(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.
(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
- (i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.
- (ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
Step-by-step solution
Idea: Multiplying both sides of a true equation by the same number keeps it true. Because k ≠ 0 we can also divide by k, so we can go back. Both directions together mean “same solutions”.
(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.
- Some solutions of 3x + 4y = 7: (1, 1), (−3, 4), (5, −2). In 6x + 8y = 14: 6 + 8 = 14 ✓, −18 + 32 = 14 ✓, 30 − 16 = 14 ✓. They match.½ mark
- Reason: if 3p + 4q = 7 for a pair (p, q), multiplying by 2 gives 6p + 8q = 14. So every solution of the first is a solution of the second.½ mark
- Conversely, if 6p + 8q = 14, dividing by 2 gives 3p + 4q = 7. So every solution of the second is a solution of the first. The two sets of solutions are the same (both equations have the same graph).½ mark
(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
- Let (p, q) be a solution of ax + by = c, so ap + bq = c. Multiply both sides by k: kap + kbq = kc. So (p, q) is a solution of the second equation.½ mark
- Let (p, q) be a solution of kax + kby = kc, so k(ap + bq) = kc. Since k ≠ 0 we may divide by k: ap + bq = c. So (p, q) solves the first equation.½ mark
- Each set is inside the other, so the sets of solutions are equal. (k ≠ 0 is needed: with k = 0 the second equation becomes 0 = 0, which every pair satisfies.)½ mark
Check: Take k = −3 in (ii) with x + y = 2: −3x − 3y = −6. The pair (5, −3) gives 2 ✓ and −15 + 9 = −6 ✓.
Answer to write in the exam
(i)
Let (p, q) satisfy 3x + 4y = 7: 3p + 4q = 7
× 2: 6p + 8q = 14 ⇒ (p, q) satisfies 6x + 8y = 14
Conversely 6p + 8q = 14 ÷ 2 ⇒ 3p + 4q = 7
∴ Both equations have the same set of solutions
(ii)
Let ap + bq = c ⇒ kap + kbq = kc ⇒ (p, q) solves kax + kby = kc
Let kap + kbq = kc ⇒ k(ap + bq) = kc ⇒ ap + bq = c (k ≠ 0) ⇒ (p, q) solves ax + by = c
∴ The two equations have the same set of solutions
Common mistakes that cost marks
- Checking a few pairs and stopping. Examples support the claim, but the argument needs the multiply-and-divide reasoning.
- Showing only one direction (first ⇒ second). “Same set” needs both directions.
- Forgetting why k ≠ 0 matters: you cannot divide by 0.
How this can come in the exam
Which equation has exactly the same solutions as 2x − y = 3?
- 4x − y = 6
- −6x + 3y = −9
- 2x + y = 3
- 4x − 2y = 3
Show answer
(B) −6x + 3y = −9
−6x + 3y = −9 is −3 × (2x − y = 3).
Try one yourself
Do x − 2y = 4 and 5x − 10y = 20 have the same solutions? Give a reason.
Show answer
Yes: the second is 5 × the first, and dividing by 5 returns it. E.g. (4, 0) and (0, −2) solve both.
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