LESSON PLAN

Linear Relations

Apothem Team

Grade 8 · Algebra & Patterning

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Two ride-share offers on the board: "GoCar: \$4 pickup + \$1.50/km. ZipRide: \$7 pickup + \$1.00/km." Ask only: "Which do YOU take, and what does your answer depend on?" The room discovers it depends on trip length — short trips favour GoCar, long ones ZipRide.

Grade 8 linear relations sharpen last year's $y = mx + b$ into a comparison tool: graphing lines together, reading intersections, and interpreting slope as a rate with units attached.

Explore

Duelling-lines lab: pairs graph both ride offers on one grid (cost vs km), locate the crossing (6 km, \$13), and write the verdict as an interval: under 6 km GoCar, over 6 km ZipRide, at 6 km flip a coin.

Then slope surgery: compute rise-over-run between marked points on four lines, including a NEGATIVE slope (a draining tank: $-2$ L/min) and a ZERO slope (a flat monthly fee — cost unchanging with usage). Each slope gets its units said aloud: "dollars per kilometre," "litres per minute." Slope without units is a number; with units it's a STORY.

Formalize

Formalize slope as measured rate and the full linear anatomy:

$m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1} \qquad y = mx + b$

The four slope personalities on one poster: positive (climbing), negative (falling), zero (flat — horizontal line), undefined (vertical — a run of zero breaks the fraction; "cost per km" makes no sense for a line where km never changes). Intersections of two lines mean: the x where both stories give the same y.

Practice

Practice: graph two line-pairs and interpret each intersection; compute four slopes from point-pairs (one negative, one zero); one story-to-graph-to-decision problem; one slope-units matching set.

Exit ticket: line through $(2, 11)$ and $(5, 2)$ — slope, and one sentence on what such a slope would mean if $x$ is hours and $y$ is battery percent. ( $m = -3$ ; losing 3% per hour.)

Exit ticket

Practice: graph two line-pairs and interpret each intersection; compute four slopes from point-pairs (one negative, one zero); one story-to-graph-to-decision problem; one slope-units matching set.

Exit ticket: line through $(2, 11)$ and $(5, 2)$ — slope, and one sentence on what such a slope would mean if $x$ is hours and $y$ is battery percent. ( $m = -3$ ; losing 3% per hour.)

TIP Force the units-sentence on every slope for two weeks ("3 dollars per shirt," "−2 litres per minute"). Slope stops being a fraction ritual the day it starts talking.

WORKED EXAMPLES

Example 1 — The gym-membership duel: intersection as decision point

FitZone: \$60 joining fee + \$25/month. PulseGym: no fee, \$40/month.

Step 1: Equations: $F = 25m + 60$ ; $P = 40m$ .

Step 2: Intersect: $40m = 25m + 60 \to 15m = 60 \to m = 4$ months (both $160).

Step 3: Verdict by interval: under 4 months, PulseGym (no fee to amortize); beyond 4, FitZone's cheaper rate wins; at exactly 4, tie.

Step 4: Graph-check: PulseGym's line starts lower (intercept 0 vs 60) but climbs steeper (40 vs 25) — steeper always catches and passes a shallower line eventually. Intercept is the head start; slope decides the long run. That single sentence is most of linear-relations literacy.

Example 2 — Slope from data: the cooling coffee

The measurements: coffee at $(0 \text{ min}, 90°)$ and $(8 \text{ min}, 66°)$ .

Step 1: Slope: $m = \frac{66 - 90}{8 - 0} = \frac{-24}{8} = -3$ degrees per minute.

Step 2: Equation (intercept given free at $t=0$ ): $T = -3t + 90$ .

Step 3: Use it: when does it hit a drinkable 60°? $-3t + 90 = 60 \to t = 10$ min.

Step 4: The model-boundary conversation, Grade 8 edition: the line predicts $T = 0°$ at 30 min and $-30°$ at 40 — but coffee stops cooling at room temperature (~21°). Linear models describe the stretch where the data lives; real cooling bends (an exponential preview, one grade early and worth the wink). The math is right; the model expires. Both facts belong in the answer.

Example 3 — From table to verdict: is this relation even linear?

The table: $x: 1, 2, 3, 4 \to y: 5, 8, 13, 20$ .

Step 1: Test constant differences: $+3, +5, +7$ — NOT constant. The relation is not linear; no line fits it, and forcing $y = mx + b$ would be fiction.

Step 2: Contrast table: $x: 1, 2, 3, 4 \to y: 2, 7, 12, 17$ — differences $+5, +5, +5$ ✓ linear: $m = 5$ , back up one step from $(1,2)$ : $b = -3$ , so $y = 5x - 3$ .

Step 3: Check the claimed equation on a far row: $x = 4$ : $20 - 3 = 17$ ✓.

Step 4: The refusal skill, formally honoured: "not linear" is a complete, correct answer — and the growing differences ( $+3, +5, +7$ ) whisper what IS going on (a square: $y = x^2 + 4$ , for the curious). Diagnosing the pattern's TYPE before fitting a formula is the mathematician's move; everything else is curve-fitting by hope.

MATERIALS

•Grid paper

•Duelling-offers cards

•Slope-personality poster

•Practice set (PDF)

WATCH FOR

!Rise and run swapped. Anchor: slope is how much y CHANGES per single step of x — the units-sentence enforces the order.

!Zero and undefined slopes confused. Flat road (zero) vs wall (undefined) — walkable vs not.

!Intersection read as "the answer" without interpretation. It's the input where the options TIE; the decision lives on either side.