Solve the pair of linear equations x + 3y = 6 and 2x − 3y = 12 graphically. Comment on the nature of the solution.
Answer: The lines meet at B(6, 0), so x = 6, y = 0; the pair has a unique solution (a1a2 = 12 ≠ b1b2 = −1).
Step-by-step solution
Idea: Draw each line from two easy points, then read off the point common to both lines. Intersecting lines mean exactly one solution.
- x + 3y = 6 ⇒ y = 6 − x3. x = 0 → y = 2; x = 6 → y = 0. Points A(0, 2), B(6, 0).
1 markx 0 6 y 2 0 - 2x − 3y = 12 ⇒ y = 2x − 123. x = 0 → y = −4; x = 3 → y = −2. Points P(0, −4), Q(3, −2).
1 markx 0 3 y −4 −2 - Plot the points and draw lines AB and PQ (see the graph). They intersect at B(6, 0), which lies on both lines.1 mark
- So the solution is x = 6, y = 0, and it is unique. This agrees with the ratios: a1a2 = 12, b1b2 = 3−3 = −1, and 12 ≠ −1.1 mark
x = 6, y = 0 (the lines intersect at (6, 0)); the pair has a unique solution.
Check: 6 + 3(0) = 6 ✓; 2(6) − 3(0) = 12 ✓.
Answer to write in the exam
x + 3y = 6:
| x | 0 | 6 |
| y | 2 | 0 |
2x − 3y = 12:
| x | 0 | 3 |
| y | −4 | −2 |
Plot A(0, 2), B(6, 0), P(0, −4), Q(3, −2); draw AB and PQ
The lines intersect at B(6, 0)
∴ x = 6, y = 0; unique solution (12 ≠ 3−3)
Common mistakes that cost marks
- Reading the meeting point from a rough sketch and not checking it in both equations.
- Plotting P at (−4, 0) instead of (0, −4).
- Not commenting on the nature of the solution, which the question asks for.
How this can come in the exam
MCQ (1 mark)
The lines x + y = 4 and x − y = 2 intersect at
- (3, 1)
- (1, 3)
- (2, 2)
- (4, 0)
Show answer
(A) (3, 1)
Adding: 2x = 6 ⇒ x = 3, y = 1.
Short answer (3 marks)
Solve graphically: x + y = 5 and 2x − y = 4.
Show answer
Points (0, 5), (5, 0) and (0, −4), (2, 0) (1 mark); draw both lines (1 mark); they meet at (3, 2), so x = 3, y = 2 (1 mark).Try one yourself
Solve graphically: x − y = 1 and x + 2y = 7.
Show answer
Points (0, −1), (1, 0) and (1, 3), (7, 0); the lines meet at (3, 2).
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