NCEA Level 2Simultaneous equationsAS 91261

Simultaneous equations: 7 exam questions solved step by step

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.
Key idea

7 questions, one idea

  • 1. Two lines: eliminate or substitute.
  • 2. Line and curve: set the ys equal.
  • 3. Find y for every x.
  • 4. The discriminant counts meeting points.
  • 5. Check in both equations.
Your turn

Question 1 · Achieved

  • Solve simultaneously:
  • 2x + 3y = 13
  • 5x − 3y = 1
Worked example

Q1: elimination

  • +3y and −3y: add.
  • Check in the other one.
  1. 2x + 3y = 13 (1)
  2. 5x − 3y = 1 (2)
  3. 7x = 14(1) + (2): y cancels
  4. x = 2÷ 7
  5. 2(2) + 3y = 13substitute in (1)
  6. 3y = 9−4 both sides
  7. y = 3÷ 3
  8. (2): 5(2) − 3(3) = 10 − 9 = 1
  9. x = 2, y = 3
Your turn

Question 2 · Achieved

  • Solve simultaneously:
  • y = 3x − 4
  • 2x + y = 11
Worked example

Q2: substitution

  • y is already the subject.
  • Brackets round what goes in.
  1. y = 3x − 4 (1)
  2. 2x + y = 11 (2)
  3. 2x + (3x − 4) = 11substitute (1) in (2)
  4. 5x − 4 = 11collect x terms
  5. 5x = 15+4 both sides
  6. x = 3÷ 5
  7. y = 3(3) − 4 = 5into (1)
  8. (2): 2(3) + 5 = 11
  9. x = 3, y = 5
Your turn

Question 3 · Merit

  • Find where the line y = 3x + 1
  • meets the curve y = x² + 2x − 5.
  • Give both coordinates of each point.
Worked example

Q3: a line and a parabola

  • Set the ys equal.
  • Find y for both points.
  1. x² + 2x − 5 = 3x + 1set the ys equal
  2. x² − x − 6 = 0−3x, −1 both sides
  3. (x − 3)(x + 2) = 0−3 × 2, −3 + 2
  4. x = 3 or x = −2null factor law
  5. x = 3: y = 3(3) + 1 = 10into the line
  6. x = −2: y = 3(−2) + 1 = −5
  7. curve: 9 + 6 − 5 = 10
  8. curve: 4 − 4 − 5 = −5
  9. (3, 10) and (−2, −5)
Your turn

Question 4 · Merit

  • How many times does the line y = 2x − 7
  • meet the curve y = x² − 4x + 2?
  • Use the discriminant, then find the point(s).
Worked example

Q4: count with Δ

  • Δ > 0: 2, Δ = 0: 1, Δ < 0: 0
  • Then find the point.
  1. x² − 4x + 2 = 2x − 7set the ys equal
  2. x² − 6x + 9 = 0−2x, +7 both sides
  3. Δ = b² − 4aca = 1, b = −6, c = 9
  4. = (−6)² − 4(1)(9)
  5. = 36 − 36 = 0
  6. (x − 3)² = 0, so x = 3a perfect square
  7. y = 2(3) − 7 = −1into the line
  8. curve: 9 − 12 + 2 = −1
  9. one point: touches at (3, −1)
Your turn

Question 5 · Merit, in context

  • A school sausage sizzle sells sausages at $3
  • and drinks at $2. It sells 150 items
  • and takes $390. How many of each did it sell?
Worked example

Q5: two facts, two equations

  • Define both unknowns.
  • One equation per fact.
  1. Let s = sausages, d = drinks
  2. s + d = 150 (1)items sold
  3. 3s + 2d = 390 (2)dollars taken
  4. 2s + 2d = 300(1) × 2
  5. s = 90(2) − this: d cancels
  6. d = 150 − 90 = 60from (1)
  7. 3(90) + 2(60) = 270 + 120 = 390
  8. 90 sausages and 60 drinks
Your turn

Question 6 · Excellence, in context

  • A rectangular vege garden has
  • a perimeter of 26 m and an area of 40 m².
  • Find its length and width.
Worked example

Q6: form the equations

  • Perimeter and area.
  • Make y the subject.
  1. Let the sides be x m and y m
  2. 2x + 2y = 26perimeter
  3. xy = 40area
  4. y = 13 − x÷ 2, then −x
  5. x(13 − x) = 40substitute
  6. 13x − x² = 40expand
  7. x² − 13x + 40 = 0rearrange to zero
Worked example

Q6: solve and interpret

  • Two solutions, one shape.
  • Answer in words, with units.
  1. x² − 13x + 40 = 0
  2. (x − 5)(x − 8) = 0−5 × −8, −5 + −8
  3. x = 5 or x = 8
  4. x = 5: y = 13 − 5 = 8
  5. x = 8: y = 13 − 8 = 5
  6. both: the same rectanglesides swapped
  7. 2(8) + 2(5) = 26, 8 × 5 = 40
  8. The garden is 8 m by 5 m.
Your turn

Question 7 · Excellence

  • The line y = kx − 1 is a tangent
  • to the curve y = x² + 3.
  • Find every value of k, and the point
  • where the line touches the curve each time.
Worked example

Q7: a tangent means Δ = 0

  • b is −k here.
  • k² = 16 has two roots.
  1. x² + 3 = kx − 1set the ys equal
  2. x² − kx + 4 = 0−kx, +1 both sides
  3. Δ = 0a tangent touches once
  4. (−k)² − 4(1)(4) = 0a = 1, b = −k, c = 4
  5. k² − 16 = 0
  6. k² = 16+16 both sides
  7. k = 4 or k = −4±√16
Worked example

Q7: both tangents, both points

  • Put each k back in.
  • Find x, then y.
  1. k = 4 or k = −4
  2. k = 4: x² − 4x + 4 = 0
  3. (x − 2)² = 0, so x = 2
  4. y = 4(2) − 1 = 7into the line
  5. k = −4: x² + 4x + 4 = 0
  6. (x + 2)² = 0, so x = −2
  7. y = −4(−2) − 1 = 7into the line
  8. curve: (±2)² + 3 = 7
  9. k = 4 at (2, 7); k = −4 at (−2, 7)
Summary

The method that solved all seven

  • Substitute, solve, find every y, check.
StepHow
Two lineseliminate or substitute
Line, curveset the ys equal, get to 0
Every xfind its y from the line
How manyΔ > 0: 2, Δ = 0: 1, Δ < 0: 0
TangentΔ = 0, then every k
Checkin both equations

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