RD Sharma Solutions: Exercise 3.4 - Pair of Linear Equations in Two Variables Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RD Sharma Solutions: Exercise 3.4 - Pair of Linear Equations in Two Variables Notes | EduRev

 Page 1


 
 
 
 
 
                                                        Exercise 3.4 
Solve each of the following systems of equations by the method of cross-multiplication: 
 
1. 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Here, 
1 1 1
2 2 2
1, 2, 1
2, 3 12
a b c
a b and c
? ? ?
? ? ? ? ?
 
By cross-multiplication, we get 
? ? ? ? ? ? ? ?
1
2 12 1 3 1 12 1 2 1 3 2 2
x y ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
24 3 12 2 3 4
x y ?
? ? ?
? ? ? ? ? ?
 
1
21 14 7
x y ?
? ? ?
? ? ?
 
Now, 
1
21 7
21
3
7
x
x
?
? ?
?
? ? ?
?
 
And, 
Page 2


 
 
 
 
 
                                                        Exercise 3.4 
Solve each of the following systems of equations by the method of cross-multiplication: 
 
1. 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Here, 
1 1 1
2 2 2
1, 2, 1
2, 3 12
a b c
a b and c
? ? ?
? ? ? ? ?
 
By cross-multiplication, we get 
? ? ? ? ? ? ? ?
1
2 12 1 3 1 12 1 2 1 3 2 2
x y ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
24 3 12 2 3 4
x y ?
? ? ?
? ? ? ? ? ?
 
1
21 14 7
x y ?
? ? ?
? ? ?
 
Now, 
1
21 7
21
3
7
x
x
?
? ?
?
? ? ?
?
 
And, 
 
1
14 7
1
14 7
14
2
7
y
y
y
?
?
??
?
??
?
? ? ? ?
 
Hence, the solution of the given system of equations is 3, 2. xy ? ? ? 
 
2. 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
3, 2, 25 a b c ? ? ? 
2 2 2
2, 1 10 a b and c ? ? ? 
By cross-multiplication, we have 
1
2 10 25 1 3 10 25 2 3 1 2 2
xy ?
? ? ?
? ? ? ? ? ? ? ? ?
 
1
20 25 30 50 3 4
1
5 20 1
xy
xy
?
? ? ?
? ? ?
?
? ? ?
? ? ?
 
Now, 
1
51
x
?
??
 
5
5
1
x
?
? ? ?
?
 
And, 
1
20 1
1
20
20
y
y
y
?
?
??
??
? ? ?
 
Hence, 5, 20 xy ? ? ? is the solution of the given system of equations. 
 
3. 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
Page 3


 
 
 
 
 
                                                        Exercise 3.4 
Solve each of the following systems of equations by the method of cross-multiplication: 
 
1. 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Here, 
1 1 1
2 2 2
1, 2, 1
2, 3 12
a b c
a b and c
? ? ?
? ? ? ? ?
 
By cross-multiplication, we get 
? ? ? ? ? ? ? ?
1
2 12 1 3 1 12 1 2 1 3 2 2
x y ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
24 3 12 2 3 4
x y ?
? ? ?
? ? ? ? ? ?
 
1
21 14 7
x y ?
? ? ?
? ? ?
 
Now, 
1
21 7
21
3
7
x
x
?
? ?
?
? ? ?
?
 
And, 
 
1
14 7
1
14 7
14
2
7
y
y
y
?
?
??
?
??
?
? ? ? ?
 
Hence, the solution of the given system of equations is 3, 2. xy ? ? ? 
 
2. 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
3, 2, 25 a b c ? ? ? 
2 2 2
2, 1 10 a b and c ? ? ? 
By cross-multiplication, we have 
1
2 10 25 1 3 10 25 2 3 1 2 2
xy ?
? ? ?
? ? ? ? ? ? ? ? ?
 
1
20 25 30 50 3 4
1
5 20 1
xy
xy
?
? ? ?
? ? ?
?
? ? ?
? ? ?
 
Now, 
1
51
x
?
??
 
5
5
1
x
?
? ? ?
?
 
And, 
1
20 1
1
20
20
y
y
y
?
?
??
??
? ? ?
 
Hence, 5, 20 xy ? ? ? is the solution of the given system of equations. 
 
3. 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
 
Sol: 
The given system of equations may be written as 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
2, 1, 35 a b c ? ? ? ? 
2 2 2
3, 4 65 a b and c ? ? ? ? 
By cross multiplication, we have 
? ? ? ? ? ? ? ?
1
1 65 35 4 2 65 35 3 2 4 1 3
xy ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
65 140 130 105 8 3
xy ?
? ? ?
? ? ? ? ?
 
1
75 25 5
xy ?
? ? ?
?
 
1
75 25 5
xy
? ? ? 
Now, 
1
25 5
25
5
5
y
y
?
? ? ?
 
Hence, 15, 5 xy ?? is the solution of the given system of equations. 
 
4. 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Here,  
1 1 1
2 2 2
2, 1, 6
1, 1 2
a b c
a b and c
? ? ? ? ?
? ? ? ? ?
 
By cross multiplication, we get 
Page 4


 
 
 
 
 
                                                        Exercise 3.4 
Solve each of the following systems of equations by the method of cross-multiplication: 
 
1. 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Here, 
1 1 1
2 2 2
1, 2, 1
2, 3 12
a b c
a b and c
? ? ?
? ? ? ? ?
 
By cross-multiplication, we get 
? ? ? ? ? ? ? ?
1
2 12 1 3 1 12 1 2 1 3 2 2
x y ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
24 3 12 2 3 4
x y ?
? ? ?
? ? ? ? ? ?
 
1
21 14 7
x y ?
? ? ?
? ? ?
 
Now, 
1
21 7
21
3
7
x
x
?
? ?
?
? ? ?
?
 
And, 
 
1
14 7
1
14 7
14
2
7
y
y
y
?
?
??
?
??
?
? ? ? ?
 
Hence, the solution of the given system of equations is 3, 2. xy ? ? ? 
 
2. 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
3, 2, 25 a b c ? ? ? 
2 2 2
2, 1 10 a b and c ? ? ? 
By cross-multiplication, we have 
1
2 10 25 1 3 10 25 2 3 1 2 2
xy ?
? ? ?
? ? ? ? ? ? ? ? ?
 
1
20 25 30 50 3 4
1
5 20 1
xy
xy
?
? ? ?
? ? ?
?
? ? ?
? ? ?
 
Now, 
1
51
x
?
??
 
5
5
1
x
?
? ? ?
?
 
And, 
1
20 1
1
20
20
y
y
y
?
?
??
??
? ? ?
 
Hence, 5, 20 xy ? ? ? is the solution of the given system of equations. 
 
3. 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
 
Sol: 
The given system of equations may be written as 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
2, 1, 35 a b c ? ? ? ? 
2 2 2
3, 4 65 a b and c ? ? ? ? 
By cross multiplication, we have 
? ? ? ? ? ? ? ?
1
1 65 35 4 2 65 35 3 2 4 1 3
xy ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
65 140 130 105 8 3
xy ?
? ? ?
? ? ? ? ?
 
1
75 25 5
xy ?
? ? ?
?
 
1
75 25 5
xy
? ? ? 
Now, 
1
25 5
25
5
5
y
y
?
? ? ?
 
Hence, 15, 5 xy ?? is the solution of the given system of equations. 
 
4. 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Here,  
1 1 1
2 2 2
2, 1, 6
1, 1 2
a b c
a b and c
? ? ? ? ?
? ? ? ? ?
 
By cross multiplication, we get 
 
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
1
1 2 6 1 2 2 6 1 2 1 1 1
1
2 6 4 6 2 1
1
4 2 1
1
42
xy
xy
xy
xy
?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
?
? ? ?
? ? ? ? ?
?
? ? ?
??
?
? ? ? ?
?
 
Now, 
? ? ? ?
1
4
4 1 4
x
x
??
?
? ? ? ? ? ?
 
And, 
? ?
1
2
12
2
2
y
y
y
y
?
??
? ? ? ? ?
? ? ? ?
??
 
Hence, 4, 2 xy ?? is the solution of the given system of the equations. 
 
5. 2
xy
xy
?
? 
6
xy
xy
?
? 
Sol: 
The given system of equations is 
? ?
2
2
11
2
11
2 ......
xy
xy
xy
xy xy
yx
i
xy
?
?
? ? ?
? ? ?
? ? ?
 
And, 
Page 5


 
 
 
 
 
                                                        Exercise 3.4 
Solve each of the following systems of equations by the method of cross-multiplication: 
 
1. 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
2 1 0
2 3 12 0
x y
x y
? ? ?
? ? ?
 
Here, 
1 1 1
2 2 2
1, 2, 1
2, 3 12
a b c
a b and c
? ? ?
? ? ? ? ?
 
By cross-multiplication, we get 
? ? ? ? ? ? ? ?
1
2 12 1 3 1 12 1 2 1 3 2 2
x y ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
24 3 12 2 3 4
x y ?
? ? ?
? ? ? ? ? ?
 
1
21 14 7
x y ?
? ? ?
? ? ?
 
Now, 
1
21 7
21
3
7
x
x
?
? ?
?
? ? ?
?
 
And, 
 
1
14 7
1
14 7
14
2
7
y
y
y
?
?
??
?
??
?
? ? ? ?
 
Hence, the solution of the given system of equations is 3, 2. xy ? ? ? 
 
2. 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equation is 
3 2 25 0
2 10 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
3, 2, 25 a b c ? ? ? 
2 2 2
2, 1 10 a b and c ? ? ? 
By cross-multiplication, we have 
1
2 10 25 1 3 10 25 2 3 1 2 2
xy ?
? ? ?
? ? ? ? ? ? ? ? ?
 
1
20 25 30 50 3 4
1
5 20 1
xy
xy
?
? ? ?
? ? ?
?
? ? ?
? ? ?
 
Now, 
1
51
x
?
??
 
5
5
1
x
?
? ? ?
?
 
And, 
1
20 1
1
20
20
y
y
y
?
?
??
??
? ? ?
 
Hence, 5, 20 xy ? ? ? is the solution of the given system of equations. 
 
3. 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
 
Sol: 
The given system of equations may be written as 
2 35 0
3 4 65 0
xy
xy
? ? ?
? ? ?
 
Here, 
1 1 1
2, 1, 35 a b c ? ? ? ? 
2 2 2
3, 4 65 a b and c ? ? ? ? 
By cross multiplication, we have 
? ? ? ? ? ? ? ?
1
1 65 35 4 2 65 35 3 2 4 1 3
xy ?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
65 140 130 105 8 3
xy ?
? ? ?
? ? ? ? ?
 
1
75 25 5
xy ?
? ? ?
?
 
1
75 25 5
xy
? ? ? 
Now, 
1
25 5
25
5
5
y
y
?
? ? ?
 
Hence, 15, 5 xy ?? is the solution of the given system of equations. 
 
4. 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Sol: 
The given system of equations may be written as 
2 6 0
20
xy
xy
? ? ?
? ? ?
 
Here,  
1 1 1
2 2 2
2, 1, 6
1, 1 2
a b c
a b and c
? ? ? ? ?
? ? ? ? ?
 
By cross multiplication, we get 
 
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
1
1 2 6 1 2 2 6 1 2 1 1 1
1
2 6 4 6 2 1
1
4 2 1
1
42
xy
xy
xy
xy
?
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
?
? ? ?
? ? ? ? ?
?
? ? ?
??
?
? ? ? ?
?
 
Now, 
? ? ? ?
1
4
4 1 4
x
x
??
?
? ? ? ? ? ?
 
And, 
? ?
1
2
12
2
2
y
y
y
y
?
??
? ? ? ? ?
? ? ? ?
??
 
Hence, 4, 2 xy ?? is the solution of the given system of the equations. 
 
5. 2
xy
xy
?
? 
6
xy
xy
?
? 
Sol: 
The given system of equations is 
? ?
2
2
11
2
11
2 ......
xy
xy
xy
xy xy
yx
i
xy
?
?
? ? ?
? ? ?
? ? ?
 
And, 
 
? ?
6
6
11
6
11
6 ......
xy
xy
xy
xy xy
yx
ii
xy
?
?
? ? ?
? ? ?
? ? ?
 
Taking 
11
, u and v
xy
?? we get 
? ? 2 2 0 ..... u v u v iii ? ? ? ? ? ? 
And, ? ? 6 6 0 ..... u v u v iv ? ? ? ? ? ? ? 
Here, 
111
2 2 2
1, 1, 2
1, 1 6
a b c
a b and c
? ? ? ?
? ? ? ?
 
By cross multiplication  
? ? ? ? ? ? ? ?
1
1 6 2 1 1 6 2 1 1 1 1 1
uv
? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ?
 
1
6 2 6 2 1 1
1
4 8 2
uv
uv
?
? ? ?
? ? ? ?
?
? ? ?
?
 
Now, 
1
42
u
?
?
 
4
2
2
u ? ? ? ?
?
 
And, 
1
82
v ?
?
?
 
8
4
2
4
4
v
v
v
? ? ? ? ?
?
? ? ? ?
??
 
Now, 
1 1 1 1
24
x and y
uv
?
? ? ? ? 
Hence, 
11
,
24
xy
?
?? is the solution of the given system of equations. 
 
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