Dear Readers,
Although there are a multiple method to solve linear equations with two variables (like substitution method, elimination method, graphical method & matrix method). But in this article we will limit ourselves to substitution method. While other methods, we will be discussing in our subsequent articles.
(Don’t forget to take the quiz at the end of this article)
Let’s understand the concept with the help of an example:
Suppose we have a linear equation in two variables as
2X + 3Y = 6
If you look at the equation carefully, you will find that there are two variables X & Y in the equation. The degree or power of X & Y is unity and the variables are not multiplied with each other. Therefore, we can call this as Linear equation in two variables.
As you can see that the above equation has two variables X & Y. So, to arrive at the values of X & Y we need one more equation having the same variables.
Solving Linear Equations with Two Variables – Substitution Method (Definition)
Before delving into the steps to solve the problem, let’s understand the meaning of the word ‘Substitution’
Substitution refers to the act of replacing one thing with another. From the meaning itself, you will get an idea that substitution method involves some sort of replacement. Basically, in this method we will represent one variable in terms of other variable after that substituting this value into the other equation.
Steps to Solve Problems Substitution Method
2X + 3Y = 6 ————–(1)
6X + 4Y = 13 ————–(2)
Step 1: Solve any one of the equation and represent any variable either X or Y in terms of other variable.
So, solving equation (1) and representing X in terms of Y.
2X + 3Y = 6
2X = 6 ― 3Y
X = 3 ― 3Y/2 ————(3)
Step 2: Now substituting the value of one of the variable (which has been represented in terms of other variable) into the other equation. This will result in one equation with one variable that we can easily solve.
In this case, substituting the value of X in equation (2) we get,
6 (3 ― 3Y/2) + 4Y = 13
18 – 9Y + 4Y = 13
5Y = 5
Y = 1
Step 3: Once we have arrived at the value of one of the variable then substitute this value to any one of the equation to get the value of the other variable.
Substituting the value of Y in equation (1) we get,
2X + 3×1 = 6
X = 3/2
Solution is ( 3/2 , 1)
Note: If you want to crosscheck just put the value of X & Y in the above equation (1) & (2) to see whether L.H.S = R.H.S or not. And, if lefthand side value is equal to the righthand side then the above obtained solution is correct.
Linear Equations with two variables  Quiz
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Question 1 of 5
1. Question
3 pointsThe sum of the present ages of father and son is 60 years. After 8 years, the age of the father becomes 3 times the age of his son. Find their present ages.
Correct
Let the present ages of father & son be X & Y years.
X + Y = 60 ———(1)
After 8 years, the age of father = (X + 8) years
age of son will be (Y + 8) years
According to question:
(X + 8) = 3 (Y+8)
X – 3Y = 16———–(2)
Substituting the value of X from equation (1) into equation (2), we get
60 Y – 3Y = 16
4Y = 44
Y = 11 years
& X = (60 – Y) = (60 11) = 49 years.
Therefore, the present age of father & son will be 49 years & 11 years.
Incorrect
Let the present ages of father & son be X & Y years.
X + Y = 60 ———(1)
After 8 years, the age of father = (X + 8) years
age of son will be (Y + 8) years
According to question:
(X + 8) = 3 (Y+8)
X – 3Y = 16———–(2)
Substituting the value of X from equation (1) into equation (2), we get
60 Y – 3Y = 16
4Y = 44
Y = 11 years
& X = (60 – Y) = (60 11) = 49 years.
Therefore, the present age of father & son will be 49 years & 11 years.

Question 2 of 5
2. Question
3 pointsSolve the system of equations:
X + 3Y = 10
2X + Y = 5
Find the value of X & Y?
Correct
X + 3Y = 10 ———–(1)
2X + Y = 5 ————(2)
Substitute the value of X from equation (1) into equation (2), we get
2(103Y) + Y = 5
20 – 6Y + Y = 5
5Y = 15
Y = 3
Now putting the value of Y in any of the above equation.
X + 3×3 = 10
X = 1
Solution (X , Y) is (1, 3)
Incorrect
X + 3Y = 10 ———–(1)
2X + Y = 5 ————(2)
Substitute the value of X from equation (1) into equation (2), we get
2(103Y) + Y = 5
20 – 6Y + Y = 5
5Y = 15
Y = 3
Now putting the value of Y in any of the above equation.
X + 3×3 = 10
X = 1
Solution (X , Y) is (1, 3)

Question 3 of 5
3. Question
3 pointsThe monthly income of Akshay & Varun are in the ratio of 3:4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 5,000 per month. Find their monthly incomes.
Correct
Let the monthly income of Akshay & Varun be 3X & 4X whereas, monthly expenditure are 5Y & 7Y respectively.
Income = Expenditure + Saving
Saving of Akshay = 3X− 5Y
3X − 5Y = 5000 —————(1)
Saving of Varun = 4X − 7Y
4X − 7Y = 5000 —————(2)
Now putting the value of X from equation (1) into equation (2)
4 {(5000 + 5Y)/3} −7Y = 5000
20000 + 20Y − 21 Y = 5000×3
Y = Rs 5000
Now putting the value of Y in equation (1) we get,
3X – 5×5000 = 5000
3X = 30000
X = Rs 10000
Therefore, monthly income of Akshay = 3X = 3×10000 = Rs 30,000
& monthly income of Varun = 4X = 4×10000 = Rs 40,000
Incorrect
Let the monthly income of Akshay & Varun be 3X & 4X whereas, monthly expenditure are 5Y & 7Y respectively.
Income = Expenditure + Saving
Saving of Akshay = 3X− 5Y
3X − 5Y = 5000 —————(1)
Saving of Varun = 4X − 7Y
4X − 7Y = 5000 —————(2)
Now putting the value of X from equation (1) into equation (2)
4 {(5000 + 5Y)/3} −7Y = 5000
20000 + 20Y − 21 Y = 5000×3
Y = Rs 5000
Now putting the value of Y in equation (1) we get,
3X – 5×5000 = 5000
3X = 30000
X = Rs 10000
Therefore, monthly income of Akshay = 3X = 3×10000 = Rs 30,000
& monthly income of Varun = 4X = 4×10000 = Rs 40,000

Question 4 of 5
4. Question
3 pointsSolve the system of equation:
Y = 6X −11
2X + 3Y = 7
Correct
Y = 6X −11 —————(1)
2X + 3Y = 7 —————(2)
Substituting the value of X from equation (1) into equation (2), we get
2X + 3(6X −11) = 7
20X = 7 + 33
X = 40/20 = 2
Now putting the value of X in any one of the above equation, we get
Y = 6×2 − 11 = 12 − 11 = 1
Solution is ( 2, 1)
Incorrect
Y = 6X −11 —————(1)
2X + 3Y = 7 —————(2)
Substituting the value of X from equation (1) into equation (2), we get
2X + 3(6X −11) = 7
20X = 7 + 33
X = 40/20 = 2
Now putting the value of X in any one of the above equation, we get
Y = 6×2 − 11 = 12 − 11 = 1
Solution is ( 2, 1)

Question 5 of 5
5. Question
3 pointsThree coffees and two samosa cost a total of Rs 30. Two coffees and four samosas cost Rs 36. What is the individual price of single coffee and a single samosa?
Correct
Let the cost of single coffee be Rs X, and
Cost of single samosa be Rs Y.
3X + 2Y = 30 ———–(1)
2X + 4 Y = 36———–(2)
Substituting the value of X from equation (2) into equation (1), we get
3 (18 – 2Y) + 2Y = 30
54 – 6Y + 2Y = 30
4Y = 24
Y = 6
Now putting the value of Y in anyone of the above equation, we get
3X + 2×6 = 30
3X = 30 – 12
X = 18/3 = 6
Therefore, the individual price of single coffee = X = Rs 6, and
the individual price of single samosa = Y = Rs 6.
Incorrect
Leaderboard: Linear Equations with two variables  Quiz
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