For what values of p does the pair of equations given below have a unique solution?
4x + py + 8 = 0; 2x + 2y + 2 = 0.
Answer: Unique solution when 42 ≠ p2, i.e. for all p ≠ 4.
Step-by-step solution
Idea: A pair has a unique solution exactly when a1a2 ≠ b1b2.
- a1a2 = 42 = 2 and b1b2 = p2. Unique solution needs 42 ≠ p2.1 mark
- 2 ≠ p2 ⇒ p ≠ 4. So the pair has a unique solution for every value of p except 4. (For p = 4: 82 = 4 ≠ 2, so the lines are parallel.)1 mark
All real values of p except p = 4.
Check: p = 4: 4x + 4y + 8 = 0 ⇒ x + y = −2, and the second is x + y = −1: parallel, no solution ✓. p = 0: x = −2, then y = 1: unique ✓.
Answer to write in the exam
Unique solution ⇔ a1a2 ≠ b1b2
42 ≠ p2 ⇒ 2 ≠ p2
∴ p ≠ 4
Common mistakes that cost marks
- Answering “p = 4″ (that is the one value that does not work).
- Using c1c2 in the condition; for a unique solution only a1a2 and b1b2 matter.
How this can come in the exam
MCQ (1 mark)
The pair kx + 3y = 5, 2x + 6y = 7 has a unique solution if
- k = 1
- k ≠ 1
- k = 3
- k ≠ 3
Show answer
(B) k ≠ 1
k2 ≠ 36 ⇒ k ≠ 1.
Try one yourself
For what k does 3x + ky = 1, 6x + 4y = 5 have a unique solution?
Show answer
36 ≠ k4 ⇒ k ≠ 2.
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