RS Aggarwal Solutions: Exercise 3B - Linear Equations in two variables Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RS Aggarwal Solutions: Exercise 3B - Linear Equations in two variables Notes | EduRev

 Page 1


Exercise  3B
1. Solve for x and y:
x + y = 3, 4x 3y = 26
Sol:
The given system of equation is:
4x 
On multiplying (i) by 3, we get: 
Page 2


Exercise  3B
1. Solve for x and y:
x + y = 3, 4x 3y = 26
Sol:
The given system of equation is:
4x 
On multiplying (i) by 3, we get: 
 
On adding (ii) and (iii), we get:
7x = 35
x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
y = (3 5) = -2
Hence, the solution is x = 5 and y = -2
2. Solve for x and y:
x y = 3, + = 6
Sol:
The given system of equations is
x y = 3                              
+ = 6
From (i), write y in terms of x to get
y = x 3
Substituting y = x 3 in (ii), we get
+ = 6
2x + 3(x 3) = 36
2x + 3x 9 = 36
x = = 9
Now, substituting x = 9 in (i), we have
9 y = 3
y = 9 3 = 6
Hence, x = 9 and y = 6.
3. Solve for x and y:
2x + 3y = 0, 3x + 4y = 5
Sol:
The given system of equation is:
On multiplying (i) by 4 and (ii) by 3, we get:
iii)
On subtracting (iii) from (iv) we get:
Page 3


Exercise  3B
1. Solve for x and y:
x + y = 3, 4x 3y = 26
Sol:
The given system of equation is:
4x 
On multiplying (i) by 3, we get: 
 
On adding (ii) and (iii), we get:
7x = 35
x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
y = (3 5) = -2
Hence, the solution is x = 5 and y = -2
2. Solve for x and y:
x y = 3, + = 6
Sol:
The given system of equations is
x y = 3                              
+ = 6
From (i), write y in terms of x to get
y = x 3
Substituting y = x 3 in (ii), we get
+ = 6
2x + 3(x 3) = 36
2x + 3x 9 = 36
x = = 9
Now, substituting x = 9 in (i), we have
9 y = 3
y = 9 3 = 6
Hence, x = 9 and y = 6.
3. Solve for x and y:
2x + 3y = 0, 3x + 4y = 5
Sol:
The given system of equation is:
On multiplying (i) by 4 and (ii) by 3, we get:
iii)
On subtracting (iii) from (iv) we get:
 
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
3y = -30
y = -10
Hence, the solution is x = 15 and y = -10.
4. Solve for x and y:
2x - 3y = 13, 7x - 2y = 20
Sol:
The given system of equation is:
2x -
7x -
On multiplying (i) by 2 and (ii) by 3, we get:
4x -
21x -
On subtracting (iii) from (iv) we get:
17x = (60 26) = 34
x = 2
On substituting the value of x = 2 in (i), we get:
4 3y = 13
3y = (4 13) = -9
y = -3
Hence, the solution is x = 2 and y = -3.
5. Solve for x and y:
3x - 5y - 19 = 0, -7x + 3y + 1 = 0
Sol:
The given system of equation is:
3x - 5y -
-
On multiplying (i) by 3 and (ii) by 5, we get:
9x -
-35x + 15y = -
On subtracting (iii) from (iv) we get:
-26x = (57 5) = 52
x = -2
On substituting the value of x = -2 in (i), we get:
Page 4


Exercise  3B
1. Solve for x and y:
x + y = 3, 4x 3y = 26
Sol:
The given system of equation is:
4x 
On multiplying (i) by 3, we get: 
 
On adding (ii) and (iii), we get:
7x = 35
x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
y = (3 5) = -2
Hence, the solution is x = 5 and y = -2
2. Solve for x and y:
x y = 3, + = 6
Sol:
The given system of equations is
x y = 3                              
+ = 6
From (i), write y in terms of x to get
y = x 3
Substituting y = x 3 in (ii), we get
+ = 6
2x + 3(x 3) = 36
2x + 3x 9 = 36
x = = 9
Now, substituting x = 9 in (i), we have
9 y = 3
y = 9 3 = 6
Hence, x = 9 and y = 6.
3. Solve for x and y:
2x + 3y = 0, 3x + 4y = 5
Sol:
The given system of equation is:
On multiplying (i) by 4 and (ii) by 3, we get:
iii)
On subtracting (iii) from (iv) we get:
 
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
3y = -30
y = -10
Hence, the solution is x = 15 and y = -10.
4. Solve for x and y:
2x - 3y = 13, 7x - 2y = 20
Sol:
The given system of equation is:
2x -
7x -
On multiplying (i) by 2 and (ii) by 3, we get:
4x -
21x -
On subtracting (iii) from (iv) we get:
17x = (60 26) = 34
x = 2
On substituting the value of x = 2 in (i), we get:
4 3y = 13
3y = (4 13) = -9
y = -3
Hence, the solution is x = 2 and y = -3.
5. Solve for x and y:
3x - 5y - 19 = 0, -7x + 3y + 1 = 0
Sol:
The given system of equation is:
3x - 5y -
-
On multiplying (i) by 3 and (ii) by 5, we get:
9x -
-35x + 15y = -
On subtracting (iii) from (iv) we get:
-26x = (57 5) = 52
x = -2
On substituting the value of x = -2 in (i), we get:
 
6 5y 19 = 0
5y = ( 6 19) = -25
y = -5
Hence, the solution is x = -2 and y = -5.
6. Solve for x and y:
2x y + 3 = 0, 3x 7y + 10 = 0
Sol:
The given system of equation is:
2x 
3x 
From (i), write y in terms of x to get
y=2x + 3
Substituting y = 2x + 3 in (ii), we get
3x 7(2x + 3) + 10 = 0
3x 14x 21 + 10 = 0
-7x = 21 10  = 11
x = 
Now substituting x = in (i), we have
y + 3 = 0
y = 3 - = -
Hence, x = and y = .
7. Solve for x and y:
9x - 2y = 108, 3x + 7y = 105
Sol:
The given system of equation can be written as:
9x -
On multiplying (i) by 7 and (ii) by 2, we get:
63x + 6x = 108 × 7 + 105 × 2
69x = 966
x = = 14
Now, substituting x = 14 in (i), we get:
9 × 14 2y = 108
2y = 126 108
Page 5


Exercise  3B
1. Solve for x and y:
x + y = 3, 4x 3y = 26
Sol:
The given system of equation is:
4x 
On multiplying (i) by 3, we get: 
 
On adding (ii) and (iii), we get:
7x = 35
x = 5
On substituting the value of x = 5 in (i), we get:
5 + y = 3
y = (3 5) = -2
Hence, the solution is x = 5 and y = -2
2. Solve for x and y:
x y = 3, + = 6
Sol:
The given system of equations is
x y = 3                              
+ = 6
From (i), write y in terms of x to get
y = x 3
Substituting y = x 3 in (ii), we get
+ = 6
2x + 3(x 3) = 36
2x + 3x 9 = 36
x = = 9
Now, substituting x = 9 in (i), we have
9 y = 3
y = 9 3 = 6
Hence, x = 9 and y = 6.
3. Solve for x and y:
2x + 3y = 0, 3x + 4y = 5
Sol:
The given system of equation is:
On multiplying (i) by 4 and (ii) by 3, we get:
iii)
On subtracting (iii) from (iv) we get:
 
x = 15
On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
3y = -30
y = -10
Hence, the solution is x = 15 and y = -10.
4. Solve for x and y:
2x - 3y = 13, 7x - 2y = 20
Sol:
The given system of equation is:
2x -
7x -
On multiplying (i) by 2 and (ii) by 3, we get:
4x -
21x -
On subtracting (iii) from (iv) we get:
17x = (60 26) = 34
x = 2
On substituting the value of x = 2 in (i), we get:
4 3y = 13
3y = (4 13) = -9
y = -3
Hence, the solution is x = 2 and y = -3.
5. Solve for x and y:
3x - 5y - 19 = 0, -7x + 3y + 1 = 0
Sol:
The given system of equation is:
3x - 5y -
-
On multiplying (i) by 3 and (ii) by 5, we get:
9x -
-35x + 15y = -
On subtracting (iii) from (iv) we get:
-26x = (57 5) = 52
x = -2
On substituting the value of x = -2 in (i), we get:
 
6 5y 19 = 0
5y = ( 6 19) = -25
y = -5
Hence, the solution is x = -2 and y = -5.
6. Solve for x and y:
2x y + 3 = 0, 3x 7y + 10 = 0
Sol:
The given system of equation is:
2x 
3x 
From (i), write y in terms of x to get
y=2x + 3
Substituting y = 2x + 3 in (ii), we get
3x 7(2x + 3) + 10 = 0
3x 14x 21 + 10 = 0
-7x = 21 10  = 11
x = 
Now substituting x = in (i), we have
y + 3 = 0
y = 3 - = -
Hence, x = and y = .
7. Solve for x and y:
9x - 2y = 108, 3x + 7y = 105
Sol:
The given system of equation can be written as:
9x -
On multiplying (i) by 7 and (ii) by 2, we get:
63x + 6x = 108 × 7 + 105 × 2
69x = 966
x = = 14
Now, substituting x = 14 in (i), we get:
9 × 14 2y = 108
2y = 126 108
 
y = = 9
Hence, x = 14 and y = 9.
8. Solve for x and y:
+ = 11, - + 7 = 0
Sol:
The given equations are:
+ = 11
and - + 7 = 0
5x 2y = -
On multiplying (i) by 2 and (ii) by 3, we get:
15x 6y = -
On adding (iii) and (iv), we get:
23x = 138
x = 6
On substituting x = 6 in (i), we get:
24 + 3y = 132
3y = (132 24) = 108
y = 36
Hence, the solution is x = 6 and y = 36.
9. Solve for x and y:
4x - 3y = 8, 6x - y = 
Sol:
The given system of equation is:
4x -
6x - y = 
On multiplying (ii) by 3, we get:
18x 
On subtracting (iii) from (i) we get:
-14x = -21
x = =
Now, substituting the value of x = in (i), we get:
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