NCEA Level 1Simultaneous linear equationsAS 91947

Simultaneous equations: 7 exam questions solved step by step

What this lesson covers

  • Two equations in two unknowns are solved by the pair of values that makes both true at once.
  • Graphically, that pair is the point where the two lines cross.
Key idea

7 questions, one method

  • 1. Label the equations (1) and (2).
  • 2. Eliminate or substitute one letter.
  • 3. Solve, then find the other letter.
  • 4. Check in both original equations.
Your turn

Question 1 · Achieved

  • Solve 2x + y = 11 (1)
  • and 3x − y = 9 (2)
Worked example

Q1: 2x + y = 11, 3x − y = 9

  • Opposite signs: add.
  • Check in both equations.
  1. +y and −y: opposite signs
  2. 5x = 20(1) + (2)
  3. x = 4÷ 5
  4. 2 × 4 + y = 11x = 4 into (1)
  5. 8 + y = 11, y = 3
  6. (1): 8 + 3 = 11
  7. (2): 12 − 3 = 9
  8. x = 4, y = 3
Your turn

Question 2 · Achieved

  • Solve y = 3x − 2 (1)
  • and 2x + y = 13 (2)
Worked example

Q2: y = 3x − 2, 2x + y = 13

  • y is given: substitute.
  • Use brackets.
  1. 2x + (3x − 2) = 13replace y in (2)
  2. 5x − 2 = 13collect x terms
  3. 5x = 15+2 both sides
  4. x = 3÷ 5
  5. y = 3 × 3 − 2 = 7x = 3 into (1)
  6. (1): 7 = 9 − 2
  7. (2): 6 + 7 = 13
  8. x = 3, y = 7
Your turn

Question 3 · Merit

  • Solve 3x + 4y = 18 (1)
  • and 2x − 3y = −5 (2)
  • Careful with the negative.
Worked example

Q3: make the x terms match

  • Multiply both equations.
  • Minus a negative is plus.
  1. 6x + 8y = 36(1) × 2
  2. 6x − 9y = −15(2) × 3
  3. 6x − 6x = 0(1) − (2): x goes
  4. 8y − (−9y) = 17yminus a negative
  5. 36 − (−15) = 51
  6. 17y = 51
  7. y = 3÷ 17
Worked example

Q3: find x, check both

  • Back into equation (1).
  • Check both equations.
  1. y = 3
  2. 3x + 4 × 3 = 18y = 3 into (1)
  3. 3x + 12 = 18
  4. 3x = 6−12 both sides
  5. x = 2÷ 3
  6. (1): 6 + 12 = 18
  7. (2): 4 − 9 = −5
  8. x = 2, y = 3
Your turn

Question 4 · Merit

  • The lines y = 2x + 1 and x + y = 7
  • Find where they cross, using algebra.
Question 4 · Merit
Worked example

Q4: where the lines cross

  • Crossing = fits both.
  • y is given: substitute.
  1. x + (2x + 1) = 7replace y
  2. 3x + 1 = 7collect x terms
  3. 3x = 6−1 both sides
  4. x = 2÷ 3
  5. y = 2 × 2 + 1 = 5x = 2 into y = 2x + 1
  6. 2 + 5 = 7
  7. they cross at (2, 5)
Your turn

Question 5 · Merit, in context

  • Hāngī tickets: adults $15, children $8.
  • 120 tickets sold, $1450 taken.
  • How many of each were sold?
Worked example

Q5: hāngī tickets

  • Define the letters.
  • Answer in words.
  1. a = adult, c = child tickets
  2. a + c = 120counts tickets
  3. 15a + 8c = 1450counts dollars
  4. 8a + 8c = 960first × 8
  5. 7a = 490, a = 70money − that, ÷ 7
  6. 70 + c = 120, c = 50
  7. 70 + 50 = 120
  8. 1050 + 400 = 1450
  9. 70 adult, 50 child tickets
Your turn

Question 6 · Excellence, in context

  • Kayak hire: a fixed fee + an hourly rate.
  • 2 hours costs $50. 5 hours costs $95.
  • How many hours can you hire for $140?
Worked example

Q6: find the fee and the rate

  • Stage 1: fee and rate.
  • Same sign: subtract.
  1. f = fee ($), r = rate ($ per h)
  2. f + 2r = 502 hours (1)
  3. f + 5r = 955 hours (2)
  4. 3r = 45(2) − (1)
  5. r = 15÷ 3
  6. f + 30 = 50, f = 20
  7. (2): 20 + 75 = 95
  8. $20 fee, $15 per hour
Worked example

Q6: hours for $140

  • Stage 2: use the rule.
  • Do not forget the fee.
  1. $20 fee, $15 per hour
  2. 140 ÷ 15 ≈ 9.3forgets the $20 fee
  3. C = 15h + 20cost of h hours
  4. 15h + 20 = 140C = 140
  5. 15h = 120−20 both sides
  6. h = 8÷ 15
  7. 15 × 8 + 20 = 140
  8. 8 hours of kayak hire
Your turn

Question 7 · Excellence

  • 3x + y = 5k (1), k a constant
  • x − y = 3k (2)
  • Solve for x and y in terms of k.
  • If x = 6, find k and the point.
Worked example

Q7: solve in terms of k

  • k is just a number.
  • Add: the y terms cancel.
  1. 4x = 8k(1) + (2)
  2. x = 2k÷ 4
  3. 2k − y = 3kx = 2k into (2)
  4. −y = k, y = −k−2k both sides
  5. (1): 6k − k = 5k
  6. (2): 2k − (−k) = 3k
  7. x = 2k, y = −k
Worked example

Q7: the lines meet at x = 6

  • Use x = 2k.
  • Check with k = 3.
  1. x = 2k, y = −k
  2. 2k = 6x = 6
  3. k = 3÷ 2
  4. y = −3y = −k
  5. (1): 18 − 3 = 15 = 5 × 3
  6. (2): 6 − (−3) = 9 = 3 × 3
  7. k = 3, point (6, −3)
Summary

The method that solved all seven

  • Label, match, eliminate, check both, and answer in words.
StepHow
Labelnumber them (1) and (2)
Matchmultiply every term
Eliminateadd if opposite, else subtract
Substitutewhen one gives y already
Back-subfind the other letter
Checkin both original equations
Contextanswer in words, with units

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