BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Book Answers

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Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 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 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 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 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbooks. These Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 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 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
Chapters	All
Provider	Hsslive

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BSEB Bihar Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions
Solution:
Given equation is y = 3x + 5
For y = 0, 3x + 5 = 0 ⇒ x = – 53
∴ (- 53, 0) is one solution.
For x = 0, y = 0 + 5 = 5
∴ (0, 5) is another solution.
For x = 1, y = 3 x 1 + 5 = 8
∴ (1, 8) is another solution.
Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with this value of y constitutes, another solution of the given solution. So, there is no end to different solutions of a linear equation in two variables.
∴ A linear equation in two variables has infinitely many solutions.

Question 2.
Write four solutions for each of the following equations :
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) The given equation can be written as y = 7 – 2x.
For x = 0, y = 7 – 2 x 0 = 7 – 0 = 7
For x = 1, y = 7 – 2 x 1 = 7 – 2 = 5
For x = 2, y = 7 – 2 x 2 = 7 – 4 = 3
For x = 3, y = 7 – 2 x 3 = 7 – 6 = 1
∴ The four solutions of the given equation are (0, 7), (1, 5), (2, 3) and (3, 1).

(ii) The given equation can be written as y = 9 – πx
For x = 0, y = 9 – 0 = 9
For x = 1, y = 9 – π
For x = 2, y = 9 – 2π
For x = 3, y = 9 – 3π
∴ The four solution of the given equation are (0, 9), (1, 9 – π), (2, 9 – π) and (3, 9 – 3π).

(iii) The given equation can be written as x = Ay
For x = 0, y = 0
For x = 1, y = 4 x 1 = 4
For x = 2, y = 4 x 2 = 8
For x = 3, y = 4 x 3 = 12
∴ The four solutions of the given equation are (0, 0), 0, 4), (2, 8) and (3, 12).

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not :
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) (2‾√, 42‾√)
(v) (1, 1)
Solution:
(i) Putting x = 0, y = 2 in L.H.S. of x – 2y = 4, we
have
L.H.S. = 0 – 2 x 2 = – 4 ≠ R.H.S.
∴ x = 0, y = 2 is not its solution.

(ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have
L.H.S. = 2 – 2 x 0 = 2 – 0 = 2 ≠ R.H.S.
∴ x = 2,y = 0 is not its solution.

(iii) putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 4 – 0 = 4 = R.H.S.
∴ x = 4, y = 0 is its solution.

(iv) Putting x = 2‾√ , y = 42‾√ in the L.H.S. of x – 2y = 4, we have
L.H.S. = 2‾√ – 2 x 42‾√ = 2‾√ – 82‾√
= – 72‾√ ≠ R.H.S.
∴ (2‾√, 42‾√) is not its solution.

(v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 1 – 2 x 1 = 1 – 2 = – 1 ≠ R.H.S.
∴ x = 1, y = 1 is not its solution.

Question 4.
Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.
∴ 2 x 2 + 3 x 1 = k ⇒ k = 4 + 3 = 7.

BSEB Textbook Solutions PDF for Class 9th

Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 exam preparation. The BSEB STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Books State Board syllabus with maximum efficiency.

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