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RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1

January 11, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.1.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 9
Subject Maths
Chapter Chapter 4
Chapter Name Linear Equations in Two Variables
Exercise Ex 4.1
Number of Questions Solved 11
Category RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1

Solve the following pair of equations graphically

Question 1.
x + 3y = 6; 2x – 3y = 12
Solution.
From the given (RBSESolutions.com) equations
x + 3y = 6 ⇒ x = 6 – 3y …(i)
when y = 1, x = 6 – 3 × 1 = 3
when y = 2, x = 6 – 3 × 2 = 0
when y = 0, x = 6 – 3 × 0 = 6
when y = – 1, x = 6 – 3 × – 1 = 9
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 1
Now plotting and joining the points (RBSESolutions.com) from table 1 and table II on the same graph paper, we find their point of intersection is (6, 0).
Hence, (6, 0) is the solution of the pair of equations.

RBSE Solutions

Question 2.
2x + y = 6; 2x – y + 2 = 0
Solution.
From the given equation
2x + y = 6 …(i)
Expressing y in terms of x, we get
y = 6 – 2x
when x = 0, y = 6 – 2 × 0 = 6
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 2
Now plotting and joining the (RBSESolutions.com) points from table I and II on the same graph paper, we find their point of intersection is (1, 4).
Hence, (1, 4) is the required solution.

Question 3.
x – 2y = 6; 3x – 6y = 0
Solution.
From the given equation
x – 2y = 6 …(i)
Expressing x in terms of y, we get
x = 2y + 6
when y = 0, x = 2 × 0 + 6 = 6
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 3
Now plotting and joining the points (RBSESolutions.com) from table (i) and (ii) on the same graph paper, we find that their is no point of intersection i.e. no solution i.e. lines goes parallel to each other.

Question 4.
x + y = 4; 2x – 3y = 3
Solution.
From the given equation
x + y = 4
⇒ y = 4 – x …(i)
when x = 0, y = 4 – 0 = 4
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 4
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 5
Now plotting and joining the (RBSESolutions.com) points from table (i) and (ii) on the same graph paper,
we find that their point of intersection is (3, 1).
Hence, (3, 1) is the required solutions.

RBSE Solutions

Question 5.
2x – 3y + 13 = 0; 3x – 2y + 12 = 0
Solution.
From the given equation
2x – 3y + 13 = 0
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 6
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 7
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 8
Now plotting and joining the points (RBSESolutions.com) from table (i) and (ii) on the same graph, we find that their point of intersection is (- 2, 3).
Hence, the required solution is (- 2, 3).

Question 6.
3x – 4y = 1; \(-2x+\frac { 8 }{ 3 }y=5\)
Solution.
From the given equation
3x – 4y = 1
=> 4y = 3x – 1
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 9
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 10
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 11
Now plotting and joining the points from (RBSESolutions.com) table (i) and (ii) on the same graph paper we find that these lines are parallel to each other it means no solution.
Hence, the system of equations has no solutions.

RBSE Solutions

Question 7.
\(2x+\frac { y }{ 2 }-5=0\); \(\frac { x }{ 2 }+y=-4\)
Solution.
From the given equation
\(2x+\frac { y }{ 2 }-5=0\)
4x + y – 10 = 0
⇒ y = – 4x + 10
⇒ y = 10 – 4x …..(i)
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 12
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 13
Plot all above points (- 2, – 3), (- 4, – 2), (- 6, – 1) and (4, – 6) on the same (RBSESolutions.com) graph paper and join them and get the graph CD (Line segment).
The point of intersection of AB and CD is P(4, – 6) therefore the required solution is x = 4 and y = – 6.

Question 8.
0.3x + 0.4y = 3.2; 0.6x + 0.8y = 2.4
Solution.
The above equations can also be written as
3x + 4y = 32 …(i)
and 6x + 8y = 24 …(ii)
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 14
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 15
Now plotting and joining the points from (RBSESolutions.com) table (i) and (ii) on the same graph paper. We find that the lines goes parallel to each other. Hence the system of equation has no solution.

RBSE Solutions

Question 9.
2x + 3y = 8; \(4x-\frac { 3 }{ 2 }y=1\)
Solution.
We have,
2x + 3y = 8
\(4x-\frac { 3 }{ 2 }y=1\)
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 16
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 17
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 18
Plot the above pair of points from table (i) and table (ii) on the (RBSESolutions.com) same graph paper. Join them. After joining we find that the point of intersection of both the lines is (1, 2).
Hence, x = 1, y = 2 is the required solution.

Question 10.
3x – y = 2; 6x – 2y = 4
Solution.
Consider the equation 3x – y = 2
Expressing y in terms of x, we get
y = 3x – 2 …(i)
when x = 0, y = 3×0 – 2 = -2
when x = 1, y = 3×1 – 2 = 1
when x = 2, y = 3×2 – 2 = 4
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 19
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 20
Now plotting and joining the points from table (i) and (ii) we (RBSESolutions.com) find that the lines overlapping each other i.e. all the points coincide with each other. It means the system of equations has infinitely many solutions.

RBSE Solutions

Question 11.
3x + 2y = 0; 2x + y = -1
Solution.
From the given equation
3x + 2y = 0
We 2y = -3x
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 21
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 22
Now plot the points (1, – 3), (- 1, 1) and (- 2, 3) on the same (RBSESolutions.com) graph paper, join these points and obtain the line CD i.e. the graph of the equation 2x + y = – 1.
From the graph of two equations we see that AB and CD intersect and the coordinates of the point of intersection is (-2, 3)
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 23
Hence, x = – 2, and y = 3 is the required solution.

We hope the given RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations Ex 4.1 in Two Variables will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1, drop a comment below and we will get back to you at the earliest.

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