# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## BSEB Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Book Answers BSEB Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Book Answers

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## Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Books Solutions

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## BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks Solutions with Answer PDF Download

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## BSEB Bihar Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

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

1. x + y = 5 and 2x – 3y = 4
2. 3x + 4y = 10 and 2x – 2y = 2
3. 3x – 5y – 4 = 0 and 9x = 2y + 7
4. 𝑥2 + 2𝑦3 = – 1 and x – 𝑦3 = 3

Solution:
1. By elimination method:
The given system of equations is
x +y = 5 ……………….. (1) and
2x – 3y = 4 ………………….. (2)
Multiplying (1) by 3, we get
3x + 3y = 15 ……………….. (3)
Adding (2) and (3), we get
5x = 19 or x = 195
Putting x = 195 in (1), we get
195 + y = 5 or y = 5 – 195 = 25−195 = 65
Hence, x = 195, y = 65.

By substitution method:
The given system of equations is
x + y = 5 ……………. (1) and
2x – 3y = 4 ……………….. (2)
From (1), y = 5 – x
Substituting y = 5 – x in (2), we get
2x – 15 + 3x = 4 or 5x = 4 + 15
or 5x = 19 or x = 195
Putting x = 195 in (1), we get
195 + y = 5 or y = 5 – 195 = 25−195 = 65
Hence, x = 195, y = 65.

2. By elimination method:
The given system of equations is
𝑥2 + 2𝑦3 = – 1 or 3x + 4y = – 6 ……………… (1)
and x – 𝑦3 = 3 or 3x – y = 9 ………………. (2)
Multiplying (2) by 4 and adding to (1), we get
15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, y = – 3
By substitution method:
The given system of equations is
𝑥2 + 2𝑦3 = – 1 or 3x + 4y = – 6 ……………….. (1)
and x – 𝑦3 = 3 or 3x – y = 9 …………….. (2)
From (2), Putting y = 3x – 9 in (1), we get y = 3x – 9
3x + 4(3x – 9) = – 6 or 3x + 12x – 36 = – 6
or 15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, y = – 3

3. By elimination method:
The given system of equations is
3x – 5y – 4 = 0 or 3x – 5y = 4 ……………….. (1)
and 9x = 2y + 7 or 9x – 2y = 7 ………………….. (2)
Multiplying (1) by 3 , we get
9x – 15y = 12 ………………. (3)
Substracting (1) by 3, we get
– 13y = 5 or y = – 513
Putting y = – 513 in (1), we get By substitution method:
The given system of equations is 4. By elimination method:
The given system of equations is Multiplying (2) by 4 and adding to (1), we get
15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3

By substitution method:
The given system of equations is From (2), putting y = 3x – 9 in (1), we get y = 3x – 9
3x + 4(3x – 9) = – 6 or 3x + 12x – 36 = – 6
or 15x = 30 or x = 2
Putting x = 2 in (2), we get
3(2) – y = 9 or – y = 9 – 6 = 3
or y = – 3
Hence, x = 2, 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:
1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 if we only add 1 to the denominator. What is the fraction?

2. 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?

3. 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.

4. 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.

5. 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 she 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.
Solution:
1. Let x be the numerator and y be the denominator of the fraction. So, the fraction is 𝑥𝑦.
By given conditions:
𝑥+1𝑦−1 = 1 or x + 1 = y – 1
or x – y = – 2 ……………….. (1)
and 𝑥𝑦+1 = 12 or 2x = y + 1
or 2x – y = 1 ……………….. (2)
Subtracting (2) from (1), we get
(x – y) – (2x – y) = – 2 – 1
or x – y – 2x + y = – 3
or – x = – 3 or x = 3
Substituting x = 3 in (1), we get
3 – y = – 2 or y = 5
Hence, the required fraction is 35.

2. Let the present age of Nuri be x years
and the present age of Sonu be y years.
Five years ago,
Nuri’s age = (x – 5) years
Sonu’s age = (y – 5) years
As per conditions,
x – 5 = 3(y – 5) or x – 5 = 3y – 15
or x – 3y = – 15 + 5
or x – 3y = – 10 ………………… (1)
Ten years later,
Nuri’s age = (x + 10) years
As per condition’s,
x + 10 = 2 (y + 10) or x + 10 = 2y + 20
or x – 2y = 20 – 10
or x – 2y = 10 ………………… (2)
Putting y = 20 in (2), we get
x – 2(20) = 10 or x = 10 + 40 = 50
Nuri’s present age = 50 years
and Sonu’s present age = 20 years

3. Let the digits in the unit’s place and ten’s place be x and y respectively.
∴ Number = 10y + x
If the digits are reversed, the new number = 10y + x
As per conditions:
x + y = 9 ……………… (1)
and 9(10y + x) = 2(10x + y)
or 90y + 9x = 20x + 2y
or 20x – 9x + 2y – 90y = 0
From (1), y = 9 – x
Putting y = 9 – x in (2), we get
11x – 88(9 – x) = 0
or 11x – 88 × 9 + 88x = 0
or 99x = 88 × 9
Putting x = 8 in (1), we get
8 + y = 9 or y = 1
Hence, the number = 10y + x = 10 × 1 + 8 = 18

4. Let the number of Rs 50 notes be x and number of Rs 100 notes be y. Then,
x + y – 25 …………….. (1)
and 50x + 100y = 2000
or x + 2y = 40 …………….. (2)
On subtracting (1) from (2), we get
y = 15
Putting y = 15 in (1), we get
x + 15 = 25 or x = 10
Hence, number of Rs 50 notes = 10
and number of Rs 100 notes = 15

5. Let the fixed charges for 3 days be Rs x and charges per day be Rs y.
∴ By the given conditions,
x + 4y = 27 ………….. (1)
and x + 2y = 21 ……………… (2)
Subtracting (2) from (1), we get
2y = 6 or y = 3 Putting y = 3 in (1), we get
x + 4 × 3 = 27 or x = 27 – 12 = 15
∴ Fixed charges = Rs 15 and
charges per day = Rs 3

## Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks for Exam Preparations

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