Consistency and the number of solutions
Compare a₁/a₂, b₁/b₂ and c₁/c₂. If the first two ratios differ, there is one solution. All three equal: infinitely many. Only the first two equal: none.
The ratio test
| Compare the ratios | Graph | Number of solutions | The pair is |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting lines | Exactly one | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident lines | Infinitely many | Consistent (dependent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel lines | None | Inconsistent |
Consistent vs inconsistent
Write both equations in the same form, such as ax + by = c, before you compare the ratios.
Example 1: test three pairs without solving
Say how many solutions each pair has: (i) 2x − 3y = 5 and 4x + y = 3 (ii) x + 2y = 4 and 3x + 6y = 12 (iii) 2x + 5y = 7 and 4x + 10y = 9.
- (i) a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = −3/1 = −3. The ratios differ, so there is exactly one solution.
- (ii) a₁/a₂ = 1/3, b₁/b₂ = 2/6 = 1/3, c₁/c₂ = 4/12 = 1/3. All equal, so infinitely many solutions.
- (iii) a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 5/10 = 1/2, c₁/c₂ = 7/9. The first two are equal but not the third, so no solution.
(i) one solution (ii) infinitely many (iii) no solution
Example 2: find k for no solution
Find the value of k for which the pair kx + 3y = 5 and 6x + 9y = 11 has no solution.
- For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
- a₁/a₂ = k/6, b₁/b₂ = 3/9 = 1/3, c₁/c₂ = 5/11.
- k/6 = 1/3 gives k = 2.
- Check the third ratio: 1/3 ≠ 5/11, so the condition holds.
k = 2
Example 3: find k for infinitely many solutions
Find the value of k for which the pair 2x + ky = 6 and 6x + 12y = 18 has infinitely many solutions.
- For infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂.
- a₁/a₂ = 2/6 = 1/3 and c₁/c₂ = 6/18 = 1/3.
- So b₁/b₂ = k/12 must also be 1/3, giving k = 4.
- Check: with k = 4 the second equation is 3 times the first.
k = 4
Two shops may charge the same price per kg of onions (an example rate) but different fixed bag fees. Their cost lines are then parallel. The bills are never equal, so the pair has no solution.
1 marks: “Without solving, state the number of solutions of 3x − 5y = 7 and 6x − 10y = 3.” a₁/a₂ = 1/2 = b₁/b₂, but c₁/c₂ = 7/3. So there is no solution.
2 marks: “Find the value of k for which x + 2y = 3 and 5x + ky = 15 has infinitely many solutions.” 1/5 = 2/k = 3/15. So 2/k = 1/5 and k = 10.
2 marks: “For what values of k does 3x + y = 1 and kx + 2y = 5 have a unique solution?” Need a₁/a₂ ≠ b₁/b₂: 3/k ≠ 1/2, so k ≠ 6. Any k except 6 works.
a and b ratios differ: one meeting. All three equal: same line. Only c differs: parallel.