Simultaneous equations: two lines, or a line and a curve
What this lesson covers
- Solving two equations together finds the points where their graphs meet.
- Substituting a linear equation into a quadratic gives a quadratic, whose discriminant counts the meeting points.
Two equations, one answer
- Each equation is a graph.
- Solving them together finds where they meet.
- Every meeting point satisfies both.
Two lines: elimination
- Match the y terms.
- Add: y cancels.
- 3x + 2y = 16 (1)
- x − y = 2 (2)
- 2x − 2y = 4(2) × 2
- 5x = 20add: y cancels
- x = 4divide by 5
- 4 − y = 2substitute in (2)
- y = 2
Check in both
- Check in both equations.
- One is not enough.
- (1): 3(4) + 2(2) = 12 + 4 = 16
- (2): 4 − 2 = 2
A line and a parabola
- y = x² − 2x − 3 and y = x + 1.
- They meet at two points.
- (−1, 0) and (4, 5).
Substitute the line into the curve
- Set the ys equal.
- Find y for each x.
- x² − 2x − 3 = x + 1set the ys equal
- x² − 3x − 4 = 0subtract x + 1
- (x − 4)(x + 1) = 0−4 × 1 = −4
- x = 4 or x = −1
- x = 4: y = 4 + 1 = 5from y = x + 1
- x = −1: y = −1 + 1 = 0
- (4, 5) and (−1, 0)
How many meeting points?
- Substituting always gives a quadratic.
- Its discriminant counts the meeting points.
- Two, one, or none.
| Discriminant | Line and curve |
|---|---|
| Δ > 0 | line cuts the curve twice |
| Δ = 0 | line touches: a tangent |
| Δ < 0 | line misses the curve |
Cut, touch, or miss
- y = 2x + 3 cuts y = x² twice.
- y = 2x − 1 just touches it.
- y = 2x − 3 misses it completely.
Using Δ = 0 to find a tangent
- A tangent: Δ = 0.
- Then find the point.
- x² = 2x + cset the ys equal
- x² − 2x − c = 0rearrange
- Δ = 0a tangent touches once
- (−2)² − 4(1)(−c) = 0a = 1, b = −2
- 4 + 4c = 0
- 4c = −4, so c = −1
- x² − 2x + 1 = (x − 1)² = 0
- x = 1, y = 1: touches at (1, 1)
From a word problem
- One equation per fact.
- Substitute, then expand.
- 2a + 3c = 58 (1)
- a + c = 24 (2)
- a = 24 − crearrange (2)
- 2(24 − c) + 3c = 58substitute in (1)
- 48 − 2c + 3c = 58expand
- 48 + c = 58, so c = 10collect
- a = 24 − 10 = 14
- adult $14, child $10
- 2(14) + 3(10) = 28 + 30 = 58
Where the marks are
- Achieved: solve two linear equations.
- Merit: a line and a quadratic.
- Excellence: the discriminant with both values.
What to take away
- Substitute, solve, then find both coordinates.
| Two lines | eliminate or substitute |
| Line and curve | substitute the line in |
| Result | a quadratic, so up to two points |
| Both values | find x, then y, for each |
| How many points | the discriminant decides |
Your turn: four questions
- Pause here and answer each one.
- Give both coordinates of each point.
- Questions 3 and 4 use the discriminant.
| Question | |
|---|---|
| 1 | Solve 2x + y = 11 and x − y = 1. |
| 2 | Where do y = x² and y = x + 6 meet? |
| 3 | Does y = 2x − 3 meet y = x² + 1? Use the discriminant. |
| 4 | Find k so y = 4x + k is a tangent to y = x². |
Show answer
1. Add: 3x = 12, x = 4, y = 3. 2. x² − x − 6 = 0, x = 3 or −2: (3, 9) and (−2, 4). 3. x² − 2x + 4 = 0, Δ = 4 − 16 = −12 < 0: they never meet. 4. x² − 4x − k = 0, Δ = 16 + 4k = 0, k = −4.
The answers: questions 1 and 2
- 1. 2x + y = 11, x − y = 1
- 3x = 12, so x = 4add: y cancels
- 4 − y = 1, so y = 3substitute
- (4, 3)
- 2. x² = x + 6set the ys equal
- x² − x − 6 = 0
- (x − 3)(x + 2) = 0factorise
- x = 3, y = 9; x = −2, y = 4y = x + 6
- (3, 9) and (−2, 4)
The answers: questions 3 and 4
- 3. x² + 1 = 2x − 3set the ys equal
- x² − 2x + 4 = 0rearrange
- Δ = (−2)² − 4(1)(4)substitute
- = 4 − 16 = −12 < 0they never meet
- 4. x² = 4x + kset the ys equal
- x² − 4x − k = 0rearrange
- Δ = (−4)² − 4(1)(−k) = 0a tangent: Δ = 0
- 16 + 4k = 0
- k = −4
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