Step-by-step solution

1. For a unique solution, the condition is a₁/a₂ ≠ b₁/b₂.
   $(k+1)/k ≠ 8/(k+3)$
2. Cross-multiply: (k+1)(k+3) ≠ 8k.
3. Expand: k² + 4k + 3 ≠ 8k.
4. Rearrange into a quadratic inequality: k² - 4k + 3 ≠ 0.
5. Factorize the quadratic: (k-1)(k-3) ≠ 0.
6. Thus, k ≠ 1 and k ≠ 3. However, there's a common mistake here. The question asks for integer values of k. Also, we must ensure that the denominators in the ratios are not zero, meaning k ≠ 0 and k+3 ≠ 0 (i.e., k ≠ -3).
7. Let's re-write the problem to fit the given options and ensure a correct derivation. I need to make sure the factors are (k-3) and (k+8) or similar.
8. New problem attempt: For what integer value of 'k' will the system of equations (k+5)x + 8y = 4k and kx + (k+3)y = 3k - 1 have a unique solution?
9. Condition: (k+5)/k ≠ 8/(k+3).
10. Cross-multiply: (k+5)(k+3) ≠ 8k.
11. Expand: k² + 8k + 15 ≠ 8k.
12. Simplify: k² + 15 ≠ 0. This is always true for real k, as k² is always non-negative. This is not good as it yields unique solution for all k, which is not an option.
13. Let's go back to the original question coefficients and re-calculate. (k+1)x + 8y = 4k and kx + (k+3)y = 3k - 1. Condition: (k+1)/k ≠ 8/(k+3).
14. (k+1)(k+3) ≠ 8k => k² + 4k + 3 ≠ 8k => k² - 4k + 3 ≠ 0 => (k-1)(k-3) ≠ 0. So k ≠ 1 and k ≠ 3.
15. The options are: A) k ≠ 3, B) k ≠ 3 and k ≠ -8, C) k = -8, D) k = 3. My derived answer is k ≠ 1 and k ≠ 3. This means the question or options are mismatched.
16. I need to ensure the options are valid for the question I'm setting. I will adjust the coefficients of the question to match one of the options, specifically 'k ≠ 3 and k ≠ -8'.
17. Let's consider the equation: (k+8)x + 8y = 4k and kx + (k+3)y = 3k - 1.
18. Condition: (k+8)/k ≠ 8/(k+3).
19. Cross-multiply: (k+8)(k+3) ≠ 8k.
20. Expand: k² + 11k + 24 ≠ 8k.
21. Rearrange: k² + 3k + 24 ≠ 0. This discriminant is 3² - 4(1)(24) = 9 - 96 = -87 < 0. So, this quadratic never equals 0 for real k. This means unique solution for all k. Still not matching.
22. Let's use the original option structure: (k+1)x + 8y = 4k and (k+3)y = 3k - 1. I need to make a coefficient 'k' appear in 'a2' or 'b2' in a way that generates the factors.
23. Let the equations be (k+1)x + 8y = 4k and (k-1)x + (k+6)y = 3k-1. This is becoming too complex to derive a matching set of options for 'k ≠ 3 and k ≠ -8'.
24. Let me simplify and ensure the question and options are consistent. I will ensure the solution fits the option 'k ≠ 3 and k ≠ -8'.
25. Consider the system: (k+5)x + (k-3)y = 10 and 2x + 4y = 5.
26. Condition for unique solution: (k+5)/2 ≠ (k-3)/4.
27. 2(k+5) ≠ k-3 => 2k + 10 ≠ k - 3 => k ≠ -13. This still doesn't match the options.
28. I will construct a new, simpler system of equations that directly leads to the options 'k ≠ 3 and k ≠ -8'.
29. System: kx + 3y = k-3 and 12x + (k+5)y = k.
30. Condition for unique solution: k/12 ≠ 3/(k+5).
31. k(k+5) ≠ 3 × 12.
32. k² + 5k ≠ 36.
33. k² + 5k - 36 ≠ 0.
34. Factorize: (k+9)(k-4) ≠ 0. So k ≠ -9 and k ≠ 4.
35. This still doesn't match the options. This is a critical error in my plan. I need to ensure the derived answer matches one of the options explicitly. I will adjust the coefficients for the 'k ≠ 3 and k ≠ -8' option to be correct.
36. So, we need k/a2 ≠ b1/b2 such that k*b2 ≠ a2*b1 leads to k² + 5k - 24 ≠ 0.
37. Let the equations be: (k+5)x + 3y = C1 and 8x + ky = C2.
38. Then (k+5)/8 ≠ 3/k.
39. k(k+5) ≠ 3 × 8.
40. k² + 5k ≠ 24.
41. k² + 5k - 24 ≠ 0.
42. Factorize: (k+8)(k-3) ≠ 0.
43. So, k ≠ -8 and k ≠ 3. This fits option B perfectly.
44. Final Question: For what integer value of 'k' will the system of equations (k+5)x + 3y = 10 and 8x + ky = 7 have a unique solution?
45. Step 1: For a unique solution, the condition is a₁/a₂ ≠ b₁/b₂.
46. Step 2: From the given equations, a₁ = (k+5), b₁ = 3, a₂ = 8, b₂ = k. So, the condition becomes (k+5)/8 ≠ 3/k.
47. Step 3: Cross-multiply: k(k+5) ≠ 3 × 8. This expands to k² + 5k ≠ 24.
48. Step 4: Rearrange the inequality: k² + 5k - 24 ≠ 0. Factorize the quadratic expression: (k+8)(k-3) ≠ 0.
49. Step 5: This implies that k cannot be -8 and k cannot be 3. Therefore, k ≠ -8 and k ≠ 3.

Answer: k ≠ 3 and k ≠ -8

Step-by-step solution

1. For infinitely many solutions, we must have a₁/a₂ = b₁/b₂ = c₁/c₂.
2. Substitute the coefficients: (k+1)/(k+3) = 8/(k+1) = 4k/(3k-1).
3. From (k+1)/(k+3) = 8/(k+1), we get (k+1)² = 8(k+3) => k² + 2k + 1 = 8k + 24 => k² - 6k - 23 = 0. This seems incorrect. Let's recheck the equation setup. (k+1)/(k+3) = 8/(k+1) => k² + 2k + 1 = 8k + 24 => k² - 6k - 23 = 0. This quadratic has non-integer roots. Let's try equating the second and third ratios first: 8/(k+1) = 4k/(3k-1).
4. From 8/(k+1) = 4k/(3k-1): 8(3k-1) = 4k(k+1) => 24k - 8 = 4k² + 4k => 4k² - 20k + 8 = 0 => k² - 5k + 2 = 0. This also has non-integer roots. Let's try the first and third ratios: (k+1)/(k+3) = 4k/(3k-1). (k+1)(3k-1) = 4k(k+3) => 3k² + 2k - 1 = 4k² + 12k => k² + 10k + 1 = 0. This also gives non-integer roots. Let me re-evaluate my algebra. The equations are: a₁ = k+1, b₁ = 8, c₁ = 4k; a₂ = k+3, b₂ = k+1, c₂ = 3k-1.
5. Let's re-solve (k+1)/(k+3) = 8/(k+1). (k+1)² = 8(k+3) => k² + 2k + 1 = 8k + 24 => k² - 6k - 23 = 0. The problem might involve a typo in my setup or the question values. Let me assume the question intends integer values. Let's test options. If k=7: (7+1)/(7+3) = 8/10 = 4/5. 8/(7+1) = 8/8 = 1. This means k=7 is not the answer based on a₁/a₂ = b₁/b₂ condition. Let me retry the problem statement and my interpretation. Okay, I have to ensure the options are correct. Let's assume the question is well-posed and I am making a calculation error.
6. Let k=7. Then the ratios should be equal. a₁/a₂ = (7+1)/(7+3) = 8/10 = 4/5. b₁/b₂ = 8/(7+1) = 8/8 = 1. These are not equal. This implies my initial question formulation or coefficients are flawed for providing an integer answer from the standard consistency conditions.
7. Let's re-design this question entirely to ensure it works. I need a system where (k+1)/(k+3) = 8/(k+1) = 4k/(3k-1) leads to an integer solution. Let's try to enforce a solution, say k=7. This implies (k+1)/(k+3) and 8/(k+1) and 4k/(3k-1) must all be equal. If k=7, then (k+1)=8. (k+3)=10. So a₁/a₂ = 8/10 = 4/5. Then b₁/b₂ must also be 4/5. So 8/(k+1) = 4/5. This means k+1 = 10, so k=9. If k=9, then a₁/a₂ = 10/12 = 5/6. b₁/b₂ = 8/10 = 4/5. Still not matching.
8. I will construct a new question where the consistency conditions lead to a clear integer solution from the given options.
9. Let the system be:
   (k+1)x + (k+2)y = 2k+3
   (k+3)x + (k+4)y = 2k+5
   For infinitely many solutions: (k+1)/(k+3) = (k+2)/(k+4) = (2k+3)/(2k+5).
10. Consider (k+1)/(k+3) = (k+2)/(k+4).
    (k+1)(k+4) = (k+2)(k+3)
    k² + 5k + 4 = k² + 5k + 6
    4 = 6, which is impossible. This means such a system cannot have infinitely many solutions for any k. This type of question design is tricky. I need to ensure the coefficients are chosen such that the equations are consistent and lead to a solution for k.
11. Let's use a standard setup that usually works: (a-b)x + (a+b)y = 2 ; (a+b)(x+y) = a²-b².
12. New question idea for infinite solutions:
    (m-1)x + 3y = 5
    2x + (m+1)y = 10
    For infinitely many solutions: (m-1)/2 = 3/(m+1) = 5/10.
13. From 3/(m+1) = 5/10 = 1/2, we get m+1 = 6, so m = 5.
14. Check with (m-1)/2 = 1/2: (5-1)/2 = 4/2 = 2. But this should be 1/2. So this setup doesn't work either. The third ratio must be 1/2. (m-1)/2 should be 1/2. So m-1=1, m=2. Then 3/(m+1) should be 1/2. 3/(2+1) = 3/3 = 1. This is not 1/2.
15. I must select the coefficients such that all three ratios are equal for one value of k from the options. Let's try option C: k=7.
    If k=7, the common ratio should be R.
    (k+1)/A = 8/B = 4k/C = R.
    Let's assume R = 1/2.
    k+1 = A/2 => 8 = A/2 => A=16. So k+1=16 => k=15. Not 7.
    Let's assume R=1.
    k+1 = A => A = k+1.
    8 = B. So B=8.
    4k = C. C=4k.
    This makes a₁/a₂ = (k+1)/ (k+1) = 1. b₁/b₂ = 8/8 = 1. c₁/c₂ = 4k/4k = 1.
    So, I can set a₂ = a₁, b₂ = b₁, c₂ = c₁ for infinitely many solutions.
    This is too simple. The challenge is usually when a₂ = p*a₁, b₂ = p*b₁, c₂ = p*c₁ for some p.
16. Let's use a common structure for such problems:
    (a-b)x + (a+b)y = a² - 2ab - b²
    (a+b)x + (a-b)y = a² + 2ab - b²
    This is for specific solution. Let's go back to the parameter 'k' and integer options.
17. Let the system be:
    (k-3)x + 4y = k
    3x + (k+1)y = 9
    For infinitely many solutions: (k-3)/3 = 4/(k+1) = k/9.
18. From (k-3)/3 = 4/(k+1):
    (k-3)(k+1) = 12
    k² - 2k - 3 = 12
    k² - 2k - 15 = 0
    (k-5)(k+3) = 0
    So, k=5 or k=-3.
19. Now check with k/9. If k=5:
    (5-3)/3 = 2/3. 4/(5+1) = 4/6 = 2/3. 5/9.
    2/3 ≠ 5/9. So k=5 is not the solution.
20. If k=-3:
    (-3-3)/3 = -6/3 = -2. 4/(-3+1) = 4/-2 = -2. -3/9 = -1/3.
    -2 ≠ -1/3. So k=-3 is not the solution.
21. This means I need to design a question carefully from scratch. I'll make sure one of the options works for the general condition. Let's target k=7 as the answer. So, the ratios must be equal when k=7.
22. Let the first equation be: (k-5)x + 2y = 14.
    If k=7, this becomes 2x + 2y = 14 => x+y=7.
23. Let the second equation be: 3x + (k-4)y = 21.
    If k=7, this becomes 3x + 3y = 21 => x+y=7.
24. So, for k=7, the system is 2x+2y=14 and 3x+3y=21, which are equivalent to x+y=7. This gives infinitely many solutions.
25. Now, let's check the general condition for infinitely many solutions for the system:
    (k-5)x + 2y = 14
    3x + (k-4)y = 21
    Conditions: (k-5)/3 = 2/(k-4) = 14/21.
26. We know 14/21 = 2/3. So, we need (k-5)/3 = 2/3 and 2/(k-4) = 2/3.
27. From (k-5)/3 = 2/3 => k-5 = 2 => k = 7.
28. From 2/(k-4) = 2/3 => k-4 = 3 => k = 7.
29. Both conditions yield k=7. This is a good question.
30. So the question will be:
    (k-5)x + 2y = 14
    3x + (k-4)y = 21

Answer: 7

Step-by-step solution

1. For the system to have no solution, the lines must be parallel and distinct. This means a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
2. From the given equations, we have:
   a₁ = k²-1, b₁ = k-1, c₁ = k+1
   a₂ = k+1, b₂ = 1, c₂ = 1
   Setting a₁/a₂ = b₁/b₂: (k²-1)/(k+1) = (k-1)/1. This simplifies to (k-1)(k+1)/(k+1) = k-1. For k ≠ -1, this is (k-1) = k-1, which is always true. So, the coefficients of x and y are proportional for all k ≠ -1.
3. Now, check the condition for no solution, which is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. So, we need (k-1) ≠ (k+1)/1. This simplifies to k-1 ≠ k+1, which means -1 ≠ 1, always true. Thus, for any k ≠ -1, the system has no solution.
4. Consider the special case k = -1. The equations become:
   ( (-1)²-1 )x + (-1-1)y = -1+1 => 0x - 2y = 0 => y = 0
   ( -1+1 )x + y = 1 => 0x + y = 1 => y = 1
   This implies 0 = 1, which is a contradiction. Therefore, for k = -1, the system has no solution.
   Consider the special case k = 1. The equations become:
   ( 1²-1 )x + (1-1)y = 1+1 => 0x + 0y = 2 => 0 = 2, which is a contradiction. Therefore, for k = 1, the system has no solution.
   Thus, the system has no solution for k = 1 or k = -1.

Answer: k = 1 or k = -1