RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 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 Ex 4.2.

Board RBSE
Class Class 9
Subject Maths
Chapter Chapter 4
Chapter Name Linear Equations in Two Variables
Exercise Ex 4.2
Number of Questions Solved 15
Category RBSE Solutions

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Solve each of the following systems of simultaneous linear (RBSESolutions.com) equations by the method of substitution. [Q1. to Q6.]

Question 1.
2x + 3y = 9
3x + 4y = 5
Solution.
The given equations are
2x + 3y = 9 ….(i)
and 3x + 4y = 5 ….(ii)
From (i) 2x = 9 – 3y
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 1
Hence, the (RBSESolutions.com) required solution is x = – 21 and y = 17.

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Question 2.
x + 2y = – 1
2x – 3y = 12
Solution
The given equations are
x + 2y = – 1 ..(i)
and 2x – 3y = 12 …(ii)
From (i) x = – 1 – 2y …(iii)
Substituting this (RBSESolutions.com) value of x in (ii), we get
2( – 1 – 2y) – 3y = 2
⇒ – 2 – 4y – 3y = 12
⇒ – 7y = 14
⇒ y = -2
Putting y = – 2 in (iii), we get
x = -1-2×(-2) = -1+4 = 3
Hence the required solution is x = 3 and y = – 2.

Question 3.
3x + 2y = 11
2x + 3y = 4
Solution
The given equations are
3x + 27 = 11 ….(i)
and 2x + 3y = 4 …(ii)
From (i) 2y = 11 – 3x
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 2
Hence the required (RBSESolutions.com) solution is x = 5 and y = – 2

Question 4.
8x + 5y = 9
3x + 2y = 4
Solution
The given equations are
8x + 5y = 9 …(i)
and 3x + 2y = 4 …(ii)
From (ii) 2y = 4 – 3x
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 3
Hence, the required (RBSESolutions.com) solution is x = – 2 and y = 5.

Question 5.
4x – 5y = 39
2x – 7y = 51
Solution
The given equations are
4x – 5y = 39 …(i)
and 2x – 7y = 51 …(ii)
From equation (ii),
2x – 51 = 7y
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 4
Hence the required (RBSESolutions.com) solution is x = 1, y = -7.

RBSE Solutions

Question 6.
5x – 2y = 19
3x + y = 18
Solution
The given equations are
5x – 2y = 19 …(i)
and 3x + y = 18 …(ii)
From (ii), y = 18 – 3x …(iii)
Substituting y from (iii) in (i), we get
5x – 2(18 – 3x) = 19
⇒ 5x – 36 + 6x = 19
⇒ 11x = 55
⇒ x = 5
Putting x = 5 in (iii), we get
y = 18 – 3 × 5
y = 18 – 15 = 3
Hence the required (RBSESolutions.com) solution is x = 5 and y = 3.

Solve each of the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.

Question 7.
2x + y = 13
5x – 3y = 16
Solution
The given equations are
2x + y = 13 …(i)
and 5x – 3y= 16 …(ii)
Multiplying equation (i) by 3 and then adding in (ii), we get
11x = 55
\(x=\frac { 55 }{ 11 }=5\)
Substituting (RBSESolutions.com) the value of x in (i), we get
2 × 5 + y = 13
⇒ 10 + y = 13
⇒ y = 13 – 10 = 3
Hence the required solution is x = 5, y = 3.

Question 8.
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Solution
The given equations are
0.4x + 0.3y = 1.7 …(i)
0.7x – 0.2y = 0.8 …(ii)
Multiplying equation (i) by 2 and (ii) by 3 and then (RBSESolutions.com) adding, we get
2.9x = 5.8
⇒ x = 2
Substituting the value of x in (i), we get
0.4 x 2 + 0.3y = 1.7
⇒ 0.3y = 1.7 – 0.8
⇒ 0.3y = 0.9
⇒ y = 3

Question 9.
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 2
Solution
The given (RBSESolutions.com) equations are
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 5
Hence the required (RBSESolutions.com) solution is x = 14 and y = 9

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Question 10.
11x + 15y = – 23
7x – 2y = 20
Solution
The given equations are
11x + 15y = – 23 …(i)
and 7x -2y = 20 …(ii)
Multiplying (RBSESolutions.com) equation (i) by 2 and (ii) by 15 and then adding, we get
127x = 254
⇒ x = 2
Putting x = 2 in (ii), we get
7 × 2 – 2y = 20
⇒ 14 – 20 = 2y
⇒ 2y = – 6
⇒ y = – 3
Hence the required solution is x = 2 and y = – 3.

Question 11.
3x – 7y + 10 = 0
y – 2x = 3
Solution
The given (RBSESolutions.com) equations are
3x – 7y + 10 = 0 …(i)
and y – 2x = 3 …(ii)
Multiplying equation (i) by 2 and (ii) by 3 and then adding, we get
– 11y = – 11
⇒ y = 1
Putting y = 1 in (ii), we get
1 – 2x = 3 ⇒ 1 – 3 = 2x
⇒ 2x = -2
⇒ x = -1
Hence the required solution is x = – 1 and y = 1

Question 12.
x + 2y = \(\frac { 3 }{ 2 }\)
2x + y = \(\frac { 3 }{ 2 }\)
Solution
The given (RBSESolutions.com) equations are
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 6

Solve each of the (RBSESolutions.com) following (Q13. to Q15.)

Question 13.
8v – 3u = 5uv
6v – 5u = – 2uv
Solution
The given equations are
8v – 3u = 5uv …(i)
and 6v – 5u = – 2uv …(ii)
Dividing equation (i) and (ii) both side by uv, we get
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 7

RBSE Solutions

Question 14.
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 8
Solution
The given equations are
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 9
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 13

Question 15.
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 11
Solution
The given equations are
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 12
Adding (v) and (vi), we get
2x = 6
⇒ x = 3 and 3 + y = 5
⇒ y = 2
Hence the required solution is x = 3 and y = 2.

RBSE Solutions

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