Mathematics • Year 8 • Unit 2 • Lesson 6
Recognising Linear Relationships
Build fluency with the first-differences test for linearity. One worked example, one guided example with blanks, then eight independent problems from quick recall to ordering and comparing.
1. I do — fully worked example
A linear relationship has a constant rate of change. The fastest way to test a table is the first-differences test.
Problem. Is this table linear? x: 1, 2, 3, 4, 5 y: 5, 8, 11, 14, 17.
Step 1 — Find the first differences between consecutive y-values.
8 − 5 = 3, 11 − 8 = 3, 14 − 11 = 3, 17 − 14 = 3
Reason: each gap shows how much y grows when x grows by 1.
Step 2 — Check that EVERY difference is the same.
Differences: 3, 3, 3, 3 — all equal.
Reason: if even one differs, the rate of change isn't constant, so the relationship is non-linear.
Step 3 — Read the gradient from the common difference.
Common difference = 3, so m = 3.
Reason: when x grows by 1 each step, the first difference IS the gradient.
Answer: Linear, m = 3.
2. We do — fill in the missing steps
Same shape as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. Is this table linear? x: 0, 1, 2, 3, 4 y: 2, 6, 10, 14, 18.
Step 1 — Find each first difference.
6 − 2 = ______, 10 − 6 = ______, 14 − 10 = ______, 18 − 14 = ______
Step 2 — Are they all equal? ____________________
Step 3 — Common difference = ______, so gradient m = ______.
Conclusion: The table is ______________ (linear / non-linear) with gradient m = ______.
3. You do — independent practice
Show your working under each problem. First four are foundation, next two are standard, last two are extension.
Foundation — quick recognise
3.1 The first differences of a table are 4, 4, 4, 4. Is it linear? If yes, what is m? 1 mark
3.2 The first differences of a table are 5, 5, 6, 5. Is it linear? Explain in one sentence. 1 mark
3.3 For y-values 3, 7, 11, 15, 19 (with x going up by 1 each step) — find the first differences and decide if it is linear. 1 mark
3.4 Is y = 2x + 7 linear? In one sentence, explain how you can tell from the equation alone. 1 mark
Standard — complete the test
3.5 Test this table for linearity. x: 1, 2, 3, 4, 5 y: 1, 4, 9, 16, 25. Show all first differences and state your conclusion. 2 marks
3.6 A line passes through (1, 4), (2, 7), (3, 10), (4, 13). Find the common first difference and state the gradient. 2 marks
Extension — apply the rule
3.7 A table has x: 0, 1, 2, 3, 4 and y: 10, 8, 6, 4, 2. Find the first differences and state whether the gradient is positive or negative, and its value. 2 marks
3.8 A plumber charges $50 call-out plus $80 per hour. Write the cost C for h hours, and explain why C vs h is linear. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do
Differences: 4, 4, 4, 4. All equal: yes. Common difference = 4, so gradient m = 4. The table is linear with m = 4.
3.1 — Differences 4, 4, 4, 4
Yes, linear. All differences equal, so m = 4.
3.2 — Differences 5, 5, 6, 5
No — the 6 breaks the pattern. A single mismatched difference means the rate of change isn't constant, so it is not linear.
3.3 — y: 3, 7, 11, 15, 19
Differences: 4, 4, 4, 4. All equal, so linear with m = 4.
3.4 — y = 2x + 7
Yes, linear. It is in the form y = mx + c (here m = 2, c = 7), and x is only to the power of 1 — no x², no x³.
3.5 — y: 1, 4, 9, 16, 25
Differences: 4 − 1 = 3, 9 − 4 = 5, 16 − 9 = 7, 25 − 16 = 9. Differences are 3, 5, 7, 9 — not all equal, so the relationship is not linear. (It is y = x², a parabola.)
3.6 — Points (1,4), (2,7), (3,10), (4,13)
Differences: 7 − 4 = 3, 10 − 7 = 3, 13 − 10 = 3. All equal, so linear with gradient m = 3.
3.7 — y: 10, 8, 6, 4, 2
Differences: 8 − 10 = −2, 6 − 8 = −2, 4 − 6 = −2, 2 − 4 = −2. All equal to −2, so linear with negative gradient m = −2.
3.8 — Plumber's fee
C = 80h + 50. This is linear because each extra hour adds exactly $80 (a constant rate of change). The equation is in the form y = mx + c with m = 80 and c = 50, with h to the power of 1.