Question
For the system of equations: (k + 1)x + 3y = (2k + 1) and 2x + ky = k
Find the value(s) of k for which the system has:
- A unique solution
- No solution
- Infinitely many solutions
Solution — Step by Step
This is a standard "condition-based" problem. We compare the ratios of coefficients to determine the type of solution.
Step 1: Write both equations in standard form ax + by + c = 0.
(k + 1)x + 3y − (2k + 1) = 0 → a₁ = k + 1, b₁ = 3, c₁ = −(2k + 1)
2x + ky − k = 0 → a₂ = 2, b₂ = k, c₂ = −k
Step 2: State the three conditions.
Conditions for Solution Types
Unique solution: a₁/a₂ ≠ b₁/b₂
No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
Step 3: Compute the ratios.
a₁/a₂ = (k + 1)/2
b₁/b₂ = 3/k
c₁/c₂ = (2k + 1)/k
Step 4: Find when the system has a unique solution.
a₁/a₂ ≠ b₁/b₂
(k + 1)/2 ≠ 3/k
Cross-multiply: k(k + 1) ≠ 6
k² + k ≠ 6
k² + k − 6 ≠ 0
(k + 3)(k − 2) ≠ 0
So k ≠ −3 and k ≠ 2.
The system has a unique solution for all k except k = −3 and k = 2.
Step 5: Find when the system has no solution.
a₁/a₂ = b₁/b₂ but a₁/a₂ ≠ c₁/c₂
From step 4, a₁/a₂ = b₁/b₂ when k = −3 or k = 2.
Check k = 2:
a₁/a₂ = 3/2, b₁/b₂ = 3/2 ✓ (equal)
c₁/c₂ = (2(2)+1)/2 = 5/2
Since 3/2 ≠ 5/2, the condition a₁/a₂ ≠ c₁/c₂ is satisfied.
k = 2 gives no solution.
Check k = −3:
a₁/a₂ = (−3+1)/2 = −2/2 = −1
b₁/b₂ = 3/(−3) = −1 ✓ (equal)
c₁/c₂ = (2(−3)+1)/(−3) = −5/(−3) = 5/3
Since −1 ≠ 5/3, the condition a₁/a₂ ≠ c₁/c₂ is satisfied.
k = −3 also gives no solution.
Step 6: Check for infinitely many solutions.
We need a₁/a₂ = b₁/b₂ = c₁/c₂.
From above, when k = 2: ratios are 3/2, 3/2, 5/2 — not all equal. When k = −3: ratios are −1, −1, 5/3 — not all equal.
There is no value of k that gives infinitely many solutions for this system.
Summary
Unique solution: k ≠ −3 and k ≠ 2
No solution: k = 2 or k = −3
Infinitely many solutions: No such value of k exists
Why This Works
The coefficient ratios a₁/a₂ and b₁/b₂ determine whether the two lines have the same slope (are parallel or coincident). If they're equal, the lines are either parallel (no solution) or the same line (infinite solutions). The third ratio c₁/c₂ distinguishes between these two cases.
Geometrically: equal slope but different y-intercept = parallel lines (no solution); equal slope and equal y-intercept = same line (infinite solutions).
Alternative Verification for k = 2
Substitute k = 2 into original equations:
- 3x + 3y = 5
- 2x + 2y = 2 → x + y = 1
From the second: x + y = 1. Multiply by 3: 3x + 3y = 3.
But first equation says 3x + 3y = 5. We have 3 = 5, which is impossible. No solution. ✓
🎯 Exam Insider
In CBSE and JEE, always check ALL three cases even if the question asks for only one. Finding k = 2 gives no solution means k = −3 should be checked too — don't stop at the first value. Also, remember that a system can fail to have infinite solutions even when it has no unique solution.
Common Mistake
⚠️ Common Mistake
Mistake: Stopping after finding k values where a₁/a₂ = b₁/b₂ and not checking c₁/c₂.
If a₁/a₂ = b₁/b₂, you still need to check whether c₁/c₂ equals these ratios. If c₁/c₂ matches → infinite solutions. If c₁/c₂ doesn't match → no solution. Many students assume "same first two ratios = infinite solutions" without verifying the third. That's a mark-losing assumption.