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

RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B)

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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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FAQs on RS Aggarwal Solutions: Linear Equations in two variables (Exercise 3B)

1. How do I solve linear equations in two variables using the substitution method?
Ans. The substitution method involves expressing one variable in terms of the other from one equation, then substituting that expression into the second equation to find its value. Once you solve for the first variable, substitute back to find the second. This technique simplifies simultaneous linear equations by reducing them to single-variable equations, making it easier to find the solution pair (x, y) systematically.
2. What's the difference between consistent and inconsistent linear equations in two variables?
Ans. Consistent linear equations have at least one solution-either a unique solution (intersecting lines) or infinitely many solutions (coincident lines). Inconsistent equations have no solution because the lines are parallel and never meet. Understanding this distinction helps students determine whether a system of linear equations will yield valid answers before attempting lengthy calculations for Class 10 CBSE exams.
3. Can I use the elimination method to solve any pair of linear equations in two variables?
Ans. The elimination method works effectively for any system of two linear equations, but coefficients must be manipulated strategically first. By multiplying equations to make coefficients of one variable equal, then adding or subtracting to eliminate that variable, students can solve for the remaining unknown. This algebraic approach is particularly useful when substitution would create complicated fractional expressions.
4. Why do some linear equations in two variables have no solution or infinite solutions?
Ans. When two linear equations represent parallel lines, they never intersect, producing no solution-this occurs when ratios of coefficients are equal but constants differ. Conversely, infinite solutions arise when equations represent the same line (all coefficients and constants have identical ratios). These scenarios reveal the geometric relationship between equations, crucial for understanding solution sets in Exercise 3B problems.
5. How do I check if my answer to a linear equation system is correct for CBSE Class 10?
Ans. Substitute both x and y values back into both original equations; if both equations are satisfied (left side equals right side), your solution is correct. This verification step prevents careless errors and builds confidence before exams. Students can also use graphical methods or cross-reference solutions using mind maps and flashcards available on EduRev to reinforce verification techniques.
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