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Number of solutions of a pair · 3 marks

Are the converses of the above statements true?

Answer: Yes. Unique solution ⇒ a1a2 ≠ b1b2; infinitely many ⇒ a1a2 = b1b2 = c1c2; no solution ⇒ a1a2 = b1b2 ≠ c1c2. The three conditions cover every pair and never overlap, so each outcome can come from only its own condition.

Step-by-step solution

Given: The pair a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0; 1. a1a2 ≠ b1b2 ⇒ a unique solution; 2. a1a2 = b1b2 = c1c2 ⇒ infinitely many solutions; 3. a1a2 = b1b2 ≠ c1c2 ⇒ no solution

Idea: Every pair satisfies exactly one of the three ratio conditions, and each condition leads to a different outcome. So if we know the outcome, we can work backwards to the condition.

  1. The three statements are: (1) a1a2 ≠ b1b2 ⇒ unique solution; (2) a1a2 = b1b2 = c1c2 ⇒ infinitely many; (3) a1a2 = b1b2 ≠ c1c2 ⇒ no solution. The converses swap “if” and “then”.½ mark
  2. Any pair falls under exactly one condition: either a1a2 ≠ b1b2, or they are equal and then c1c2 is either equal or not. So the three conditions cover all cases with no overlap.1 mark
  3. Converse of (1): suppose a pair has a unique solution. It cannot satisfy condition (2) (that would give infinitely many) or (3) (that would give none). So it satisfies (1): a1a2 ≠ b1b2 ✓.½ mark
  4. The same argument works for the others: infinitely many solutions rules out (1) and (3), so (2) holds; no solution rules out (1) and (2), so (3) holds. All three converses are true.1 mark
Yes, all three converses are true, because the three ratio conditions cover every pair without overlap and give three different outcomes.

Check: x + y = 2, x − y = 0 has the unique solution (1, 1), and indeed 11 ≠ 1−1 ✓.

Answer to write in the exam

Every pair satisfies exactly one of: a1a2 ≠ b1b2; a1a2 = b1b2 = c1c2; a1a2 = b1b2 ≠ c1c2

These give unique, infinitely many, no solution respectively

Unique solution ⇒ not the 2nd or 3rd condition ⇒ a1a2 ≠ b1b2

Similarly for the other two

∴ All the converses are true

Common mistakes that cost marks

  • Assuming a converse is true because the statement is true. Here it is true, but only because the cases cover everything and do not overlap; this needs to be said.
  • Forgetting the “no overlap” part of the argument.

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): If px + 3y = 4 and 2x + 6y = 8 have infinitely many solutions, then p = 1.
Reason (R): Infinitely many solutions ⇒ a1a2 = b1b2 = c1c2.

  1. Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
  2. Both Assertion (A) and Reason (R) are true, but R is not the correct explanation of A.
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Show answer

(A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
p2 = 36 = 48 = 12 ⇒ p = 1; R (a converse) is what we used.

Try one yourself

A pair 2x + ky = 3, 4x + 6y = 5 has no solution. Find k.

Show answer

No solution ⇒ 24 = k6 ⇒ k = 3 (and 35 ≠ 12 ✓).

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