**Question 1:Â ****kx + 2y = 5,Â ****3x + y = 1**

**Solution:**

kx + 2y – 5 = 0 â€¦â€¦..(1)

3x + y – 1 = 0 â€¦â€¦.. (2)

a_{1 }= k, b_{1} = 2, c_{1 }= -5

a_{2} = 3, b_{2} = 1, c_{2 }= -1

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{a_{1}}{a_{2}} Â \neq \frac{b_{1}}{b_{2}} i.e \frac{k}{3} \neq \frac{2}{1} \)

\( k \neq 6 \)

Thus, for all real values of k other than 6, the given system of equations will have a unique solution.

**Question 2:Â ****x – 2y = 3,Â ****3x + ky = 1**

**Solution:**

x – 2y – 3 = 0 â€¦â€¦.. (1)

3x + ky – 1 = 0 â€¦â€¦.. (2)

a_{1 }= 1,Â b_{1} = -2, c_{1 }= -3

a_{2} = 3, b_{2} = k, c_{2 }= -1

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{1}{3} Â \neq \frac{-2}{k} \)

\( k \neq -6 \)

Thus, for all real values of k other than -6, the given system of equations will have a unique solution.

**Question 3:Â ****kx + 3y = (k – 3),Â ****12x + ky = k **

**Solution:**

kx + 3y – (k – 3) = 0 â€¦â€¦..(1)

12x + ky – k = 0 â€¦â€¦.. (2)

a_{1 }= k, b_{1} = 3, c_{1 }= -(k – 3)

a_{2} = 12, b_{2} = k, c_{2 }= -k

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{k}{12} \neq \frac{3}{k} \)

\( k^{2} \neq 36 \Rightarrow k \neq \pm 6 \)

Thus, for all real values of k other than \( \pm 6 \)

**Question 4:Â ****4x – 5y = k,Â ****2x – 3y = 12**

**Solution:**

4x – 5y -k = 0 â€¦â€¦.. (1)

2x – 3y – 12 = 0 â€¦â€¦.. (2)

a_{1 }= 4, b_{1} = -5,Â c_{1 }= -k

a_{2} = 2, b_{2} = -3, c_{2 }= -12

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{4}{2} \neq \frac{-5}{-3} \)

\( 2 \neq \frac{5}{3} \Rightarrow 6 \neq 5 \)

Thus, for all real values of k, the given system of equations will have a unique solution.

**Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:**

**Question 5:Â ****2x + 3y = 7 ,Â ****(k – 1)x + (k + 2)y = 3k**

**Solution:**

2x + 3y – 7 = 0 â€¦â€¦.. (1)

(k – 1)x + (k + 2)y – 3k = 0 â€¦â€¦..(2)

a_{1 }= 2, b_{1} = 3, c_{1 }= -7

a_{2} = (k – 1), b_{2} = (k + 2), c_{2 }= -3k

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{2}{k – 1} = \frac{3}{k + 2} = \frac{-7}{-3k} \)

\( \frac{2}{k – 1} = \frac{3}{k + 2} = \frac{7}{3k} \)

Now, the following cases arises

**Case 1:**

\( \frac{2}{k – 1} = \frac{3}{k + 2} \)

\( \Rightarrow 2(k + 2) = 3(k – 1) \Rightarrow 2k + 4 = 3k -3 \)

\( \Rightarrow k = 7 \)

**Case 2:**

\( \frac{3}{k + 2} = \frac{7}{3k} \)

\( \Rightarrow 7(k + 2) = 9k \Rightarrow 7k + 14 = 9k \)

\( \Rightarrow k = 7 \)

**Case 3:**

\( \frac{2}{k – 1} = \frac{7}{3k} \)

\( \Rightarrow 7k – 7 = 6k \)

\( \Rightarrow k = 7 \)

For k = 7, there are infinitely many solutions of the given system of equations.

**Question 6:Â ****2x + (k – 2)y = k,Â ****6x + (2k – 1)y = (2k + 5)**

**Solution:**

2x + (k – 2)y – k = 0 â€¦â€¦..(1)

6x + (2k – 1)y – (2k + 5) = 0 â€¦â€¦..(2)

a_{1 }= 2, b_{1} = (k – 2), c_{1 }= -k

a_{2} = 6, b_{2} = (2k – 1), c_{2 }= -(2k + 5)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{2}{6} = \frac{k – 2}{2k – 1} = \frac{-k}{-(2k + 5)} \)

\( \frac{1}{3} = \frac{k – 2}{2k – 1} = \frac{k}{2k + 5} \)

Now, the following cases arises

**Case 1:**

\( \frac{1}{3} = \frac{k – 2}{2k – 1} \)

\( \Rightarrow 2k – 1 = 3k – 6 \)

\( \Rightarrow k = 5 \)

**Case 2:**

\( \frac{k – 2}{2k – 1} = \frac{k}{2k + 5} \)

\( \Rightarrow 2k^{2} + 5k – 4k -10 2k^{2} – k \Rightarrow k + k = 10 \)

\( \Rightarrow k = 5 \)

**Case 3:**

\( \frac{1}{3} = \frac{k}{2k + 5} \)

\( \Rightarrow 2k + 5 = 3k \)

\( \Rightarrow k = 5 \)

For k = 5, there are infinitely many solutions of the given system of equations.

**Question 7:Â ****kx + 3y = (2k + 1),Â ****2(k + 1)x + 9y = (7k + 1)**

**Solution:**

kx + 3y – (2k + 1) = 0 â€¦â€¦..(1)

2(k + 1)x + 9y – (7k + 1) = 0 â€¦â€¦..(2)

a_{1 }= k, b_{1} = 3, c_{1 }= -(2k + 1)

a_{2} = 2(k + 1), b_{2} = 9, c_{2 }= -(7k + 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{k}{2(k + 1)} = \frac{3}{9} = \frac{-(2k + 1)}{-(7k + 1)} \)

\( \frac{k}{2(k + 1)} = \frac{1}{3} = \frac{2k +1 }{7k + 1} \)

Now, the following cases arises

**Case 1:**

\( \frac{k}{2(k + 1)} = \frac{1}{3} \)

\( \Rightarrow 2(k + 1) = 3k \Rightarrow 2k + 2 = 3k \)

\( \Rightarrow k = 2 \)

**Case 2:**

\( \frac{1}{3} = \frac{2k + 1}{7k + 1} \)

\( \Rightarrow 7k + 1 = 6k + 3 \Rightarrow 7k – 6k = 3 – 1 \)

\( \Rightarrow k = 2 \)

**Case 3:**

\( \frac{k}{2(k + 1)} = \frac{2k + 1}{7k + 1} \)

\( \Rightarrow k(7k + 1) = 2(2k + 1)(k + 1) \)

\( \Rightarrow 7k^{2} + k = 2(2k^{2} + 2k + k + 1) \)

7k^{2} + k = 4k^{2} + 6k + 2

7k^{2} – 4k^{2} + k – 6k – 2 = 0

3k^{2} – 5k – 2 = 0

3k^{2} – (6k – 1k) – 2 = 0

3k(k – 2) + 1(k – 2) = 0

(k – 2) (3k + 1) = 0

\( k = 2 or k = \frac{-1}{3} \)

Thus, for k = 2, there are infinitely many solutions of the given system of equations.

**Question 8:Â ****5x + 2y = 2k,Â ****2(k + 1)x + ky = (3k + 4)**

**Solution:**

5x + 2y – 2k = 0 â€¦â€¦.. (1)

2(k + 1)x + ky – (3k + 4) = 0 â€¦â€¦..(2)

a_{1 }= 5, b_{1} = 2, c_{1 }= -2k

a_{2} = 2(k + 1), b_{2} = k, c_{2 }= -(3k + 4)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{5}{2(k + 1)} = \frac{2}{k} = \frac{-2k}{-(3k + 4)} \)

\( \frac{5}{2(k + 1)} = \frac{2}{k} = \frac{2k}{(3k + 4)} \)

Now, the following cases arises

**Case 1:**

\( \frac{5}{2(k + 1)} = \frac{2}{k} \)

\( \Rightarrow 5k = 4(k + 1) \Rightarrow 5k = 4k + 4 \)

\( \Rightarrow k = 4 \)

**Case 2:**

\( \frac{2}{k} = \frac{2k}{(3k + 4)} \)

\( \Rightarrow 2(3k + 4) = 2k^{2} \Rightarrow 6k + 8 = 2k^{2} \)

\( \Rightarrow 2k^{2} – 6k – 8 = 0 \)

On solving the above quadratic equation, we get k = 4 or k = -1

**Case 3:**

\( \frac{5}{2(k + 1)} = \frac{2k}{(3k + 4)} \)

\( \Rightarrow 15k + 20 = 4k^{2} = 4k \)

\( \Rightarrow 4k^{2} + 4k – 15 – 20 = 0 \)

On solving the above quadratic equations we get, \( k = 4 or k = \frac{-5}{4} \)

Thus, k = 4 is a common value for which there are infinitely many solutions.

**Question 9:Â ****x+ (k – 1) y = 5,Â ****(k + 1)x + 9y = -(8k – 1)**

**Solution:**

x + (k + 1)y – 5 = 0 â€¦â€¦..(1)

(k + 1)x + 9y – (8k – 1) = 0 â€¦â€¦..(2)

a_{1 }= 1, b_{1} = (k + 1), c_{1 }= -5

a_{2} = (k + 1), b_{2} = 9, c_{2 }= -(8k – 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{1}{k + 1} = \frac{k + 1}{9} = \frac{-5}{-(8k – 1)} \)

\( \frac{1}{k + 1} = \frac{k + 1}{9} = \frac{5}{(8k – 1)} \)

Now, the following cases arises:

**Case 1:**

\( \frac{1}{k + 1} = \frac{k + 1}{9} \)

\( \Rightarrow (k + 1)^{2} = 9 Â \Rightarrow (k + 1) = \pm 3 \)

K + 1 = 3 or k + 1 = -3

K = 2 or k = -4

**Case 2:**

\( \frac{k + 1}{9} = \frac{-5}{-(8k – 1)} \)

\( \Rightarrow (k + 1)(8k – 1) = 45 \Rightarrow 8k^{2} + 7k – 46 = 0 \)

On solving the above equation,

\( \Rightarrow k = \frac{-23}{8} or k = 2 \)

**Case 3:**

\( \frac{1}{k + 1} = \frac{-5}{-(8k – 1)} \)

\( \Rightarrow 8k – 1 = 5(k + 1) \)

\( \Rightarrow 3k = 6 \Rightarrow k = 2 \)

Thus, k = 2 is the common value for which there are infinitely many solutions

**Question 10:Â ****(k – 1)x – y = 5,Â ****(k + 1)x + (1 – k)y = (3k + 1)**

**Solution:**

(k – 1)x – y – 5 = 0 â€¦â€¦..(1)

(k + 1)x + (1 – k)y – (3k + 1) = 0 â€¦â€¦..(2)

a_{1 }= (k – 1), b_{1} = -1), c_{1 }= -5

a_{2} = (k + 1), b_{2} = (1 – k), c_{2 }= -(3k + 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{k – 1}{k + 1} = \frac{-1}{-(k – 1)} = \frac{-5}{-(3k + 1)} \)

\( \frac{k – 1}{k + 1} = \frac{1}{(k – 1)} = \frac{5}{(3k + 1)} \)

Now, the following cases arise:

**Case 1:**

\( \frac{k – 1}{k + 1} = \frac{1}{(k – 1)} \)

\( \Rightarrow (k – 1)^{2} = k + 1 Â \Rightarrow k^{2} = 3k \)

k = 3

**Case 2:**

\( \frac{k – 1}{k + 1} = \frac{5}{(3k + 1)} \)

\( \Rightarrow (k + 1)(3k + 1) = 5(k + 1) \Rightarrow 3k^{2} – 7k – 6 = 0 \)

On solving the above equation,

\( \Rightarrow k = 3 \)

**Case 3:**

\( \frac{k – 1}{k + 1} = \frac{5}{(3k + 1)} \)

\( \Rightarrow (k – 1)(3k + 1) = 5(k + 1) \)

\( \Rightarrow 3k^{2} – 7k – 6 = 0 \Rightarrow k = \frac{-2}{3} or k = 3 \)

K = 3 is the common value for which the number of solution is infinitely many.

**Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:**

**Question 11:Â ****(2a – 1)x + 3y = 5,Â ****3x + (b – 1)y = 2**

**Solution:**

(2a – 1)x + 3y – 5 = 0 â€¦â€¦..(1)

3x + (b – 1)y – 2 = 0 â€¦â€¦..(2)

a_{1 }= (2a – 1), b_{1} = 3, c_{1 }= -5

a_{2} = 3, b_{2} = (b – 1), c_{2 }= -2

These holds only when,

\(\frac{2a – 1}{3}\)

\(\frac{2a – 1}{3}\)

= \( \frac{2a – 1}{3} = \frac{5}{2} \;\;and \;\; \frac{3}{b – 1} = \frac{5}{2} \)

4a – 2 = 15 and 5(b – 1) = 6

4a = 17 and 5b – 5 = 6

\( a = \frac{17}{4}\)

Therefore, \( a = \frac{17}{4}\)