Mathematics • Year 8 • Unit 2 • Lesson 20
Unit 2 Capstone — Mixed Challenge
Pull the WHOLE unit together: gradients from points, the equation y = mx + c, intercepts and sketching, choice of method for a system, and a real-world model. Six mixed problems, one "find the mistake" on rearranging to gradient–intercept form, and one open-ended challenge that asks you to design a real-life scenario from a given system.
1. Mixed problems — use the whole Unit 2 toolkit
Each question uses a different concept from Unit 2. Show all working. 3 marks each
1.1 Find the equation of the line with gradient 3 that passes through (1, 5). Give your answer in the form y = mx + c.
1.2 Find the x-intercept and y-intercept of 2x + 3y = 12, then sketch the line on a small set of axes (use the back of the page if needed). State your sketch method.
1.3 Solve y = 3x and y = x + 4 simultaneously. State the solution as an ordered pair (x, y) and verify in both equations.
1.4 Rearrange 2x + 5y = 10 into the form y = mx + c, then state its gradient and y-intercept.
1.5 Solve 2x + 3y = 13 and 5x − 3y = 1 by elimination. State your choice (add or subtract) and why.
1.6 Real-world model. The cost C (dollars) of hiring a paddleboard is $20 plus $8 per hour, h. (a) Write the linear model. (b) Find the cost for a 3-hour hire. (c) If a customer paid $52, how long did they hire for?
2. Find the mistake
A student is asked to rearrange 3x + 4y = 20 into y = mx + c form and state the gradient. Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then redo correctly. 3 marks
Student's working:
Line 1: 3x + 4y = 20
Line 2: 4y = 20 − 3x (subtract 3x from both sides)
Line 3: y = 5 − 3x (divide both sides by 4)
Line 4: y = mx + c with m = −3 and c = 5.
Line 5: Gradient = −3.
(a) Which line contains the mistake?
(b) Explain in one or two sentences exactly what went wrong on that line.
(c) Re-do the working with the correction. State the correct gradient and y-intercept.
Stuck? When you divide BOTH sides by 4, every term gets divided by 4 — not just one of them. 20/4 = 5, but −3x / 4 = −(3/4)x, not −3x.3. Open-ended challenge — invent the story
This question has more than one valid answer. 4 marks
3.1 Below is a system of two simultaneous equations:
x + y = 30
2x + 5y = 90
(a) Solve the system by elimination, showing all working.
(b) INVENT a real-world story where this system would naturally arise. Define what x and y represent (including units), and explain in one or two sentences what each equation means in your story.
(c) Translate your final solution back into your story: in plain English, what do the values of x and y actually mean?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Line with m = 3 through (1, 5)
Use y = mx + c with m = 3: y = 3x + c. Sub (1, 5): 5 = 3(1) + c → c = 2. Equation: y = 3x + 2. Check: 3(1) + 2 = 5 ✓.
1.2 — Intercepts of 2x + 3y = 12
x-intercept (y = 0): 2x = 12 → x = 6 → (6, 0). y-intercept (x = 0): 3y = 12 → y = 4 → (0, 4). Method: intercept method — plot both intercepts and draw the straight line through them.
1.3 — y = 3x and y = x + 4
Set equal: 3x = x + 4 → 2x = 4 → x = 2. Then y = 3(2) = 6. (2, 6). Check: y = x + 4 → 6 = 2 + 4 ✓.
1.4 — Rearrange 2x + 5y = 10
Subtract 2x: 5y = −2x + 10. Divide every term by 5: y = −(2/5)x + 2. Gradient m = −2/5; y-intercept c = 2.
1.5 — 2x + 3y = 13 and 5x − 3y = 1
y-terms are +3y and −3y — opposite signs → ADD. Add: 7x = 14 → x = 2. Back-sub into Eq 1: 2(2) + 3y = 13 → 3y = 9 → y = 3. (2, 3). Check Eq 2: 5(2) − 3(3) = 10 − 9 = 1 ✓.
1.6 — Paddleboard hire
(a) C = 8h + 20.
(b) At h = 3: C = 8(3) + 20 = 24 + 20 = $44.
(c) 52 = 8h + 20 → 8h = 32 → h = 4 hours.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) When dividing both sides by 4, every term must be divided by 4 — not just the 20. 20/4 = 5 (correct), but −3x ÷ 4 = −(3/4)x, NOT −3x. The student divided the constant but forgot to divide the x-term.
(c) Correct: 4y = 20 − 3x → y = (20 − 3x) / 4 = 5 − (3/4)x = y = −(3/4)x + 5. Gradient = −3/4; y-intercept = 5.
3 — Invent the story
(a) Solve. Multiply Eq 1 by 2: 2x + 2y = 60. Subtract from Eq 2: 3y = 30 → y = 10. Back-sub into Eq 1: x + 10 = 30 → x = 20. Solution: (20, 10). Check Eq 2: 2(20) + 5(10) = 40 + 50 = 90 ✓.
(b) Sample student story. "A jar contains $2 coins and $5 notes, 30 items in total. The total value is $90." Let x = number of $2 coins and y = number of $5 notes. Eq 1 (count): x + y = 30. Eq 2 (value): 2x + 5y = 90.
(c) Solution (20, 10) means there are 20 × $2 coins and 10 × $5 notes. Check: 20 + 10 = 30 items ✓; value = 20(2) + 10(5) = $40 + $50 = $90 ✓.
Other valid stories include: tickets at $2 and $5; muffins at $2 and cakes at $5; 30 students split into "x walkers at 2 km" and "y cyclists at 5 km" totalling 90 km, etc.
Marking: 1 mark for the correct system solution (20, 10) with working; 1 mark for clearly defining x and y with units in the story; 1 mark for a coherent story whose two equations match the system; 1 mark for translating the solution back into a sensible context sentence.