Open In App

Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.4

Last Updated : 01 May, 2024
Improve
Improve
Like Article
Like
Save
Share
Report

Question 1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4 

Solution:

Here, the two given eqn. are as follows:

x + y = 5 ……….(I)

2x – 3y = 4 ………..(II)

ELIMINATION METHOD:

Multiply equation (I) by 2, and then subtract (II) from it, we get

5y = 6

y = 6/5

Now putting y=6/5 in eqn. (I), we get

x + 6/5 = 5

x = (5–(6/5))

x = 19/5

SUBSTITUTION METHOD:

From (I), we get

y=5–x…….(III)

Now substituting the value of y in eqn. (II), we get

2x – 3(5–x) = 4 

2x – 15+3x = 4 

5x = 4+15

x = 19/5

As, putting x = 19/5, in eqn. (III), we get

y = 5 – 19/5

y = 6/5

Hence, by elimination method and substitution method we get,

x = 19/5 and y = 6/5.

(ii) 3x + 4y = 10 and 2x – 2y = 2

Solution:

Here, the two given eqn. are as follows:

3x + 4y = 10 ……….(I)

2x – 2y = 2 ………..(II)

ELIMINATION METHOD:

Multiply equation (II) by 2, and then add it to (I), we get

7x = 14

x = 14/7 

x = 2

Now putting x = 2 in eqn. (I), we get

3(2) + 4y = 10

4y = 10 – 6

y = 4/4

y = 1

SUBSTITUTION METHOD:

From (II), we get

x = (2+2y)/2 

x = y+1 …….(III)

Now substituting the value of x in eqn. (I), we get

3(y+1) + 4y = 10

3y + 3 + 4y = 10

7y = 10 – 3

y = 7/7

y = 1

As, putting y = 1, in eqn. (III), we get

x = 1+1

x = 2

Hence, by elimination method and substitution method we get,

x = 2 and y = 1.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 

Solution:

Here, the two given eqn. are as follows:

3x – 5y – 4 = 0

9x = 2y + 7

By rearranging we get,

3x – 5y = 4 ……….(I)

9x – 2y = 7 ………..(II)

ELIMINATION METHOD:

Multiply equation (I) by 3, and then subtract (II) from it, we get

–13y = 5

y = -5/13

Now putting y = – 5/13 in eqn. (I), we get

3x – 5(– 5/13) = 4

3x  = 4 – (25/13)

3x = 27/13

x = 9/13

SUBSTITUTION METHOD:

From (I), we get

3x – 5y = 4

x = (4+5y)/3 …….(III)

Now substituting the value of x in eqn. (II), we get

9((4+5y)/3) – 2y = 7

3(4+5y) – 2y = 7

12+15y – 2y = 7

13y = – 5

y = – 5/13

As, putting y = – 5/13, in eqn. (III), we get

x=(4+5(– 5/13))/3

x = 9/13

Hence, by elimination method and substitution method we get,

x=9/13 and y=5/13.

(iv) x/2 + 2y/3 = -1 and x – y/3 = 3

Solution:

Here, the two given eqn. are as follows:

x/2 + 2y/3 = –1 ………..(A)

x – y/3 = 3 ……………….(B)

by rearranging (multiply (A) by 6 and multiply (B) by 3) we get,

3x + 4y = – 6 ……….(I)

3x – y = 9  ………..(II)

ELIMINATION METHOD:

Subtract (II) from (I), we get

5y = – 15

y = – 3

Now putting y = -3 in eqn. (II), we get

3x – (– 3) = 9 

3x = 9 – 3

x = 6/3

x = 2

SUBSTITUTION METHOD:

From (II), we get

3x – y = 9

y = 3x – 9 …….(III)

Now substituting the value of y in eqn. (I), we get

3x + 4(3x – 9) = – 6 

3x + 12x – 36 = – 6 

15x = – 6 + 36

x = 30/15

x = 2

As, putting x = 2, in eqn. (III), we get

y = 3(2) – 9

y = – 3

Hence, by elimination method and substitution method we get,

x=2 and y=3.

Question 2. 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 ½ if we only add 1 to the denominator. What is the fraction?

Solution:

Let the fraction be p/q, where p is numerator and q is denominator.

Here, According to the given condition,

(p+1)/(q – 1) = 1 ………………..(A)

and,

p/(q+1) = 1/2 …………………..(B)

Solving (A), we get

(p+1) = q – 1

p q = 2 ……………………….(I)

Now, solving (B), we get

2p = (q+1)

2p q = 1 ……………………….(II)

When equation (I) is subtracted from equation (II) we get,

p = 3

Now putting p = 3 in eqn. (I), we get

3 – q = – 2

q = 3+2

q = 5

So, p = 3 and q = 5.

Hence, the fraction p/q 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?

Solution:

Let us assume, present age of Nuri is x

And present age of Sonu is y.

Here, According to the given condition, the equation formed will be as follows :

x – 5 = 3(y – 5)

x – 3y = 10 …………………………………..(I)

Now,

x + 10 = 2(y +10)

x – 2y = 10 …………………………………….(II)

Subtract eqn. (I) from (II), we get

y = 20

Now putting y = 20 in eqn. (II), we get

x – 2(20) = 10

x = 10+40

x = 50

Hence, 

Age of Nuri is 50 years

Age of Sonu 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.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number = 10y + x

And, reverse number = 10x + y

eg: 23 

x = 3 and y = 2

So, 23 can be represented as = 10(2) + 3 = 23

Here, According to the given condition

x + y = 9 …………………….(I)

and,

9(10y + x) = 2(10x + y)

90y + 9x = 20x + 2y

88y = 11x

x = 8y 

x 8y = 0 ………………………………………………………….. (II)

Subtract eqn. (II) from (I) we get,

9y = 9

y = 1

Now putting y = 1 in eqn. (II), we get

x – 8(1) = 0

x = 8

Hence, the number is 10y + x 

=10 × 1 + 8 

Number = 18

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

Solution:

Let the number of ₹ 50 notes be x and the number of ₹100 notes be y

Here, According to the given condition

x + y = 25 ……………………………….. (I)

50x + 100y = 2000 ………………………………(II)

Divide (II) by 50 and then subtract (I) from it.

y = 15

Now putting y = 15 in eqn. (I), we get

x + 15 = 25

x = 10

Hence, Manna has 10 notes of ₹ 50 and 15 notes of ₹ 100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. 

Solution:

Let the fixed charge for the first three days be ₹ x and,

The charge for each day extra be ₹ y.

Here, According to the given condition,

x + 4y = 27 …….…………………………. (I)

x + 2y = 21 ………………………………………….. (II)

Subtract (II) from (I), we get

2y = 6

y = 3

Now putting y = 3 in eqn. (II), we get

x + 4(3) = 27

x = 27 – 12

x = 15

Hence, the fixed charge is ₹15

And the Charge per day is ₹ 3



Previous Article
Next Article

Similar Reads

NCERT Solutions Class 10 - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1
Question 1. Form the pair of linear equations in the following problems, and find their solutions graphically.(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.Solution: Let's take, Number of girls = x Number of boys = y A
11 min read
NCERT Solutions Class 10 - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2
Question 1. Solve the following pair of linear equations by the substitution method(i) x + y = 14 and x – y = 4Solution: x + y = 14 ........... (1) x – y = 4 .............. (2) x = 14 – y Substitute x in (2) (14 – y) – y = 4 14 – 2y = 4 2y = 10 Transposing 2 y = 10/2 y = 5 x = 14 – y x = 9 Therefore, x = 9 and y = 5. (ii) s – t = 3 and (s/3) + (t/2
8 min read
NCERT Solutions Class 10 - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3
Question 1. Solve the following pair of linear equations by the elimination method and the substitution method:(i) x + y = 5 and 2x – 3y = 4 Solution: Here, the two given eqn. are as follows: x + y = 5 ..........(I) 2x – 3y = 4 ...........(II) ELIMINATION METHOD: Multiply equation (I) by 2, and then subtract (II) from it, we get 5y = 6 y = 6/5 Now
8 min read
Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.1
Question 1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. Solution: Present age of Aftab = x And, the present age of his daughter = y Seven years ago, A
3 min read
Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.2
Question 1. Form the pair of linear equations in the following problems, and find their solutions graphically.(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Solution: Let's take, Number of girls = x Number of boys = y
11 min read
Class 10 NCERT Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.3
Question 1. Solve the following pair of linear equations by the substitution method(i) x + y = 14 and x – y = 4 Solution: x + y = 14 ........... (1) x – y = 4 .............. (2) x = 14 – y Substitute x in (2) (14 – y) – y = 4 14 – 2y = 4 2y = 10 Transposing 2 y = 10/2 y = 5 x = 14 – y x = 9 Therefore, x = 9 and y = 5. (ii) s – t = 3 and (s/3) + (t/
8 min read
Class 10 NCERT Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.5
Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0 Solution: Here, a1 = 1, b1 = -3, c1 = -3 a2 = 3, b2 = -9, c2 = -2 So, [Tex]\frac{a_1}{a_2} = \frac{1}{3}[/Te
9 min read
Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.6
Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations:(i) [Tex]\frac{1}{2x} + \frac{1}{3y} = 2[/Tex] [Tex]\frac{1}{3x} + \frac{1}{2y}= \frac{13}{6}[/Tex] Solution: Lets, take 1/x = a and 1/y = b Here, the two given equation will be as follows: [Tex]\frac{a}{2}[/Tex] + [Tex]\frac{b}{3}[/Tex] = 2 Multiply i
8 min read
Class 10 NCERT Solutions- Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.7
Question 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju. Solution: The age difference between Ani and Biju = 3 yrs. Case 1: Either Biju is 3 years older than that of Ani,
11 min read
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables- This article curated by the GeeksforGeeks experts, contains free NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables to help students develop an easy approach to solving problems related to this chapter. Class 10 Maths NCERT So
59 min read