Are (1, 92), (0, 6), (4, 1) and (12, 214) solutions of the equation 3x + 2y = 12?
Can you find any other solution? How many solutions can you find?
- 1. Are (1, 92), (0, 6), (4, 1) and (12, 214) solutions of the equation 3x + 2y = 12?
- 2. Can you find any other solution? How many solutions can you find?
Step-by-step solution
Idea: Substitute each pair. To make new solutions, choose any value of x, put it in, and solve the one-variable equation for y. Every choice of x gives a new solution, so there is no end to them.
1. Are (1, 92), (0, 6), (4, 1) and (12, 214) solutions of the equation 3x + 2y = 12?
- (1, 92): 3(1) + 2(92) = 3 + 9 = 12 ✓ solution.½ mark
- (0, 6): 3(0) + 2(6) = 0 + 12 = 12 ✓ solution.½ mark
- (4, 1): 3(4) + 2(1) = 12 + 2 = 14 ≠ 12 ✗ not a solution.½ mark
- (12, 214): 3(12) + 2(214) = 32 + 212 = 242 = 12 ✓ solution.½ mark
2. Can you find any other solution? How many solutions can you find?
- Rewrite as y = 12 − 3x2 and pick values of x: x = 2 → y = 3; x = 4 → y = 0; x = −2 → y = 9; x = 6 → y = −3. So (2, 3), (4, 0), (−2, 9), (6, −3) are more solutions.½ mark
- Any real value of x gives exactly one matching y, and there are infinitely many real numbers to choose. So there are infinitely many solutions.½ mark
Check: (6, −3): 18 − 6 = 12 ✓. (−2, 9): −6 + 18 = 12 ✓.
Answer to write in the exam
1.
(1, 92): 3(1) + 2(92) = 12 ✓
(0, 6): 3(0) + 2(6) = 12 ✓
(4, 1): 3(4) + 2(1) = 14 ≠ 12 ✗
(12, 214): 32 + 212 = 12 ✓
∴ All except (4, 1) are solutions
2.
3x + 2y = 12 ⇒ y = 12 − 3x2
x = 2: y = 3; x = 4: y = 0; x = −2: y = 9
∴ (2, 3), (4, 0), (−2, 9) are solutions
Each value of x gives a solution ⇒ infinitely many solutions
Common mistakes that cost marks
- Doubling 92 wrongly: 2 × 92 = 9, not 94.
- Believing only whole-number pairs count. Fractions and negatives such as (12, 214) are solutions too.
- Saying the equation has “two solutions” because two points were found. A linear equation in two variables has infinitely many.
How this can come in the exam
How many solutions does 5x − 2y = 3 have?
- Only one
- Exactly two
- None
- Infinitely many
Show answer
(D) Infinitely many
Every value of x gives one value of y, so there are infinitely many.
Find the value of k if (13, k) is a solution of 6x + 4y = 10. Then give one more solution with x = −1.
Show answer
6(13) + 4k = 10 ⇒ 2 + 4k = 10 ⇒ k = 2 (1 mark). x = −1: −6 + 4y = 10 ⇒ y = 4, so (−1, 4) (1 mark).Try one yourself
Which of (2, 1), (−1, 5), (12, 72) are solutions of 4x + 2y = 10?
Show answer
(2, 1): 8 + 2 = 10 ✓. (−1, 5): −4 + 10 = 6 ✗. (12, 72): 2 + 7 = 9 ✗. Only (2, 1).
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- In the equations shown below, a and b are unknown numbers.
3ax + 4y = −2
2x + by = 14.
If (−3, 4) is the solution of both equations, find the values of a and b. - Four friends — Ranju, Meena, Farhan, and Toshi — are solving problems.
Ranju: I noticed something! If c = 0 in the standard form of a line ax + by + c = 0, then the line must pass through the origin. Look, if I substitute x = 0 and y = 0, in equation ax + by = 0, the equation is satisfied. So, the origin lies on the line! We can also say that the line passes through the origin.
Farhan: Let us try for the equation 2x + 3y = 0. Here a = 2, b = 3 but c = 0. If we substitute x = 0, then we get 3y = 0 or y = 0. This means (0, 0) lies on the line. So yes, this line passes through the origin.
Toshi: Suppose our equation has b = c = 0, say, 5x = 0. This becomes x = 0 which is the equation of the y-axis. And the y-axis passes through the origin.
Meena: And if we take a = c = 0, say, 7y = 0, that means y = 0, which is the equation of the x-axis that also passes through the origin.
So, they conclude: Whenever c = 0, the line ax + by + c = 0 will always pass through the origin, irrespective of the values of a or b. Do you agree with them?
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