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

Thursday, June 23, 2022

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 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 books are published by the Bihar Board Publishers. These Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 book solutions pdf online from this page.

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

Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks. These Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

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

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
Chapters	All
Provider	Hsslive

How to download Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions Answers PDF Online?

Visit our website - Hsslive
Click on the Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Answers.
Look for your Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks PDF.
Now download or read the Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions for PDF Free.

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

Find below the list of all BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

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:

x + y = 5 and 2x – 3y = 4
3x + 4y = 10 and 2x – 2y = 2
3x – 5y – 4 = 0 and 9x = 2y + 7
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

BSEB Textbook Solutions PDF for Class 10th

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

Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 exam preparation. The BSEB STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Books State Board syllabus with maximum efficiency.

FAQs Regarding Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Solutions

How to get BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Answers??

Students can download the Bihar Board Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Answers PDF from the links provided above.

Can we get a Bihar Board Book PDF for all Classes?

Yes you can get Bihar Board Text Book PDF for all classes using the links provided in the above article.

Important Terms

Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4, BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks, Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4, Bihar Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook solutions, BSEB Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks Solutions, Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4, BSEB STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks, Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4, Bihar Board STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook solutions, BSEB STD 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbooks Solutions,

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

Hsslive.co.in: Kerala Higher Secondary News, Plus Two Notes, Plus One Notes, Plus two study material, Higher Secondary Question Paper.

Thursday, June 23, 2022