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Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B)

RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) | Mathematics (Maths) Class 10 PDF Download

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 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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Document Description: RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) have been prepared according to the Class 10 exam syllabus. Information about RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) covers topics like and RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B).
Introduction of RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B) | Mathematics (Maths) Class 10
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