Find the values of a and b for which the following system of equations has infinitely many solutions:
(a + b)x − 2by = 5a + 2b + 1; 3x − y = 14.
Answer: a = 5, b = 1 (the first equation becomes 6x − 2y = 28, which is 2 × (3x − y = 14)).
Step-by-step solution
Idea: Infinitely many solutions ⇔ a1a2 = b1b2 = c1c2. Use two of these equalities to get two equations in a and b.
- Ratios: a + b3 = −2b−1 = 5a + 2b + 114.½ mark
- First and second: a + b3 = 2b ⇒ a + b = 6b ⇒ a = 5b … (1).1 mark
- Second and third: 2b = 5a + 2b + 114 ⇒ 28b = 5a + 2b + 1 ⇒ 5a − 26b = −1 … (2).½ mark
- Put (1) in (2): 25b − 26b = −1 ⇒ b = 1, so a = 5. a = 5, b = 1.1 mark
a = 5 and b = 1.
Check: With a = 5, b = 1: 6x − 2y = 25 + 2 + 1 = 28, i.e. 2(3x − y) = 2 × 14 ✓: the same line.
Answer to write in the exam
Infinitely many solutions ⇒ a + b3 = −2b−1 = 5a + 2b + 114
a + b3 = 2b ⇒ a = 5b … (1)
2b = 5a + 2b + 114 ⇒ 5a − 26b = −1 … (2)
(1) in (2): −b = −1 ⇒ b = 1, a = 5
∴ a = 5, b = 1
Common mistakes that cost marks
- Writing −2b−1 as −2b; minus divided by minus is plus.
- Using only one equality and stopping with a = 5b.
How this can come in the exam
MCQ (1 mark)
For infinitely many solutions of 2x + 3y = 7 and (k + 1)x + 6y = 14, k =
- 1
- 2
- 3
- 4
Show answer
(C) 3
2k + 1 = 36 ⇒ k + 1 = 4 ⇒ k = 3; and 714 = 12 ✓.
Try one yourself
Find a, b so that 2x + 3y = 7 and (a − b)x + (a + b)y = 21 have infinitely many solutions.
Show answer
a − b = 6, a + b = 9 ⇒ a = 152, b = 32.
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