Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5 | Set 1
Question 1. In each of the following systems of equation determine whether the system has a unique solution, no solution, or infinite solutions. In case there is a unique solution:
x − 3y − 3 = 0,
3x − 9y − 2 = 0
Solution:
Given that,
x − 3y − 3 = 0 …(1)
3x − 9y − 2 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y – c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = −3, c1 = −3
a2 = 3, b2 = −9, c2 = −2
Let’s check the equation’s,
a1/a2 = 1/3
b1/b2 = -3/-9 = 1/3
c1/c2 = -3/-9 = 3/2
a1/a2 = b1/b2 ≠ c1/c2
Hence, the given set of equations has no solution.
Question 2. In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution:
2x + y − 5 = 0,
4x + 2y − 10 = 0
Solution:
Given that,
2x + y − 5 = 0 …(1)
4x + 2y − 10 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 1, c1 = −5 and
a2 = 4, b2 = 2, c2 = −10
Lets check the equation’s,
a1/a2 = 2/4 = 1/2
b1/b2 = 1/2
and c1/c2 = -5/-10 = 1/2
Therefore, a1/a2 = b1/b2 = c1/c2
Hence, the given set of equations has infinitely many solutions.
Question 3. In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution:
3x − 5y = 20,
6x − 10y = 40
Solution:
Given that,
3x − 5y = 20 …(1)
6x − 10y = 40 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 3, b1 = −5, c1 = − 20
a2 = 6, b2 = −10, c2 = − 40
Lets check the equation’s,
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 – 1/2 and
c1/c2 = -20/-40 = 1/2
Therefore, a1/a2 = b1/b2 = c1/c2
Hence, the given set of equations has infinitely many solutions.
Question 4. In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution:
x − 2y − 8 = 0,
5x − 10y − 10 = 0
Solution:
Given that,
x − 2y − 8 = 0 …(1)
5x − 10y − 10 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = −2, c1 = −8
a2 = 5, b2 = −10, c2 = −10
Lets check the equation’s,
a1/a2 = 1/5
b1/b2 = -2/-10 and
c1/c2 = -8/-10
Therefore, a1/a2 = b1/b2 ≠ c1/c2
Hence, the given set of equations has no solution.
Question 5. Find the value of k for each of the following system of equations which have a unique solution:
kx + 2y − 5 = 0,
3x + y − 1 = 0
Solution:
Given that,
kx + 2y − 5 = 0 …(1)
3x + y − 1 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = k, b1 = 2, c1 = −5
a2 = 3, b2 = 1, c2 = −1
For unique solution,
a1/a2 ≠ b1/b2
k/3 ≠ 2/1
k ≠ 6
So, the given set of equations will have unique solution for all real values of k other than 6.
Question 6. Find the value of k for each of the following system of equations which have a unique solution:
4x + ky + 8 = 0,
2x + 2y + 2 = 0
Solution:
Given that,
4x + ky + 8 = 0 …(1)
2x + 2y + 2 = 0 …(2)
So, the given equations are in the form of:
a1x +b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = k, c1 = 8
a2 = 2, b2 = 2, c2 = 2
For unique solution,
a1/a2 ≠ b1/b2
4/2 ≠ k/2
k ≠ 4
So, the given set of equations will have unique solution for all real values of k other than 4.
Question 7. Find the value of k for each of the following system of equations which have a unique solution:
4x − 5y = k,
2x − 3y = 12
Solution:
Given that,
4x − 5y − k = 0 …(1)
2x − 3y − 12 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = −5, c1 = −k
a2 = 2, b2 = -3, c2 = -12
For unique solution,
a1/a2 ≠ b1/b2
4/2 ≠ -5/-3
Here, k can have any real values.
Hence, the given set of equations will have unique solution for all real values of k.
Question 8. Find the value of k for each of the following system of equations which have a unique solution:
x + 2y = 3,
5x + ky + 7 = 0
Solution:
Given that,
x + 2y = 3 …(1)
5x + ky + 7 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 1, b1 = 2, c1 = −3
a2 = 5, b2 = k, c2 = 7
For unique solution,
a1/a2 ≠ b1/b2
1/5 ≠ 2/k
k ≠ 10
So, the given set of equations will have unique solution for all real values of k other than 10.
Question 9. Find the value of k for which each of the following system of equations having infinitely many solutions:
2x + 3y − 5 = 0,
6x − ky − 15 = 0
Solution:
Given that,
2x + 3y − 5 = 0 …(1)
6x − ky − 15 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = 3, c1 = −5
a2 = 6, b2 = k, c2 = −15
For unique solution,
We have
a1/a2 = b1/b2 = c1/c2
2/6 = 3/k
k = 9
Hence, when k = 9 the given set of equations will have infinitely many solutions.
Question 10. Find the value of k for which each of the following system of equations having infinitely many solutions:
4x + 5y = 3,
x + 15y = 9
Solution:
Given that,
4x + 5y = 3 …(1)
kx +15y = 9 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 4, b1 = 5, c1 = 3
a2 = k, b2 = 15, c2 = 9
For unique solution,
We have
a1/a2 = b1/b2 = c1/c2
4/k = 5/15 = -3/-9
4/k = 1/3
k = 12
Hence, when k = 12 the given set of equations will have infinitely many solutions.
Question 11. Find the value of k for which each of the following system of equations having infinitely many solutions:
kx − 2y + 6 = 0,
4x + 3y + 9 = 0
Solution:
Given that,
kx − 2y + 6 = 0 …(1)
4x + 3y + 9 = 0 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = k, b1 = −2, c1 = 6
a2 = 4, b2 = −3, c2 = 9
For unique solution
We have
a1/a2 = b1/b2 = c1/c2
k/4 = -2/-3 = 2/3
k = 8/3
Hence, when k = 8/3 the given set of equations will have infinitely many solutions.
Question 12. Find the value of k for which each of the following system of equations having infinitely many solutions:
8x + 5y = 9,
kx + 10y = 19
Solution:
Given that,
8x + 5y = 9 …(1)
kx + 10y = 19 …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 8, b1 = 5, c1 = −9
a2 = k, b2 = 10, c2 = −19
For unique solution
We have
a1/a2 = b1/b2 = c1/c2
8/k = 5/10 = k = 16
Hence, when k = 16 the given set of equations will have infinitely many solutions.
Question 13. Find the value of k for which each of the following system of equations having infinitely many solutions:
2x − 3y = 7,
(k + 2)x − (2k + 1)y = 3(2k − 1)
Solution:
Given that,
2x − 3y = 7 …(1)
(k + 2)x − (2k + 1)y = 3(2k − 1) …(2)
So, the given equations are in the form of:
a1x + b1y − c1 = 0 …(3)
a2x + b2y − c2 = 0 …(4)
On comparing eq (1) with eq(3) and eq(2) with eq (4), we get
a1 = 2, b1 = −3, c1 = −7
a2 = k, b2 = − (2k + 1), c2 = −3(2k − 1)
Now, for unique solution
We have
a1/a2 = b1/b2 = c1/c2
= 2/(k + 2) = -3/-(2k + 1) = -7/-3(2k – 1)
= 2/(k + 2) = -3/-(2k + 1) and -3/-(2k + 1) = -7/-3(2k – 1)
= 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1)
= 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7
= k = 4 and 4k = 16
= k = 4
Hence, when k = 4 the given set of equations will have infinitely many solutions.