Number of solutions of a pair · 3 marks

Give 3 more examples of pairs of equations that have no solution. Can you find a simple rule to check when this will happen?

Answer: Examples: x + y = 2 & x + y = 5; 2x − 3y = 1 & 4x − 6y = 7; x + 2y = 3 & 3x + 6y = 1. Rule: a₁a₂ = b₁b₂ ≠ c₁c₂.

Step-by-step solution

Idea: Keep the x– and y-coefficients in the same ratio (same “left side” after scaling) but break the ratio for the constant.

(a) x + y = 2 and x + y = 5: the same expression cannot be 2 and 5.½ mark
(b) 2x − 3y = 1 and 4x − 6y = 7: doubling the first gives 4x − 6y = 2 ≠ 7.½ mark
(c) x + 2y = 3 and 3x + 6y = 1: tripling the first gives 3x + 6y = 9 ≠ 1.½ mark

x + y = 2, x + y = 5; 2x − 3y = 1, 4x − 6y = 7; x + 2y = 3, 3x + 6y = 1

For a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0: no solution when a₁ = ka₂, b₁ = kb₂ but c₁ ≠ kc₂, i.e. a₁a₂ = b₁b₂ ≠ c₁c₂.1 mark
Test (b): 24 = −3−6 = 12 but −1−7 = 17 ≠ 12 ✓. Graphically these are parallel lines.½ mark

a₁a₂ = b₁b₂ ≠ c₁c₂

e.g. x + y = 2 & x + y = 5; 2x − 3y = 1 & 4x − 6y = 7; x + 2y = 3 & 3x + 6y = 1. Rule: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Check: In each example, eliminating x removes y as well and leaves a false statement (0 = 3, 0 = 5, 0 = 8) ✓.

x + y = 2 and x + y = 5

2x − 3y = 1 and 4x − 6y = 7 (2 × first gives 4x − 6y = 2 ≠ 7)

x + 2y = 3 and 3x + 6y = 1 (3 × first gives 3x + 6y = 9 ≠ 1)

Rule: a₁a₂ = b₁b₂ ≠ c₁c₂ ⇒ no solution

e.g. 2x − 3y − 1 = 0, 4x − 6y − 7 = 0: 24 = −3−6 = 12, −1−7 = 17

∴ a₁a₂ = b₁b₂ ≠ c₁c₂ ⇒ no solution

Giving a pair like x + y = 2, 2x + 2y = 4, which is the same line (infinitely many solutions).
Checking only a₁a₂ = b₁b₂ and forgetting to compare the constants.

MCQ (1 mark)

For which value of k does kx + 6y = 5, 2x + 3y = 4 have no solution?

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(B) 4
k2 = 63 ⇒ k = 4, and 54 ≠ 2.

Write a pair with no solution that includes 5x − y = 2.

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e.g. 10x − 2y = 7 (coefficients doubled, constant not).