In linear algebra, the substitution method can be used to solve systems of two linear equations with two unknowns. The whole idea is, solve for one of the variables from two equations, and put the result into the other to get the value of second variable. No matter which equation we pick. We also provide a set of NCERT Solutions for class 6 to Class 12 contains solution of all linear equations using substitution method for better learning. The quality of CBSE NCERT Solutions is way beyond and help you to score more in your exams and make you ready for the other competitive exams as well. And if you feel you are very good in algebraic concepts. You may try practice questions and evaluate your own competencies and abilities.  If you want to know more about the topic then you should probably go through this section.

What is Substitution Method

A method of solving algebraic linear equations with 2 variables. This method works by solving one of the equations for any of the one variables, and then substitute the value back into the second equation to find the result for second one.

How to Solve Linear Equations by Substitution

Follow below steps how to use the substitution method for solving linear systems. 

Steps to Solve:

Step 1: Number the two given equations as (1) and (2).
Step 2: Find the value of one of the variable from any one of the equation in terms of other variable., say y from equation(1)(i.e, make one of the variable as subject of the formula )
Step 3: Substitute the value of y from step 2 in the equation (2), and reduce the equation (2) into the equation of single variable which will be x.
Step 4: solve for this variable x, since we have a linear equation in one variable x.
Step 5: Substitute this value of x, in the value of y, obtained from Step 2.
Step 6: Now we have the linear equation of one variable y, we can solve for y.
Step 7: Write the final solution in the form (x,y) = (  ___, ___ )


By substituting the one variable value with the other. We're going to explain this by using examples.
Example 1: Solve linear equations using substitution
3x + 2y = 4,
x - y = 3
Step 1:
3x + 2y = 4  ...equation (1)
x - y = 3       ...equation (2)

Step 2: Solve equation (2) for x
x - y = 3
 x = y + 3   .....equation (3)

Step 3: Substitute y + 3 for x in equation (1)
3x + 2y = 4
3(y + 3) + 2y = 4
5y + 9 = 4  (Simplify both sides)
5y = -5
or y = -1

Step 4: Put y = -1 in equation (3)
x = -1 + 3
x = 2
The solution is (x, y) = (2, -1)

Example 2: A school planned to conduct admission tests for class A and Class B. The admission basic fee for class A is Rs. 5 and Rs. 6 for class B. On a specific day, 2000 children attended the test and Rs. 11000 collected by school. How many children appeared for the written test for class A and class B.

Solution: Let us choose two different variables for the two different unknowns.
Number of children appeared for written test for class A is x, and for class B is y.

Total number of children = 2000
i.e. x + y = 2000  .........(1)
Total amount collected by school = 11000
i.e. 5x + 6y = 11000  .........(2)

Now solve for x and y
Apply substitution method, to find the value of x and y
(1) => x = 2000 - y
Substitute the value of x in equation (2), we get
5(2000 - y) + 6y = 11000 
10000 - 5y + 6y = 11000
y = 1000, again substitute this value in equation (1), we obtain
x + 1000 = 2000  or x = 1000

Therefore, 1000 children appeared for each class.

Practice Problems

Problem 1: Solve linear equations by substitution
Problem 2: Solve by substitution 2x + 3y = 12 and 4x - y = 2
Problem 3: Solve systems of linear equations by substitution