Chapter 3 - Pair Of Linear Equations In Two Variables, RD Sharma Solutions - (Part-16) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 3.74

Q.22. Find the value of k for which each of the following system of equations have no solution :
3x − 4y + 7 = 0
kx + 3y − 5 = 0
Ans. GIVEN:
3x − 4y + 7 = 0
kx + 3y − 5 = 0
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For no solution

Here,

Hence for the system of equation has no solution.

Q.23. Find the value of k for which each of the following system of equations have no solution :
2x − ky + 3 = 0
3x + 2y − 1 = 0
Ans. GIVEN:
2x − ky + 3 = 0
3x + 2y − 1 = 0
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For no solution

Here,

Hence for the system of equation has no solution.

Q.24. Find the value of k for which each of the following system of equations have no solution :
2x + ky = 11
5x − 7y = 5
Ans. GIVEN:
2x + ky = 11
5x − 7y = 5
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For no solution

Here,

Hence for the system of equation has no solution.

Q.25. Find the value of k for which each of the following system of equations have no solution :
cx + 2y = 3
12x + cy = 6
Ans. GIVEN:
cx + 2y = 3
12x + cy = 6
To find: To determine for what value of c the system of equation has no solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For no solution

Here,

Hence for c = ±6 the system of equation has no solution.

Q.26. For what value of k the following system of equations will be inconsistent?
4x + 6y = 11
2x + ky = 7
Ans. GIVEN:
4x + 6y = 11
2x + ky = 7
To find: To determine for what value of k the system of equation will be inconsistent
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c<sub>2
For the system of equation to be inconsistent

Here,

Hence for k = 3 the system of equation will be inconsistent.</sub>

Q.27. For what value of α, the system of equations will have no solution?
αx + 3y = α − 3
12x + αy = α
Ans. GIVEN:
αx + 3y = α − 3
12x + αy = α
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For no solution

Here,

Consider the following for α

Now consider the following

Hence the common value of α is − 6
Hence for α = -6 the system of equation has no solution

Q.28. Find the value of k for which the system has (i) a unique solution, and (ii) no solution.
kx + 2y = 5
3x + y = 1
Ans. GIVEN:
kx + 2y = 5
3x + y = 1
To find: To determine for what value of k the system of equation has
(1) Unique solution
(2) No solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
(1) For Unique solution

Here,

Hence for k ≠ 6 the system of equation has unique solution.
(2) For no solution

Hence for k= 6 the system of equation has no solution

Q.29. (i) Prove that there is a value of c(≠ 0) for which the system
6x + 3y = c − 3
12x + cy = c
has infinitely many solutions. Find this value.
(ii) Find c if the system  of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions?
Ans.
(i) 6x + 3y = c − 3
12x + cy = c
To find: To determine for what value of c the system of equation has infinitely many solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Now consider the following for c

But it is given that c ≠ 0. Hence c = 6
Hence for c = 6 the system of equation have infinitely many solutions.
(ii) If the system of equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 has infinitely many solutions, then
Since, the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions
Therefore,

Hence, it holds for c = 6.
For c=−6,

Hence, it does not holds for c=−6.
Hence, the value of c is 6.

Page No 3.74

Q.30. Find the values of k for which the system will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?
2x + ky = 1
3x − 5y = 7
Ans. GIVEN:
2x + ky = 1
3x − 5y = 7
To find: To determine for what value of k the system of equation has
(1) Unique solution
(2) No solution
(3) Infinitely many solution
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
(1) For Unique solution

Here,

Hence for  the system of equation has unique solution
(2) For no solution

Here,

Hence for  the system of equation has no solution
(3) For infinitely many solution

Here,

But since here
Hence the system does not have infinitely many solutions.

Q.31. For what value of k, the following system of equations will represent the coincident lines?
x + 2y + 7 = 0
2x + ky + 14 = 0
Ans. GIVEN:
x + 2y + 7 = 0
2x + ky + 14 = 0
To find: To determine for what value of k the system of equation will represents coincident lines
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For the system of equation to represent coincident lines we have the following relation

Here,

Hence for k = 4 the system of equation represents coincident lines

Q.32. Obtain the condition for the following system of linear equations to have a unique solution
ax + by = c
lx + my = n
Ans. GIVEN:
ax + by = c
lx + my = n
To find: To determine the condition for the system of equation to have a unique equation
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For unique solution

Here

Hence for the system of equation have unique solution.

Q.33. Determine the values of a and b so that the following system of linear equations have infinitely many solutions :
(2a − 1) x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
Ans.
GIVEN:
(2a − 1) x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here


Again consider



Hence for the system of equation has infinitely many solution.

Page No 3.75

Q.34. Find the values of a and b for which the following system of linear equations has infinite number of solutions :
2x − 3y = 7
(a + b) x − (a + b − 3) y = 4a + b
Ans. GIVEN:
2x − 3y = 7
(a + b) x − (a + b − 3) y = 4a + b
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Again

Multiplying eq. (2) by 4 and adding eq. (1)
9a + 45 = 0
a = -5
Putting the value of a in eq. (2)
- 5 + b + 6 = 0
b = -1
Hence for a = -5 and b = -1  the system of equation has infinitely many solution.

Q.35. Find the values of p and q for which the following system of linear equations has infinite number of solutions:
2x + 3y = 9
(p + q) x + (2p − q)y = 3(p + q + 1)
Ans. GIVEN:
2x + 3y = 9
(p + q)x + (2p − q)y = 3(p + q + 1)
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Again consider

Multiplying eq. (2) by 3 and subtracting from eq. (1)
3p - 6q - 3 - 3p + 15q = 0
9q = 3
q = 1/3
Putting the value of q in eq. (2)

Hence for the system of equation has infinitely many solution.

Q.36. Find the values of a and b for which the following system of equations has infinitely many solutions:
(i) (2a − 1)x − 3y = 5
3x + (b − 2) y = 3
(ii) 2x − (2a + 5)y = 5
(2b + 1)x − 9y = 15
(iii) (a − 1)x + 3y = 2
6x + (1 − 2b)y = 6
(iv) 3x + 4y = 12
(a + b)x + 2(a − b)y = 5a − 1
(v) 2x + 3y = 7
(a − b)x + (a + b)y = 3a + b − 2
(vi) 2x + 3y − 7 = 0
(a − 1) x + (a + 1)y = (3a − 1)
(vii) 2x + 3y = 7
(a − 1)x + (a + 2)y = 3a
(viii) x + 2y = 1
(a − b)x + (a + b)y = a + b − 2 
(ix) 2x + 3y = 7
2ax + ay = 28 − by
Ans. (i) GIVEN:
(2a − 1)x − 3y = 5
3x + (b − 2) y = 3
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider

Again consider

Hence for a = 3 and  the system of equation has infinitely many solution.
(ii) GIVEN:
2x − (2a + 5)y = 5
(2b + 1)x − 9y = 15
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Again consider

Hence for a = - 1 and  the system of equation has infinitely many solution.
(iii) GIVEN:
(a − 1)x + 3y = 2
6x + (1 − 2b)y = 6
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Again consider

Hence for a = 3 and b = -4  the system of equation has infinitely many solution.
(iv) GIVEN:
3x + 4y = 12
(a + b)x + 2(a − b)y = 5a − 1
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Again consider

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. 2
2a = 10
a = 5
Substituting the value of ‘a’ in eq. (2) we get
15 - 12b = 3
- 12b = -12
b = 1
Hence for a = 5 and b = 1 the system of equation has infinitely many solution.
(v) GIVEN:
2x + 3y = 7
(a − b)x + (a + b)y = 3a + b − 2
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

9a + 3b - 6 = 7a + 7b
2a - 4b = 6  ...(1)
Again consider the following

6a + 2b - 4 = 7a - 7b
a - 9b = - 4  ...(2)
Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. (2)
- 14b = - 14
b = 1
Substituting the value of b in eq. (2) we get
a - 9 = - 4
a = 5
Hence for a = 5 and b = 1 the system of equation has infinitely many solution.
(vi) GIVEN:
2x + 3y - 7 = 0
(a - 1) x + (a + 1) y = 3a - 1
To find: To determine for what value of k the system of equation has infinitely many solutions
Rewrite the given equations
2x + 3y - 7 = 0
(a - 1) x + (a + 1) y = 3a - 1
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Hence for a = 5 the system of equation have infinitely many solutions.
(vii) GIVEN :
2x + 3y = 7
(a - 1)x + (a + 2) y = 3a
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
For infinitely many solution

Here

Consider the following

Hence for a = 7 the system of equation have infinitely many solutions.
(viii) Given:
x + 2y = 1
(a − b)x + (a + b) y = a + b − 2
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
has infinitely many solutions if

So,

⇒ a + b = 2a − 2b and 2a + 2b − 4 = a + b
⇒ a = 3b and a + b = 4
⇒ a − 3b = 0 and a + b = 4
Solving these two equations, we get
−4b = −4
⇒ b = 1
Putting b = 1 in a + b = 4, we get
a = 3
(ix) Given:
2x + 3y = 7
2ax + ay = 28 − by
⇒ 2ax + (a + b)y = 28
We know that the system of equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
has infinitely many solutions if

Now,

⇒ a + b = 12     .....(2)
Solving (1) and (2), we get
a = 4 and b = 8