194a-NCERT New Syllabus Grade 10 Pair of Linear Equations in Two Variables Ex-3.3

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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 focuses on exploring 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

Multiply both equations by appropriate constants to make the coefficients of either x or y the same.

Step 2: Eliminate One Variable

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

solution and is inconsistent.

Step 3: Solve for One Variable

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 = 10

2x – 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 + 4

5x = 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 – x

y = 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 = 10

4x – 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 – 3

7y = 7

y = 1 ------------ equation 9

10) Put the value of y = 1 from equation 9 in equation 8, we get

x = y + 1

x = 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 = 12

9x –  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 = – 2

x –  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 x = 1 and

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, respectively. So, the number of Rs 50 notes is 10 and the number of Rs 100 notes is 15.

(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 

(in Rs) and y is the additional charge (in Rs) per day. So, the fixed charges for the first 3 days are Rs 15 and the additional charges per extra day are Rs 3.

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. By learning how to graph, calculate, and interpret their solutions, you’re not just honing your mathematical skills but also building your problem-solving mindset. 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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