Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically.
- (i) x + y = 5, 2x + 2y = 10
- (ii) x − y = 8, 3x − 3y = 16
- (iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
- (iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
Step-by-step solution
Idea: Check the ratios first to know what to expect, then draw the lines for the pairs that have solutions and read off the common points.
(i) x + y = 5, 2x + 2y = 10
- 12 = 12 = 510: the second equation is 2 × the first.½ mark
- Both pass through (0, 5) and (5, 0): the lines coincide, so there are infinitely many solutions, every point of the line, e.g. (0, 5), (2, 3), (5, 0).1 mark
(ii) x − y = 8, 3x − 3y = 16
- 13 = −1−3 = 13 but 816 = 12 ≠ 13.½ mark
- The lines are parallel ((8, 0), (0, −8) and (163, 0), (0, −163)): no solution.½ mark
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
- 24 = 12 ≠ 1−2, so the lines intersect: a unique solution.½ mark
- 2x + y = 6: (0, 6), (3, 0). 4x − 2y = 4: (0, −2), (1, 0). Drawing them, they meet at (2, 2): x = 2, y = 2.1 mark
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
- 24 = −2−4 = 12 but −2−5 = 25 ≠ 12.½ mark
- The lines (y = x − 1 and y = x − 54) are parallel: no solution.½ mark
Check: (iii) 2(2) + 2 − 6 = 0 ✓ and 4(2) − 2(2) − 4 = 0 ✓.
Answer to write in the exam
(i)
a1a2 = b1b2 = c1c2 = 12
Both lines pass through (0, 5) and (5, 0) ⇒ coincident
∴ Infinitely many solutions, e.g. (0, 5), (2, 3), (5, 0)
(ii)
a1a2 = b1b2 = 13, c1c2 = 816 = 12
a1a2 = b1b2 ≠ c1c2 ⇒ parallel
∴ No solution
(iii)
a1a2 = 24 ≠ b1b2 = 1−2 ⇒ unique solution
2x + y = 6: (0, 6), (3, 0); 4x − 2y = 4: (0, −2), (1, 0)
The lines meet at (2, 2)
∴ x = 2, y = 2
(iv)
a1a2 = b1b2 = 12, c1c2 = 25
a1a2 = b1b2 ≠ c1c2 ⇒ parallel
∴ No solution
Common mistakes that cost marks
- Cancelling signs wrongly in (iv): −2−5 = 25 (positive), not −25.
- Saying (i) has “no solution” because the two equations look different.
- Plotting (0, −2) for the line 4x − 2y = 4 as (−2, 0).
How this can come in the exam
Which pair has no solution?
- x + y = 2, x − y = 0
- x + 2y = 3, 2x + 4y = 6
- 3x − y = 1, 6x − 2y = 5
- x = 1, y = 1
Show answer
(C) 3x − y = 1, 6x − 2y = 5
36 = −1−2 ≠ 15.
Try one yourself
Does x + 3y = 6, 2x + 6y = 15 have a solution?
Show answer
12 = 36 ≠ 615 ⇒ parallel lines: no solution.
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(Hint: The given problem can be modelled as
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