solve the following pair of equations by substitution method:
7x – 15y = 2 \ \ \ \  (1)
x + 2y = 3 \ \ \ \  (2)
Class 10 Chapter 3 Example 7
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) = 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.
To understand the substitution method more clearly, let us consider it stepwise:

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.
Q 3179378216

Solve Q.1 of Exercise 3.1 by the method of substitution.
Class 10 Chapter 3 Example 8
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.
Verify this answer by checking if it satisfies the conditions of the given problems.
Q 3129478311

Let us consider Example 2 in Section 3.3, i.e., the cost of 2 pencils and
3 erasers is Rs. 9 and the cost of 4 pencils and 6 erasers is Rs. 18. Find the cost of each
pencil and each eraser.
Class 10 Chapter 3 Example 9
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 - 6 y + 6 y = 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 bcause both the given equations are the same. Therefore, Equations (1) and (2)
have infinitely many solutions. Observe that we have obtained the same solution
graphically also. (Refer to Fig. 3.3, Section 3.2.) We cannot find a unique cost of a
pencil and an eraser, because there are many common solutions, to the given situation.
Q 3139478312

Let us consider the Example 3 of Section 3.2. Will the rails cross each
other?
Class 10 Chapter 3 Example 10
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.