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NCERT New Syllabus Mathematics
Class: 10
Exercise 3.3
Topic: Pair of Linear Equations in Two Variables
Exploring the Concept of Pair of Linear Equations in Two Variables: NCERT Class 10
Linear equations are everywhere in our daily lives, from calculating expenses to solving practical problems. In this chapter, we dive deeper into understanding how pairs of linear equations work in two variables. Whether finding the meeting point of two straight lines or determining the solution to complex problems, this topic opens doors to various real-world applications.
This chapter explores how two linear equations with two variables can be represented graphically and algebraically and how to find their solutions using different methods. We will explore everything from graphical representations to algebraic solutions such as substitution, elimination, and cross-multiplication.
By this topic's end, you'll understand how to interpret, solve, and apply linear equations in two variables effectively!
Elimination Method Steps:
Step 1: Equalize the Coefficients
Step 2: Eliminate One VariableMultiply both equations by appropriate constants to make the coefficients of either x or y the same.
Add or subtract the equations to cancel out one variable. Depending on the result:
If you obtain an equation in one variable, move to Step 3. If a true statement without variables is obtained (e.g., 0 = 0), the
system has infinitely many solutions.
If a false statement is obtained (e.g., 0 = 5), the system has no
Step 3: Solve for One Variablesolution and is inconsistent.
Solve the resulting equation to find the value of x or y.
EXERCISE 3.3
Q1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) (x/2) + (2y/3) = – 1, x – (y/3) = 3
Explanation:
1) Two equations in two variables are provided.
2) Equalize the coefficients of either x or y by multiplying both equations by
suitable non-zero constants.
3) Eliminate the variable by adding or subtracting one equation from the other.
4) This process is called the Elimination Method.
Solution:
(i) x + y = 5 and 2x – 3y = 4
a) Elimination method
1) Given equations are
x + y = 5 ---------------equation 1
2x – 3y = 4 ---------------equation 2
2) Here multiply equation 1 by 2 to get the coefficient of x same.
2(x + y) = 2(5)
2x + 2y = 10 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
2x + 2y = 102x – 3y = 4
(–) (+) (–)
---------------------------
5y = 6
y = 6/5 ---------------equation 4
4) Put the value of y =6/5 from equation 4 in equation 1, we get,
x + y = 5
x + (6/5) = 5
x = 5 – (6/5)
x = (25 – 6)/5
x = (19)/5
x = 19/5 -------------------------- equation 5.
5) Therefore, x = 19/5 and y = 6/5.
b) Substitution method
6) x + y = 5 ------------ equation 6
7) 2x – 3y = 4 ------------ equation 7
8) Simplify equation 6, and we get
x + y = 5
y = 5 – x ------------ equation 8
9) Substitute the value of y = (5 – x) from equation 8 in equation 7, we get
2x – 3y = 4
2x – 3(5 – x) = 4
2x – 15 + 3x = 4
5x – 15 = 4
5x = 15 + 45x = 19
x = 19/5 ------------ equation 9
10) Put the value of x = 19/5 from equation 9 in equation 8, and we get
y = 5 – xy = 5 – (19/5)
y = (25 – 19)/5
y = 6/5 ------------ equation 10
11) The value of x = 19/5 and the value of y = 6/5.
(ii) 3x + 4y = 10 and 2x – 2y = 2
a) Elimination method
1) Given equations are
3x + 4y = 10 ---------------equation 1
2x – 2y = 2 ---------------equation 2
2) Here multiply equation 2 by 2 to get the coefficient of y same.
2(2x – 2y) = 2(2)
4x – 4y = 4 ---------------equation 3
3) Adding equation 1 and equation 3, we get,
------------------------3x + 4y = 104x – 4y = 4
7x = 14
x = 14/7
x = 2 ---------------equation 4
4) Put the value of x = 2 from equation 4 in equation 2, we get,
2x – 2y = 2
2(2) – 2y = 2
2y = 4 – 2
2y = 2
y = 1 -------------------------- equation 5.
5) Therefore, x = 2 and y = 1.
b) Substitution method
6) 3x + 4y = 10 ------------ equation 6
7) 2x – 2y = 2 ------------ equation 7
8) Simplify equation 7, we get
2x – 2y = 2
x – y = 1
x = y + 1 ------------ equation 8
9) Substitute the value of x =(y + 1) from equation 8 in equation 6, we get
3x + 4y = 10
3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y + 3 = 10
7y = 10 – 37y = 7
y = 1 ------------ equation 9
10) Put the value of y = 1 from equation 9 in equation 8, we get
x = y + 1x = 1 + 1
x = 2 ------------ equation 10
11) Therefore, x = 2 and y = 1.
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
a) Elimination method
1) Given equations are
3x – 5y – 4 = 0
3x – 5y = 4 ---------------equation 1
9x = 2y + 7
9x – 2y = 7 ---------------equation 2
2) Here multiply equation 1 by 3 to get the coefficient of x as same.
3(3x – 5y) = 3(4)
9x – 15y = 12 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
9x – 15y = 129x – 2y = 7
(–) (+) (–)
------------------------–13 y = 5
y = – (5/13) ---------------equation 4
4) Put the value of y = – (5/13) from equation 4 in equation 1, we get,
3x – 5y = 4
3x – 5(– 5/13) = 4
3x + (25/13) = 4
3x = 4 – (25/13)
3x = (52 – 25)/13
3x = (27)/13
x = (27)/(13 x 3)
x = 9/13 -------------------------- equation 5.
5) Therefore, x = 9/13 and y = - (5/13).
b) Substitution method
6) 3x – 5y = 4 ------------ equation 6
7) 9x – 2y = 7 ------------ equation 7
8) Simplify equation 7, we get
3x – 5y = 4
3x = 5y +4
x = (5y + 4)/3 ------------ equation 8
9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get
9x – 2y = 7
9(5y + 4)/3 – 2y = 7
3(5y + 4) – 2y = 7
15y + 12 – 2y = 7
13y + 12 = 7
13y = 7 – 12
13y = – 5
y = – (5/13) ------------ equation 9
10) Put the value of y = – (5/13) from equation 9 in equation 8, we get
x = (5(– 5/13) + 4)/3
x = (– 25/13) + 4)/3
x = (– 25 + 52)/(13(3))
x = (27)/(13(3))x = (9)/(13)
x = 9/13 ------------ equation 10
11) Therefore, x = 9/13 and y = – (5/13).
(iv) (x/2) + (2y/3) = – 1, x – (y/3) = 3
a) Elimination method
1) Given equations are
(x/2) + (2y/3) = – 1 ---------------equation 1
x – (y/3) = 3 ---------------equation 2
2) Here multiply equation 1 by 2 to get the coefficient of x as same.
2(x/2) + 2(2y/3) = 2(– 1)
x + (4y/3) = – 2 ---------------equation 3
3) Subtracting equation 2 from equation 3, we get,
x + 4y/3 = – 2x – y/3 = 3
(–) (+) (–)
--------------------------5 y/3 = – 5
y/3 = – 1
y = – 3 ---------------equation 4
4) Put the value of y = – 3 from equation 4 in equation 1, we get,
(x/2) + (2y/3) = – 1
(x/2) + (2(– 3)/3) = – 1
(x/2) + (– 2) = – 1
(x/2) = – 1 + 2
(x/2) = 1
x = 2 -------------------------- equation 5.
5) Therefore, x = 2 and y = – 3.
b) Substitution method
6) 3x – 5y = 4 ------------ equation 6
7) 9x – 2y = 7 ------------ equation 7
8) Simplify equation 7, we get
3x – 5y = 4
3x = 5y +4
x = (5y + 4)/3 ------------ equation 8
9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get
9x – 2y = 7
9(5y + 4)/3 – 2y = 7
3(5y + 4) – 2y = 7
15y + 12 – 2y = 7
13y + 12 = 7
13y = 7 – 12
13y = – 5
y = – (5/13) ------------ equation 9
10) Put the value of y = – (5/13) from equation 9 in equation 8, we get
x = (5(– 5/13) + 4)/3
x = (– 25/13) + 4)/3
x = (– 25 + 52)/(13(3))
x = (27)/(13(3))x = (9)/(13)
x = 9/13 ------------ equation 10
11) Therefore, x = 9/13 and y = – (5/13).
Q2. Form the pair of linear equations in the following problems, and find their
solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator,
a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later,
Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times
this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the
cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and
an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Explanation:
1) Two equations in two variables are provided.
2) Equalize the coefficients of either x or y by multiplying both equations by
suitable non-zero constants.
3) Eliminate the variable by adding or subtracting one equation from the other.
4) This process is called the Elimination Method.
Solution:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
1) Let the numerator be x and the denominator be y.
2) According to the first condition, we have,
(x + 1)/(y – 1) = 1
(x + 1) = (y – 1)
x – y = – 2 ------------ equation 1
3) According to the second condition, we have,
(x)/(y + 1) = 1/2
2x = y + 1
2x – y = 1 ------------ equation 2
4) Subtract equation 1 from equation 2, and we get,
2x – y = 1
x – y = – 2
(–) (+) (–)
----------------------------x = 3
x = 3 ---------------equation 3
5) Put the value of x = 3 from equation 3 in equation 1, we get,
x – y = – 2
3 – y = – 2
y = 3 + 2
y = 5 -------------------------- equation 4.
6) The equations are x – y = – 2 and 2x – y = 1, where x and y are the numerator
and denominator of the fraction is 3/5.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
1) Let Nuri's present age be x and Sonu's present age be y.
2) 5 years ago their ages were (x – 5) and (y – 5).
3) According to the first relation given in the problem,
(x – 5) = 3(y – 5)
(x – 5) = 3y – 15
x – 3y = – 15 + 5
x – 3y = – 10 ---------------------- equation 1
4) 10 years later, their ages will be (x + 10) and (y + 10).
5) According to the relation given in the problem,
(x + 10) = 2(y + 10)
(x + 10) = 2y + 20
x – 2y = 20 – 10
x – 2y = 10 ---------------------- equation 2
6) Subtract equation 1 from equation 2, and we get,
x – 2y = 10
x – 3y = – 10
(–) (+) (+)
----------------------------
y = 20
y = 20 ---------------equation 3
7) Put the value of y = 20 from equation 3 in equation 2, we get,
x – 2y = 10
x – 2(20) = 10
x = 10 + 40
x = 50 -------------------------- equation 4.
8) The equations are x – 3y = – 10 and x – 2y = 10, where x and y are the ages
(in years) of Nuri and Sonu respectively. So Nuri's present age is 50 years and Sonu's present age is 20 years.
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this
number is twice the number obtained by reversing the order of the digits. Find the number.
1) Let the ones place digit be x and the tens place digit be y.
2) According to the first condition, we have,
x + y = 9
x + y = 9 ------------ equation 1
3) The original number: 10 x + y
4) The number obtained by reversing the digits: 10 y + x
5) According to the second condition, we have,
9(10 x + y) = 2(10 y + x)
90 x + 9 y = 20 y + 2 x
90 x – 2 x + 9 y – 20 y = 0
88 x – 11 y = 0
11(8 x – y) = 0
8 x – y = 0 ------------ equation 2
4) Add equation 1 to equation 2, and we get,
8 x – y = 0
x + y = 9
----------------------------
9x = 9
x = 1 ---------------equation 3
5) Put the value of x = 1 from equation 3 in equation 1, we get,
x + y = 9
1 + y = 9
y = 9 – 1
y = 8 -------------------------- equation 4.
6) The equations are x + y = 9 and 8 x – y = 0, Where x and y represent the tens
and units digits of the number, respectively. Solving these, we get
y = 8, which means the number is 18.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give
her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
1) Let the number of Rs 50 be x and the number of Rs 100 be y.
2) According to the first condition, we have,
x + y = 25
x + y = 25 ------------ equation 1
3) According to the second condition, we have,
50 x + 100 y = 2000
50 (x + 2 y) = 50 (40)
x + 2 y = 40
x + 2 y = 40 ------------ equation 2
4) Subtract equation 1 from equation 2, and we get,
x + 2 y = 40
x + y = 25
(–) (–) (–)
----------------------------
y = 15
y = 15 ---------------equation 3
5) Put the value of y = 15 from equation 3 in equation 1, we get,
x + y = 25
x + 15 = 25
x = 25 – 15
x = 10 -------------------------- equation 4.
6) The equations are x + y = 25 and x + 2 y = 40, Where x and y represent the
number of Rs 50 notes and Rs 100 notes
(v) A lending library has a fixed charge for the first three days and an
additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.
1) Let the fixed charges for the first 3 days be Rs x.
2) Let the additional charges per extra day be Rs y.
3) Saritha paid Rs 27 for 7 days, so she paid Rs x for the first 3 days and the remaining for
4 days, so we have,
x + 4y = 27
x + 4y = 27 ------------ equation 1
4) Susy paid Rs 21 for 5 days, so she paid Rs x for the first 3 days and the remaining for
2 days, so we have,
x + 2y = 21
x + 2y = 21 ------------ equation 2
5) Subtract equation 2 from equation 1, and we get,
x + 4 y = 27
x + 2 y = 21
(–) (–) (–)
----------------------------
2 y = 6
y = 3 ---------------equation 3
5) Put the value of y = 3 from equation 3 in equation 2, we get,
x + 2 y = 21
x + 2(3) = 21
x + 6 = 21
x = 21 – 6
x = 15 -------------------------- equation 4.
6) The equations are x + 4y = 27 and x + 2y = 21, where x is the fixed charge
So,
Conclusion: Mastering Pair of Linear Equations in Two Variables
As we reach the end of this fascinating topic on Pair of Linear Equations in Two Variables, it’s clear how these equations form the foundation for solving real-world problems involving relationships between two variables. You're honing your mathematical skills and building your problem-solving mindset by learning how to graph, calculate, and interpret their solutions. Remember, mastering this concept will open doors to more advanced mathematical techniques in the future. So keep practicing, explore different substitution, elimination, and graphical representation methods, and enjoy the learning journey!
Stay curious and keep solving!
As we reach the end of this fascinating topic on Pair of Linear Equations in Two Variables, it’s clear how these equations form the foundation for solving real-world problems involving relationships between two variables. You're honing your mathematical skills and building your problem-solving mindset by learning how to graph, calculate, and interpret their solutions. Remember, mastering this concept will open doors to more advanced mathematical techniques in the future. So keep practicing, explore different substitution, elimination, and graphical representation methods, and enjoy the learning journey!
Stay curious and keep solving!
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