RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables Exercise  3D Class 10 Maths
Chapter Name  RS Aggarwal Chapter 3 Linear Equations in Two Variables 
Book Name  RS Aggarwal Mathematics for Class 10 
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Exercise 3D Solutions
1. Show that the following system of equations has a unique solution:
3x + 5y = 12,
5x + 3y = 4.
Also, find the solution of the given system of equations.
Solution
The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 3, b_{1} = 5, c_{1} = 12 and a_{2} = 5, b_{2} = 3, c_{2} = 4
For a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2 }i.e., 3/5 ≠ 5/3
Hence, the given system of equations has a unique solution.
Again, the given equations are:
3x + 5y = 12 ...(i)
5x + 3y = 4 ...(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36 ….(iii)
25x + 15y = 20 …(iv)
On subtracting (iii) from (iv), we get:
16x = 16
⇒ x = 1
On substituting x = 1 in (i), we get:
3(1) + 5y = 12
⇒ 5y = (12 + 3) = 15
⇒ y = 3
Hence, x = 1 and y = 3 is the required solution.
2. Show that the following system of equations has a unique solution:
2x  3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
Solution
The given system of equations is:
2x  3y  17 = 0 ….(i)
4x + y  13 = 0 …..(ii)
The given equations are of the form
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1 }= 2, b_{1}= 3, c_{1} = 17 and a_{2} = 4, b_{2} = 1, c_{2 }= 13
Now,
a_{1}/a_{2} = 2/4 = 1/2 and b_{1}/b_{2} = 3/1 =  3
Since, a_{1}/a_{2} ≠ b_{1}/b_{2}, therefore the system of equations has unique solution.
Using cross multiplication method, we have
x/(b_{1}c_{2} – b_{2}c_{1}) = y/(c_{1}a_{2} – c_{2}a_{1}) = 1/(a_{1}b_{2} – a_{2}b_{1})
⇒ x/(3(13) – 1 × (17) = y/(17 × 4 – (13) × 2 = 1/(2 × 1 – 4 × (3))
⇒ x/(39 + 17) = y/(68 + 26) = 1/(2 + 12)
⇒ x/56 = y/42 = 1/14
⇒ x = 56/14, y = −42/14
⇒ x = 4, y = 3
Hence, x = 4 and y = 3.
3. Show that the following system of equations has a unique solution:
x/3 + y/2 = 3, x – 2y = 2.
Also, find the solution of the given system of equations.
Solution
The given system of equations is:
x/3 + y/2 = 3
⇒ (2x + 3y)/6 = 3
2x + 3y = 18
⇒ 2x + 3y – 18 = 0 ….(i)
and
x – 2y = 2
x – 2y – 2 = 0 ...(ii)
These equations are of the forms:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2 }= 0
where, a_{1} = 2, b_{1}= 3, c_{1} = 18 and a_{2 }= 1, b_{2} = 2, c = 2
For a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}, i.e., 2/1 ≠ 3/2
Hence, the given system of equations has a unique solution. Again, the given equations are:
2x + 3y – 18 = 0 ….(iii)
x – 2y – 2 = 0 ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y – 36 = 0 ….(v)
3x  6y – 6 = 0 …(vi)
On adding (v) from (vi), we get:
7x = 42
⇒ x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒3y = (18  12) = 6
⇒ y = 2
Hence, x = 6 and y = 2 is the required solution.
4. Find the value of k for which the system of equations has a unique solution:
2x + 3y = 5,
kx  6y = 8.
Solution
The given system of equations are
2x + 3y – 5 = 0
kx  6y  8 = 0
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 2, b_{1 }= 3, c_{1} = 5 and a_{2} = k, b_{2} = 6, c_{2} = 8
Now, for the given system of equations to have a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}
⇒ 2/k ≠ 3/6
⇒ k ≠ 4
Hence, k ≠ 4
5. Find the value of k for which the system of equations has a unique solution:
x – ky = 2,
3x + 2y + 5 = 0
Solution
The given system of equations are
x  ky – 2 = 0
3x + 2y + 5 = 0
This system of equations is of the form:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 1, b_{1}= k, c_{1} = 2 and a_{2} = 3, b_{2} = 2, c_{2} = 5
Now, for the given system of equations to have a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}
⇒ 1/3 ≠  k/2
⇒ k ≠ 2/3
Hence, k ≠ 2/3
6. Find the value of k for which the system of equations has a unique solution:
5x – 7y = 5,
2x + ky = 1.
Solution
The given system of equations are
5x  7y – 5 = 0 ….(i)
2x + ky  1 = 0 …(ii)
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 5, b_{1}= 7, c_{1} = 5 and a_{2} = 2, b_{2} = k, c_{2} = 1
Now, for the given system of equations to have a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}
⇒ 5/2 ≠ 7/k
⇒ k ≠ (14/5)
Hence, k ≠ 14/5.
7. Find the value of k for which the system of equations has a unique solution:
4x + ky + 8 = 0,
x + y + 1 = 0.
Solution
The given system of equations are
4x + ky + 8 = 0
x + y + 1 = 0
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 4, b_{1}= k, c_{1} = 8 and a_{2} = 1, b_{2} = 1, c_{2} = 1
For the given system of equations to have a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2 }
⇒ 4/1 ≠ k/1
⇒ k ≠ 4
Hence, k ≠ 4.
8. Find the value of k for which the system of equations has a unique solution:
4x  5y = k,
2x  3y = 12.
Solution
The given system of equations are
4x  5y = k
⇒ 4x  5y  k = 0 ….(i)
And, 2x  3y = 12
⇒ 2x  3y  12 = 0 …(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Here, a_{1} = 4, b_{1}= 5, c_{1} = k and a_{2} = 2, b_{2} = 3, c_{2} = 12
For a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}
i.e., 4/2 ≠ −5/−3
⇒ 2 ≠ 5/3 ⇒ 6 ≠ 5
Thus, for all real values of k, the given system of equations will have a unique solution.
9. Find the value of k for which the system of equations has a unique solution:
kx + 3y = (k – 3),
12x + ky = k
Solution
The given system of equations:
kx + 3y = (k – 3)
⇒ kx + 3y – (k  3) = 0 ….(i)
And, 12x + ky = k
⇒ 12x + ky  k = 0 …(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Here, a_{1} = k, b_{1} = 3, c_{1} = (k – 3) and a_{2} = 12, b_{2} = k, c_{2} = k
For a unique solution, we must have:
a_{1}/a_{2} ≠ b_{1}/b_{2}
i.e., k/12 ≠ 3/k
⇒ k^{2 }≠ 36
⇒ k ≠ ±6
Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.
10. Show that the system equations
2x  3y = 5,
6x  9y = 15 has an infinite number of solutions
Solution
The given system of equations:
2x  3y = 5
⇒ 2x  3y – 5 = 0 ….(i)
6x  9y = 15
⇒ 6x  9y  15 = 0 …(ii)
These equations are of the following forms:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Here, a_{1} = 2, b_{1}= 3, c_{1} = 5 and a_{2} = 6, b_{2} = 9, c_{2 }= 15
∴ a_{1}/a_{2} = 2/6 = 1/3, b_{1}/b_{2} = 3/9 = 1/3 and c_{1}/c_{2} = 5/15 = 1/3
Thus, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
Hence, the given system of equations has an infinite number of solutions.
11. Show that the system of equations
6x + 5y = 11,
9x + 15/2y = 21 has no solution.
Solution
The given system of equations can be written as
6x + 5y – 11 = 0 ….(i)
⇒ 9x + 15/2y  21 = 0 …(ii)
This system is of the form
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
Here, a_{1} = 6, b_{1} = 5, c_{1} = 11 and a_{2 }= 9, b_{2} = 15/2, c_{2} = 21
Now,
a_{1}/a_{2} = 6/9 = 2/3
b_{1}/b_{2} = 5/(15/2) = 2/3
Thus, a_{1}/a_{2} = b_{1}/b_{2 }≠ c_{1}/c_{2}, therefore the given system has no solution.
12. For what value of k, the system of equations
kx + 2y = 5,
3x  4y = 10 has (i) a unique solution, (ii) no solution?
Solution
The given system of equations:
kx + 2y = 5
⇒ kx + 2y  5 = 0 ….(i)
3x  4y = 10
⇒ 3x  4y  10 = 0 …(ii)
These equations are of the forms:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = k, b_{1} = 2, c_{1} = 5 and a_{2} = 3, b_{2} = 4, c_{2} = 10
(i) For a unique solution, we must have:
∴ a_{1}/a_{2} ≠ b_{1}/b_{2} i.e., k/3 ≠ 2/4 ⇒ k ≠ −3/2
Thus for all real values of k other than −3/2, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
⇒ k/3 = 2/4 ≠ 5/10
⇒ k/3 = 2/4 and k/3 ≠ 1/2
⇒ k = −3/2, k ≠ 3/2
Hence, the required value of k is −3/2.
13. For what value of k, the system of equations
x + 2y = 5,
3x + ky + 15 = 0
has (i) a unique solution, (ii) no solution?
Solution
The given system of equations:
x + 2y = 5
⇒ x + 2y  5 = 0 ….(i)
3x + ky + 15 = 0 …(ii)
These equations are of the forms:
a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 1, b_{1}= 2, c_{1} = 5 and a_{2} = 3, b_{2} = k, c_{2} = 15
(i) For a unique solution, we must have:
∴ a_{1}/a_{2} ≠ b_{1}/b_{2} i.e., 1/3 ≠ 2/k ⇒ k ≠ 6
Thus for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
⇒ a_{1}/a_{2} ≠ b_{1}/b_{2 }≠ c_{1}/c_{2 }
⇒ 1/3 = 2/k ≠ 5/15
⇒ 1/3 = 2/k and 2/k ≠ −5/15
⇒ k = 6, k ≠ 6
Hence, the required value of k is 6.
14. For what value of k, the system of equations
x + 2y = 3,
5x + ky + 7 = 0
Have (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equation has infinitely namely solutions
Solution
The given system of equations:
x + 2y = 3
⇒ x + 2y  3 = 0 ….(i)
And, 5x + ky + 7 = 0 …(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 1, b_{1} = 2, c_{1} = 3 and a_{2} = 5, b_{2} = k, c_{2} = 7
(i) For a unique solution, we must have:
∴ a_{1}/a_{2} ≠ b_{1}/b_{2} i.e., 1/5 ≠ 2/k ⇒ k ≠ 10
Thus for all real values of k other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
⇒ 1/5 ≠ 2/k ≠ −3/7
⇒ 1/5 ≠ 2/k and 2/k ≠ −3/7
⇒ k = 10, k ≠ 14/−3
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.
15. Find the value of k for which the system of linear equations has an infinite number of solutions:
2x + 3y = 7,
(k – 1)x + (k + 2)y = 3k.
Solution
The given system of equations:
2x + 3y = 7,
⇒ 2x + 3y  7 = 0 ….(i)
And, (k – 1)x + (k + 2)y = 3k
⇒ (k – 1)x + (k + 2)y  3k = 0 …(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 2, b_{1} = 3, c_{1} = 7 and a_{2} = (k – 1), b_{2} = (k + 2), c_{2} = 3k
For an infinite number of solutions, we must have:
⇒ a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
2/(k – 1) = 3/(k + 2) = 7/3k
⇒ 2/(k – 1) = 3/(k + 2) = 7/3k
Now, we have the following three cases:
Case I:
2/(k – 1) = 3/(k + 2)
⇒ 2(k + 2) = 3(k – 1) ⇒ 2k + 4 = 3k – 3 ⇒ k = 7
Case II:
3/(k + 2) = 7/3k
⇒ 7(k + 2) = 9k ⇒ 7k + 14 = 9k ⇒ 2k = 14 ⇒ k = 7
Case III:
2/(k – 1) = 7/3k
⇒ 7k – 7 = 6k ⇒ k = 7
Hence, the given system of equations has an infinite number of solutions when k is equal to 7.
16. Find the value of k for which the system of linear equations has an infinite number of solutions:
2x + (k – 2)y = k,
6x + (2k  1)y = (2k + 5).
Solution
The given system of equations:
2x + (k – 2)y = k
⇒ 2x + (k – 2)y – k = 0 ...(i)
And, 6x + (2k – 1)y = (2k + 5)
⇒ 6x + (2k – 1)y – (2k + 5) = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1} = 2, b_{1} = (k – 2), c_{1} =  k and a_{2} = 6, b_{2} = (2k – 1), c_{2} = (2k + 5)
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
2/6 = (k – 2)/(2k – 1) = k/(2k + 5)
⇒ 1/3 = (k – 2)/(2k – 1) = k(2k + 5)
Now, we have the following three cases:
Case I:
1/3 = (k – 2)/(2k – 1)
⇒ (2k – 1) = 3(k – 2)
⇒ 2k – 1 = 3k – 6 ⇒ k = 5
Case II:
(k – 2)/(2k – 1) = k/(2k + 5)
⇒ (k – 2)(2k + 5) = k(2k – 1)
⇒ 2k^{2} + 5k – 4k – 10 = 2k^{2} – k
⇒ k + k = 10 ⇒ 2k = 10 ⇒ k = 5
Case III:
1/3 = k/(2k + 5)
⇒ 2k + 5 = 3k ⇒ k = 5
Hence the given system of equations has an infinite number of solutions when k is equal to 5.
17. Find the value of k for which the system of linear equations has an infinite number of solutions:
kx + 3y = (2k + 1),
2(k + 1)x + 9y = (7k + 1)
Solution
The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y – (2k + 1) = 0 ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y – (7k + 1) = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1 }= k, b_{1} = 3, c_{1} = (2k + 1) and a_{2} = 2(k + 1), b_{2} = 9, c_{2} = (7k + 1)
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
i.e., k/2(k + 1) = 3/9 = (2k + 1)/(7k + 1)
⇒ k/2(k + 1) = 1/3 = (2k + 1)/(7k + 1)
Now, we have the following three cases:
Case I:
k/2(k + 1) = 1/3
⇒ 2(k + 1) = 3k
⇒ 2k + 2 = 3k
⇒ k = 2
Case II:
1/3 = (2k + 1)/(7k + 1)
⇒ (7k + 1) = 6k + 3
⇒ k = 2
Case III:
k/2(k + 1) = (2k + 1)/(7k + 1)
⇒ k(7k + 1) = (2k + 1) × 2(k + 1)
⇒ 7k^{2} + k = (2k + 1)(2k + 2)
⇒ 7k^{2} + k = 4k^{2} + 4k + 2k + 2
⇒ 3k^{2} – 5k – 2 = 0
⇒ 3k^{2} – 6k + k – 2 = 0
⇒ 3k(k – 2) + 1(k – 2) = 0
⇒ (3k + 1)(k – 2) = 0
⇒ k = 2 or k = 1/3
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.
18. Find the value of k for which the system of linear equations has an infinite number of solutions.
5x + 2y = 2k,
2(k + 1)x + ky = (3k + 4).
Solution
The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y – 2k = 0 ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k = 1)x + ky – (3k + 4) = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{2}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1} = 5, b_{1} = 2, c_{1} =  2k and a_{2} = 2(k + 1), b_{2} = k, c_{2} = (3k + 4)
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
5/2(k + 1) = 2/k = 2k/(3k + 4)
⇒ 5/2(k + 1) = 2/k = 2k/(3k + 4)
Now, we have the following three cases:
Case I:
5/2(k + 1) = 2/k
⇒ 2 × 2(k + 1) = 5k
⇒ 4(k + 1) = 5k
⇒ 4k + 4 = 5k
⇒ k = 4
Case II:
2/k = 2k/(3k + 4)
⇒ 2k^{2} = 2 × (3k + 4)
⇒ 2k^{2} = 6k + 8 ⇒ 2k^{2} – 6k – 8 = 0
⇒ 2(k^{2} – 3k – 4) = 0
⇒ k^{2 }– 4k + k – 4 = 0
⇒ k(k – 4) + 1(k – 4) = 0
⇒ (k + 1)(k – 4) = 0
⇒ (k + 1) = 0 or (k – 4) = 0
⇒ k =  1 or k = 4
Case III:
5/2(k + 1) = 2k/(3k + 4)
⇒ 15k + 20 = 4k^{2} + 4k
⇒ 4k^{2 }– 11k – 20 = 0
⇒ 4k^{2} – 16k + 5k – 20 = 0
⇒ 4k(k – 4) + 5(k – 4) = 0
⇒ (k – 4)(4k + 5) = 0
⇒ k = 4 or k = 5/4
Hence, the given system of equations has an infinite number of solutions when k is equal to 4.
19. Find the value of k for which the system of linear equations has an infinite number of solutions:
(k – 1)x – y = 5,
(k + 1)x + (1 – k)y = (3k + 1).
Solution
The given system of equations:
(k – 1)x – y = 5
⇒ (k – 1)x – y – 5 = 0 ...(i)
And, (k + 1)x + (1 – k)y = (3k + 1)
⇒ (k + 1)x + (1 – k)y – (3k + 1) = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = (k – 1), b_{1} = 1, c_{1} =  5, a_{2} = (k + 1), b_{2} = (1 – k), c_{2} = (3k + 1)
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
i.e., (k – 1)/(k + 1) = 1/(k – 1) = 5/(3k + 1)
⇒ (k – 1)/(k + 1) = 1/(k – 1) = 5/(3k + 1)
Now, we have the following three cases:
Case I:
(k – 1)/(k + 1) = 1/(k – 1)
⇒ (k – 1)^{2} = (k + 1)
⇒ k^{2 }+ 1 – 2k = k + 1
⇒ k^{2} – 3k = 0 ⇒ k(k – 3) = 0
⇒ k = 0 or k = 3
Case II:
1/(k – 1) = 5/(3k + 1)
⇒ 3k + 1 = 5k – 5
⇒ 2k = 6 ⇒ k = 3
Case III:
(k – 1)/(k + 1) = 5/(3k + 1)
⇒ (3k + 1)(k – 1) = 5(k + 1)
⇒ 3k^{2} + k – 3k – 1 = 5k + 5
⇒ 3k^{2} – 2k – 5k – 1 – 5 = 0
⇒ 3k^{2 } 7k – 6 = 0
⇒ 3k^{2} – 9k + 2k – 6 =0
⇒ 3k(k – 3) = 2(k – 3) = 0
⇒ (k – 3)(3k + 2) = 0
⇒ (k – 3) = 0 or (3k + 2) = 0
⇒ k = 3 or k = 2/3
Hence, the given system of equations has an infinite number of solutions when k is equal to 3.
20. Find the value of k for which the system of linear equations has a unique solution:
(k – 3)x + 3y – k, kx + ky – 12 = 0
Solution
The given system of equations can be written as
(k – 3)x + 3y – k = 0
kx + ky – 12 = 0
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1} = k, b_{1} = 3, c_{1} = k and a_{2} = k, b_{2} = k, c_{2} = 12
For the given system of equations to have a unique solution, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ (k – 3)/k = 3/k = k/12
⇒ k – 3 = 3 and k^{2} = 36
⇒ k = 6 and k ± 6
⇒ k = 6
Hence, k = 6
21. Find the values of a and b for which the system of linear equations has an infinite number of solutions:
(a – 1)x + 3y = 2, 6x + (1 – 2b) y = 6
Solution
The given system of equations can be written as
(a – 1) x + 3y = 2
⇒ (a – 1) x + 3y – 2 = 0 ...(i)
and 6x + (1 – 2b) y = 6
6x + (1 – 2b) y  6 = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = (a – 1), b_{1} = 3, c_{1} =  2 and a_{2} = 6, b_{2} = (1 – 2b), c_{2} =  6
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ (a – 1)/6 = 3/(1 – 2b) = 2/6
⇒ (a – 1)/6 = 3/(1 – 2b) = 1/3
⇒ (a – 1)/6 = 1/3 and 3/(1 – 2b) = 1/3
⇒ 3a – 3 = 6 and 9 = 1 – 2b
⇒ 3a = 9 and 2b =  8
⇒ a = 3 and b = 4
∴ a = 3 and b =  4
22. Find the values of a and b for which the system of linear equations has an infinite number of solutions:
(2a – 1)x + 3y = 5, 3x + (b – 1)y = 2
Solution
The given system of equations can be written as
(2a – 1)x + 3y = 5
⇒ (2a – 1)x + 3y – 5 = 0 ...(i)
and 3x + (b – 1)y – 2 = 0
⇒ 3x + (b – 1)y – 2 = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = (2a – 1), b_{1} = 3, c_{1} =  5 and a_{2} = 3, b_{2} = (b – 1), c_{2} =  2
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ (2a – 1)/3 = 3/(b – 1) =  5/2
⇒ (2a – 1)/6 = 3/(b – 1) = 5/2
⇒ (2a – 1)/6 = 5/2 and 3/(b – 1) = 5/2
⇒ 2(2a – 1) = 15 and 6 = 5(b – 1)
⇒ 4a – 2 = 15 and 6 = 5b – 5
⇒ 4a = 17 and 5b = 11
∴ a = 17/4 and b = 11/5
23. Find the values of a and b for which the system of linear equations has an infinite number of solutions:
2x – 3y = 7, (a + b)x – (a + b – 3)y = 4a + b.
Solution
The given system of equations can be written as
2x – 3y = 7
⇒ 2x – 3y – 7 = 0 ...(i)
and (a + b)x – (a + b – 3)y = 4a + b
⇒ (a + b)x – (a + b – 3)y – 4a + b = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1 }= 0, a_{2}x + b_{2}y + c_{2} = 0
Here, a_{1} = 2, b_{1} =  3, c_{1} =  7 and a_{2} = (a + b), b_{2} = (a + b – 3), c_{2} = (4a + b)
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
2/(a + b) = 3/(a + b – 3) = 7/(4a + b)
⇒ 2/(a + b) = 3/(a + b – 3) = 7/(4a + b)
⇒ 2/(a + b) = 7/(4a + b) and 3/(a + b  3) = 7/(4a + b)
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b – 3)
⇒ 8a + 2b = 7a + 5b and 12a + 3b = 7a + 7b – 21
⇒ 4a = 17 and 5b = 11
∴ a = 5b ...(iii)
And 5a = 4b – 21 ....(iv)
On substituting a = 5b in (iv), we get;
25b = 4b – 21
⇒ 21b =  21
⇒ b =  1
On substituting b =  1 in (iii), we get;
a = 5(1) =  5
∴ a =  5 and b =  1.
24. Find the values of a and b for which the system of linear equations has an infinite number of solutions:
2x + 3y = 7, (a + b + 1)x – (a + 2b + 2)y = 4(a + b) + 1
Solution
The given system of equations can be written as
2x + 3y = 7
⇒ 2x + 3y – 7 = 0 ...(i)
and (a + b + 1)x – (a + 2b + 2)y = 4(a + b) + 1
(a + b + 1)x – (a + 2b + 2)y – [4(a + b) + 1] = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1 }= 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 2, b_{1} = 3, c_{1} = 7 and a_{2} = (a + b + 1), b_{2} = (a + 2b + 2), c_{2} = [4(a + b) + 1]
For an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2 }= c_{1}/c_{2}
2/(a + b + 1) = 3/(a + 2b + 2) = 7/[4(a + b) + 1]
⇒ 2/(a + b + 1) = 3/(a + 2b + 2) = 7/[4(a + b) + 1]
⇒ 2/(a + b + 1) = 3/(a + 2b + 2) and 3/(a + 2b + 2) = 7/[4(a + b) + 1]
⇒ 2(a + 2b + 2) = 3(a + b + 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a – b – 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a – b = 1 and 5a  2b = 11
a = (b + 1) ...(iii)
5a – 2b = 11 ...(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) – 2b = 11
⇒ 5b + 5 – 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴ a = 3 and b = 2
25. Find the values of a and b for which the system of linear equations has infinite number of solutions:
2x +3y = 7, (a + b)x + (2a – b)y = 21.
Solution
The given system of equations can be written as
2x + 3y – 7 = 0 ...(i)
(a + b)x + (2a – b)y – 21 = 0 ...(ii)
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 2, b_{1} = 3, c_{1} =  7 and a_{2} = a + b, b_{2} = 2a – b, c_{2} =  21
For the given system of linear equations to have an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
⇒ 2/(a + b) = 3/(2a – b) = 7/21
⇒ 2/(a + b) = 7/21 = 1/3 and 3/(2a – b) = 7/21 = 1/3
⇒ a + b = 6 and 2a – b = 9
Adding a + b = 6 and 2a – b = 9, we get
3a = 15 ⇒ a = 15/3 = 3
Now substituting a = 5 in a + b = 6, we have
5 + b = 6 ⇒ b = 6 – 5 = 1
Hence, a = 5 and b = 1.
26. Find the values of a and b for which the system of linear equations has an infinite number of solutions:
2x + 3y = 7, 2ax + (a + b)y = 28.
Answer:
The given system of equations can be written as
2x + 3y – 7 = 0 ...(i)
2ax + (a + b)y – 28 = 0 ...(ii)
This system is of the form:
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = 2, b_{1} = 3, c_{1} =  7 and a_{2} = 2a, b_{2 }= a + b, c_{2} =  28
For the given system of linear equations to have an infinite number of solutions, we must have:
a_{1}/a_{2} = b_{1}/b_{2 }= c_{1}/c_{2}
⇒ 2/2a = 3/(a + b) = 7/28
⇒ 2/2a = 7/28 = 1/4 and 3/(a + b) = 7/28 = 1/4
⇒ a = 4 and a + b = 12
Substituting a = 4 in a + b = 12, we get
4 + b = 12 ⇒ b = 12 – 4 = 8
Hence, a = 4 and b = 8.
27. Find the value of k for which the systems of equations
8x + 5y = 9, kx + 10y = 15 has a nonzero solution.
Solution
The given system of equations:
8x + 5y = 9
8x + 5y – 9 = 0 ...(i)
kx + 10y = 15
kx + 10y – 15 = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
where a_{1} = 8, b_{1} = 5, c_{1} =  9 and a_{2} = k, b_{2} = 10, c_{2} =  15
In order that the given system has no solution, we must have:
a_{1}/a_{2} = b_{1}/b_{2 }≠ c_{1}/c_{2}
⇒ 8/k = 5/10 ≠ 9/15
⇒ 8/k = 1/2 ≠ 3/5
⇒ 8/k = 1/2 and 8/k ≠ 3/5
⇒ k = 16 and k ≠ 40/3
Hence, the given system of equations has no solutions when K is equal to 16.
28. Find the value of k for which the system of equations kx + 3y = 3, 12x + ky = 6 has no solution.
Solution
The given system of equations:
kx + 3y = 3
⇒ kx + 3y – 3 = 0 ...(i)
12x + ky = 6
⇒ 12x + ky – 6 = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where a_{1} = k, b_{1} = 3, c_{1} =  3 and a_{2} = 12, b_{2} = k, c_{2} =  6
In order that the given system has no solution, we must have:
a_{1}/a_{2 }= b_{1}/b_{2 }≠ c_{1}/c_{2 }
i.e., k/12 = 3/k ≠ 3/6
k/12 = 3/k and 3/k ≠ 1/2
⇒ k^{2} = 36 and k ≠ 6
⇒ k = ± 6 and k ≠ 6
Hence, the given system of equations has no solution when k is equal to – 6.
29. Find the value of k for which the system of equations
3x – y = 5, 6x – 2y = k has no solution.
Solution
The given system of equations:
3x – y – 5 = 0 ...(i)
And, 6x – 2y + k = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1} = 3, b_{1} =  1, c_{1} =  5 and a_{2} = 6, b_{2} =  2, c_{2} = k
In order that the given system has no solution, we must have:
a_{1}/a_{2 }= b_{1}/b_{2} ≠ c_{1}/c_{2}
i.e., 3/6 = 1/2 ≠ 5/k
⇒ 1/2 ≠ 5/k ⇒ k ≠  10
Hence, equations (i) and (ii) will have no solution if k ≠  10.
30. Find the value of k for which the system of equations
kx + 3y + 3 – k = 0, 12x + ky – k = 0 has no solution.
Solution
The given system of equations can be written as
kx + 3y + 3 – k = 0 ...(i)
12x + ky – k = 0 ...(ii)
This system of the form:
a_{1}x + b_{1}y + c_{1} = 0,
a_{2}x + b_{2}y + c_{2} = 0
where, a_{1} = k, b_{1} = 3, c_{1} = 3 – k and a_{2} = 12, b_{2} = k, c_{2} =  k
For the given system of linear equations to have no solution, we must have:
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
⇒ k/12 = 3/k ≠ (3 – k)/k
⇒ k/12 = 3/k and 3/k ≠ (3 – k)/k
⇒ k^{2} = 36 and 3 ≠ 3 – k
⇒ k ± 6 and k ≠ 6
⇒ k =  6
Hence, k =  6.
31. Find the value of k for which the system of equations 5x – 3y = 0, 2x + ky = 0
has a nonzero solution.
Solution
The given system of equations:
5x – 3y = 0 ...(i)
2x + ky = 0 ...(ii)
These equations are of the following form:
a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0
Where, a_{1} = 5, b_{1} =  3, c_{1} = 0 and a_{2} = 2, b_{2} = k, c_{2} = 0
For a nonzero solution, we must have:
a_{1}/a_{2} = b_{1}/b_{2}
⇒ 5/2 = 3/k
⇒ 5k =  6 ⇒ k = 6/5
Hence, the required value of k is 6/5.