RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables EX 3D

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RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D

Question 1:

Question 2:

Question 3:
3x – 5y – 7 = 0
6x – 10y – 3 = 0

Hence the given system of equations is inconsistent
Question 4:
2x – 3y – 5 = 0, 6x – 9y – 15 = 0
These equations are of the form

Hence the given system of equations has infinitely many solutions
Question 5:
kx + 2y – 5 = 0
3x – 4y – 10 = 0
These equations are of the form

This happens when
(kneq frac { -3 }{ 2 })
Thus, for all real value of k other that , the given system equations will have a unique solution
(ii) For no solution we must have

Hence, the given system of equations has no solution if (k=frac { -3 }{ 2 } )
Question 6:
x + 2y – 5 = 0
3x + ky + 15 = 0
These equations are of the form of

Thus for all real value of k other than 6, the given system of equation will have unique solution
(ii) For no solution we must have

Therefore k = 6
Hence the given system will have no solution when k = 6.
Question 7:
x + 2y – 3 = 0, 5x + ky + 7 = 0
These equations are of the form

(i) For a unique solution we must have

Thus, for all real value of k other than 10
The given system of equation will have a unique solution.
(ii) For no solution we must have

Hence the given system of equations has no solution if

For infinite number of solutions we must have

This is never possible since

There is no value of k for which system of equations has infinitely many solutions
Question 8:
8x + 5y – 9 = 0
kx + 10y – 15 = 0
These equations are of the form

Clearly, k = 16 also satisfies the condition

Hence, the given system will have no solution when k = 16.
Question 9:
kx + 3y – 3 = 0 —-(1)
12x + ky – 6 = 0 —(2)

These equations are of the form

Hence, the given system will have no solution when k = -6
Question 10:
3x + y – 1 = 0
(2k – 1)x + (k – 1)y – (2k + 1) = 0
These equations are of the form

Thus,

Hence the given equation has no solution when k = 2
Question 11:
(3k + 1)x + 3y – 2 = 0

these equations are of the form

Thus, k = -1 also satisfy the condition

Hence, the given system will have no solution when k = -1
Question 12:
The given equations are
3x – y – 5 = 0 —(1)
6x – 2y + k = 0—(2)

Equations (1) and (2) have no solution, if

Question 13:
kx + 2y – 5 = 0
3x + y – 1 = 0
These equations are of the form

Thus, for all real values of k other than 6, the given system of equations will have a unique solution
Question 14:
x – 2y – 3 = 0
3x + ky – 1 = 0
These equations are of the form of

Thus, for all real value of k other than -6, the given system of equations will have a unique solution
Question 15:
kx + 3y – (k – 3) = 0
12x + ky – k = 0
These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solution
Question 16:
4x – 5y – k = 0, 2x – 3y – 12 = 0
These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solution
Question 17:
2x + 3y – 7 = 0
(k – 1)x + (k + 2)y – 3k = 0
These are of the form

This hold only when

Now the following cases arises
Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equations
Question 18:
2x + (k – 2)y – k = 0
6x + (2k – 1)y – (2k + 5) = 0
These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutions
Question 19:
kx + 3y – (2k +1) = 0
2(k + 1)x + 9y – (7k + 1) = 0
These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise
Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutions
Question 20:
5x + 2y – 2k = 0
2(k +1)x + ky – (3k + 4) = 0
These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.
Question 21:
x + (k + 1)y – 5 = 0
(k + 1)x + 9y – (8k – 1) = 0
These are of the form

For infinitely many solutions, we must have

Question 22:
(k – 1)x – y – 5 = 0
(k + 1)x + (1 – k)y – (3k + 1) = 0
These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely many
Question 23:
(a – 1)x + 3y – 2 = 0
6x + (1 – 2b)y – 6 = 0
These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4
Question 24:
(2a – 1)x + 3y – 5 = 0
3x + (b – 1)y – 2 = 0
These equations are of the form

These holds only when

Question 25:
2x – 3y – 7 = 0
(a + b)x + (a + b – 3)y – (4a + b) = 0
These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1
Question 27:
The given equations are
2x + 3y = 7 —-(1)
a(x + y) – b(x – y) = 3a + b – 2 —(2)
Equation (2) is
ax + ay – bx + by = 3a + b – 2
(a – b)x + (a + b)y = 3a + b -2
Comparing with the equations

There are infinitely many solution

2a + 2b = 3a – 3b and 3(3a + b – 2) = 7(a + b)
-a = -5b and 9a + 3b – 6 = 7a + 7b
a = 5b and 9a – 7a + 3b – 7b = 6
or 2a – 4b = 6
or a – 2b = 3
thus equation in a, b are
a = 5b —(3)
a – 2b = 3 —(4)
putting a = 5b in (4)
5b – 2b = 3 or 3b = 3 Þ b = 1
Putting b = 1 in (3)
a = 5 and b = 1
Question 28:
We have 5x – 3y = 0 —(1)
2x + ky = 0 —(2)
Comparing the equation with

These equations have a non – zero solution if

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