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Exercise 3.5 - Pair of Linear Equations in Two Variables RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

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 Page 1


 
  
  
  
  
  
  
 
Exercise 3.5 
In each of the following systems of equations determine whether the system has a unique 
solution, no solution or infinitely many solutions. In case there is a unique solution, find it: 
(1 -4) 
1. 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
Page 2


 
  
  
  
  
  
  
 
Exercise 3.5 
In each of the following systems of equations determine whether the system has a unique 
solution, no solution or infinitely many solutions. In case there is a unique solution, find it: 
(1 -4) 
1. 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 3, 3 a b c ? ? ? ? ? 
And 
2 2 2
3, 9, 2 a b c ? ? ? ? ? 
We have, 
1
2
1
2
1
3
31
93
a
a
b
b
?
?
? ?
?
 
And, 
1
2
33
22
c
c
?
? ?
?
 
Clearly, 
1 1 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solutions. 
 
2. 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
2, 1, 5 a b c ? ? ? ? 
And 
222
4, 2, 10 a b c ? ? ? ? 
We have, 
1
2
1
2
21
42
1
2
a
a
b
b
? ?
?
 
And, 
1
2
51
10 2
c
c
?
? ?
?
 
Page 3


 
  
  
  
  
  
  
 
Exercise 3.5 
In each of the following systems of equations determine whether the system has a unique 
solution, no solution or infinitely many solutions. In case there is a unique solution, find it: 
(1 -4) 
1. 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 3, 3 a b c ? ? ? ? ? 
And 
2 2 2
3, 9, 2 a b c ? ? ? ? ? 
We have, 
1
2
1
2
1
3
31
93
a
a
b
b
?
?
? ?
?
 
And, 
1
2
33
22
c
c
?
? ?
?
 
Clearly, 
1 1 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solutions. 
 
2. 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
2, 1, 5 a b c ? ? ? ? 
And 
222
4, 2, 10 a b c ? ? ? ? 
We have, 
1
2
1
2
21
42
1
2
a
a
b
b
? ?
?
 
And, 
1
2
51
10 2
c
c
?
? ?
?
 
 
Clearly, 
111
222
a b c
a b c
?? 
So, the given system of equation has infinity many solutions. 
 
3. 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Sol: 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Compare it with 
1 1 1
1 2 2
0
0
a x by c
a x by c
? ? ?
? ? ?
 
We get 
1 3, 1 5 1 20
2 6, 2 10 2 40
a b and c
a b and c
? ? ? ? ?
? ? ? ? ?
 
1 1 1
2 2 2
3 5 20
,
6 10 40
a b c
and
a b c
? ?
? ? ?
??
 
Simplifying it we get 
1 1 1
2 2 2
1 1 1
,
2 2 2
a b c
and
a b c
? ? ? 
Hence 
111
222
a b c
a b c
?? 
So both lines are coincident and overlap with each other 
So, it will have infinite or many solutions 
 
4. 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
The given system if equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 2, 8 a b c ? ? ? ? ? 
And, 
2 2 2
5, 10, 10 a b c ? ? ? ? ? 
Page 4


 
  
  
  
  
  
  
 
Exercise 3.5 
In each of the following systems of equations determine whether the system has a unique 
solution, no solution or infinitely many solutions. In case there is a unique solution, find it: 
(1 -4) 
1. 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 3, 3 a b c ? ? ? ? ? 
And 
2 2 2
3, 9, 2 a b c ? ? ? ? ? 
We have, 
1
2
1
2
1
3
31
93
a
a
b
b
?
?
? ?
?
 
And, 
1
2
33
22
c
c
?
? ?
?
 
Clearly, 
1 1 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solutions. 
 
2. 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
2, 1, 5 a b c ? ? ? ? 
And 
222
4, 2, 10 a b c ? ? ? ? 
We have, 
1
2
1
2
21
42
1
2
a
a
b
b
? ?
?
 
And, 
1
2
51
10 2
c
c
?
? ?
?
 
 
Clearly, 
111
222
a b c
a b c
?? 
So, the given system of equation has infinity many solutions. 
 
3. 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Sol: 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Compare it with 
1 1 1
1 2 2
0
0
a x by c
a x by c
? ? ?
? ? ?
 
We get 
1 3, 1 5 1 20
2 6, 2 10 2 40
a b and c
a b and c
? ? ? ? ?
? ? ? ? ?
 
1 1 1
2 2 2
3 5 20
,
6 10 40
a b c
and
a b c
? ?
? ? ?
??
 
Simplifying it we get 
1 1 1
2 2 2
1 1 1
,
2 2 2
a b c
and
a b c
? ? ? 
Hence 
111
222
a b c
a b c
?? 
So both lines are coincident and overlap with each other 
So, it will have infinite or many solutions 
 
4. 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
The given system if equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 2, 8 a b c ? ? ? ? ? 
And, 
2 2 2
5, 10, 10 a b c ? ? ? ? ? 
 
We have, 
1
2
1
2
1
5
21
10 5
a
a
b
b
?
?
??
?
 
And, 
1
2
84
10 5
c
c
?
? ?
?
 
Clearly, 
1 2 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solution. 
 
5. 
2 5 0
3 1 0
kx y
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 5 0
3 1 0
kx y
xy
? ? ?
? ? ?
 
The system of equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
, 2, 5 a k b c ? ? ? ? 
And, 
2 2 2
3, 1, 1 a b c ? ? ? ? 
For a unique solution, we must have 
11
22
2
31
6
ab
ab
k
k
?
? ?
??
 
So, the given system of equations will have a unique solution for all real values of k other 
than 6. 
 
6. 4x + ky + 8 = 0 
2x + 2y + 2 = 0 
Sol: 
Here 
1 2 1 2
4, , 2, 2 a a k b b ? ? ? ? 
Now for the given pair to have a unique solution: 
11
22
ab
ab
? 
Page 5


 
  
  
  
  
  
  
 
Exercise 3.5 
In each of the following systems of equations determine whether the system has a unique 
solution, no solution or infinitely many solutions. In case there is a unique solution, find it: 
(1 -4) 
1. 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
 
3 3 0
3 9 2 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 3, 3 a b c ? ? ? ? ? 
And 
2 2 2
3, 9, 2 a b c ? ? ? ? ? 
We have, 
1
2
1
2
1
3
31
93
a
a
b
b
?
?
? ?
?
 
And, 
1
2
33
22
c
c
?
? ?
?
 
Clearly, 
1 1 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solutions. 
 
2. 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 5 0
4 2 10 0
xy
xy
? ? ?
? ? ?
 
The given system of equations is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
2, 1, 5 a b c ? ? ? ? 
And 
222
4, 2, 10 a b c ? ? ? ? 
We have, 
1
2
1
2
21
42
1
2
a
a
b
b
? ?
?
 
And, 
1
2
51
10 2
c
c
?
? ?
?
 
 
Clearly, 
111
222
a b c
a b c
?? 
So, the given system of equation has infinity many solutions. 
 
3. 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Sol: 
3 5 20
6 10 40
xy
xy
? ?
? ?
 
Compare it with 
1 1 1
1 2 2
0
0
a x by c
a x by c
? ? ?
? ? ?
 
We get 
1 3, 1 5 1 20
2 6, 2 10 2 40
a b and c
a b and c
? ? ? ? ?
? ? ? ? ?
 
1 1 1
2 2 2
3 5 20
,
6 10 40
a b c
and
a b c
? ?
? ? ?
??
 
Simplifying it we get 
1 1 1
2 2 2
1 1 1
,
2 2 2
a b c
and
a b c
? ? ? 
Hence 
111
222
a b c
a b c
?? 
So both lines are coincident and overlap with each other 
So, it will have infinite or many solutions 
 
4. 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation may be written as 
2 8 0
5 10 10 0
xy
xy
? ? ?
? ? ?
 
The given system if equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 2, 8 a b c ? ? ? ? ? 
And, 
2 2 2
5, 10, 10 a b c ? ? ? ? ? 
 
We have, 
1
2
1
2
1
5
21
10 5
a
a
b
b
?
?
??
?
 
And, 
1
2
84
10 5
c
c
?
? ?
?
 
Clearly, 
1 2 1
2 2 2
a b c
a b c
?? 
So, the given system of equation has no solution. 
 
5. 
2 5 0
3 1 0
kx y
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 5 0
3 1 0
kx y
xy
? ? ?
? ? ?
 
The system of equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
, 2, 5 a k b c ? ? ? ? 
And, 
2 2 2
3, 1, 1 a b c ? ? ? ? 
For a unique solution, we must have 
11
22
2
31
6
ab
ab
k
k
?
? ?
??
 
So, the given system of equations will have a unique solution for all real values of k other 
than 6. 
 
6. 4x + ky + 8 = 0 
2x + 2y + 2 = 0 
Sol: 
Here 
1 2 1 2
4, , 2, 2 a a k b b ? ? ? ? 
Now for the given pair to have a unique solution: 
11
22
ab
ab
? 
 
i.e., 
4
22
k
? 
i.e., 4 k ? 
Therefore, for all values of k, except 4, the given pair of equations will have a unique 
solution. 
 
7. 
45
2 3 12
x y k
xy
??
? ?
 
Sol: 
The given system of equation is 
4 5 0
2 3 12 0
x y k
xy
? ? ?
? ? ?
 
The system of equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
4, 5, a b c k ? ? ? ? ? 
And, 
2 2 2
2, 3, 12 a b c ? ? ? ? ? 
For a unique solution, we must have 
11
22
45
23
ab
ab
?
?
? ?
?
 
k ? is any real number. 
So, the given system of equations will have a unique solution for all real values of k. 
 
8. 
23
5 7 0
xy
x ky
? ?
? ? ?
 
Sol: 
The given system of equation is 
2 3 0
5 7 0
xy
x ky
? ? ?
? ? ?
 
The system of equation is of the form 
1 1 1
2 2 2
0
0
a x b y c
a x b y c
? ? ?
? ? ?
 
Where, 
1 1 1
1, 2, 3 a b c ? ? ? ? 
And, 
2 2 2
5, , 7 a b k c ? ? ? 
For a unique solution, we must have
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Document Description: RD Sharma Solutions: Exercise 3.5 - Pair of Linear Equations in Two Variables for Class 10 2024 is part of Extra Documents, Videos & Tests for Class 10 preparation. The notes and questions for RD Sharma Solutions: Exercise 3.5 - Pair of Linear Equations in Two Variables have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Exercise 3.5 - Pair of Linear Equations in Two Variables covers topics like and RD Sharma Solutions: Exercise 3.5 - Pair of Linear Equations in Two Variables Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Exercise 3.5 - Pair of Linear Equations in Two Variables.
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