Offline Classroom Activities — Mathematics 10
Term 1 · Week 7 · Regions of Solutions · CG Quarter 1, competencies 9–10
Five activities that need no electricity, no devices and no internet. Three are individual and two are group. Every one of them can be run with chalk, paper and a ruler.
| Learning Area | Mathematics |
| Grade Level | Grade 10 |
| Competencies | Solve quadratic inequalities in two variables; determine the region of solutions of a linear or quadratic inequality in two variables |
| Class size assumed | 45–55 learners at fixed desks |
Why no graphing tools
A graphing calculator or a phone app will shade the region for you. That is the entire decision this week is about, so a tool that makes it is a tool that removes the lesson. The competencies are worded as determine and solve, not display.
The same is not true of the parabola itself. A learner who cannot sketch y = x² − 4 has a Grade 9 gap, and making them fight the curve while also learning the region teaches neither. Pre-draw the boundary for anyone who needs it — the competency is what happens next.
Squared paper
Days 2–4 want squares. If there is none:
- A ruled exercise book turned sideways gives a grid from the lines plus a ruled margin
- One sheet of squared paper photocopied onto the back of last term's handouts
- Or draw the axes and the boundary on the board and have learners work orally, which is what Activities I-1 and G-1 are designed for
INDIVIDUAL ACTIVITIES
Three options. Pick by what the class needs, not by what comes next in the list.
I-1 · In, Out, or On
Time 20 minutes · Grouping Individual · Use Day 1, replaces the digital game
No graphing at all. Learners are given an inequality and a point and must say whether the point is in the region, out of it, or on the boundary. That is the whole competency stripped of drawing, and a learner who cannot do this cannot honestly shade anything.
Instructions
Write one inequality on the board and leave it there:
y > x² − 4
Read out points one at a time. Learners write IN, OUT, or ON — and, beside it, the substitution that decided it.
| # | Point | Substitution | Answer |
|---|---|---|---|
| 1 | (0, 0) | 0 > −4 | IN |
| 2 | (0, −10) | −10 > −4 | OUT |
| 3 | (2, 0) | 0 > 0 | ON |
| 4 | (3, 6) | 6 > 5 | IN |
| 5 | (3, 4) | 4 > 5 | OUT |
| 6 | (1, −3) | −3 > −3 | ON |
| 7 | (−2, 5) | 5 > 0 | IN |
| 8 | (−4, 8) | 8 > 12 | OUT |
| 9 | (−3, 5) | 5 > 5 | ON |
| 10 | (5, 21) | 21 > 21 | ON |
The question to ask while circulating
"Which of the three answers can you give without substituting?"
None of them. That is the point. Items 3, 6, 9 and 10 sit exactly on the curve, and no amount of looking at the inequality will tell you so.
Why four of the ten are ON
Because learners left to themselves will decide that ON is a rare curiosity and stop checking for it. Forty per cent forces the habit. Items 9 and 10 are the two that catch people — (−3, 5) because the negative x is squared, and (5, 21) because the numbers are large enough that learners estimate instead of substituting.
Differentiation
- Struggling — items 1, 2, 4, 5 only, and say the substitution aloud before writing
- Ahead — after item 10, ask for a point of their own that is ON, and a second one in the third quadrant
Assessment — 10 points
One point per item. Award the point only if the substitution is written; an unsupported IN/OUT/ON is worth nothing here, because the guess rate is one in three.
I-2 · The Same Region, Written Four Ways
Time 20 minutes · Grouping Individual · Use Day 2 or Day 3
Four inequalities. Three of them describe the same region and one does not. Learners find the odd one out — by testing, not by looking.
The set
A. y > x² − 4
B. x² − y < 4
C. x² − 4 − y < 0
D. y < x² − 4
Instructions
- Test (0, 0) in all four. Write each substitution.
- Test (0, −10) in all four. Write each substitution.
- Circle the odd one out.
- In one sentence, say what makes it different.
Answer key
| Inequality | At (0, 0) | At (0, −10) | ||
|---|---|---|---|---|
| A | y > x² − 4 | 0 > −4 · TRUE | −10 > −4 · FALSE | same region |
| B | x² − y < 4 | 0 < 4 · TRUE | 10 < 4 · FALSE | same region |
| C | x² − 4 − y < 0 | −4 < 0 · TRUE | 6 < 0 · FALSE | same region |
| D | y < x² − 4 | 0 < −4 · FALSE | −10 < −4 · TRUE | the odd one out |
D is the odd one. It is the complement — the other side of the same boundary. All four share a boundary; only D shades the opposite side.
Why this beats a worked example
A learner who has memorised "> means above" will mark A as the odd one out, because it is the only one of the four whose sign is >. That mistake is worth more than a correct answer arrived at by luck, and it is visible in thirty seconds of marking.
Assessment — 8 points
| 4 | one per inequality correctly tested at both points, with substitutions shown |
| 2 | odd one out correctly identified |
| 2 | the sentence names the side, not the sign |
I-3 · Where Can You Stand?
Time 25 minutes · Grouping Individual · Use Day 3 or Day 4, extension
Six inequalities. For each, the learner must choose a legal test point and say why the obvious one will not do. Three of the six pass through the origin.
The set
| # | Inequality | Origin usable? | A point that works |
|---|---|---|---|
| 1 | y > x² − 4 | yes | (0, 0) |
| 2 | y > −x | no — the origin is on it | (1, 1) |
| 3 | y ≤ 2x + 1 | yes | (0, 0) |
| 4 | y < x² | no — the origin is on it | (0, 1) → false, so shade the other side |
| 5 | x + y ≥ 6 | yes | (0, 0) → false, so shade the far side |
| 6 | y ≥ 3x | no — the origin is on it | (0, 1) |
Instructions
For each inequality:
- Write the boundary.
- Substitute (0, 0) and say whether the origin is on the boundary.
- If it is, choose another point and mark it clearly.
- State which side gets shaded.
Why the origin fails on three of them
Every boundary of the form y = mx passes through (0, 0), because m × 0 = 0. So does y = x². A learner who substitutes anyway gets a false statement — not an error message — and shades the wrong half with clean-looking working. There is nothing in their page to catch it.
This is the only failure mode this week that is invisible in the learner's own work. Every other mistake leaves a trace. This one does not, which is why it gets its own activity.
The rule worth writing down
If substituting the origin gives you 0 on both sides, the origin is on the boundary. Pick another point.
Assessment
| 6 | one per inequality: correct boundary and a legal test point |
| 3 | the three origin-on-boundary cases identified as such |
| 3 | correct side named for items 4, 5 and 6 — the three where the test comes out false |
12 points. Items 4, 5 and 6 are separated out because a false test result requires the learner to shade the opposite side, and that reversal is where the marks go.
GROUP / TEAM ACTIVITIES
Two options. Both work at fixed desks with no room to move.
G-1 · The Test Point Relay
Time 35 minutes · Groups of 5 · Use Day 3, consolidation
One inequality per group. Each member does exactly one step and passes the paper on. Nobody may do two steps, and nobody may correct the step before theirs without saying so out loud.
Preparation
Eight strips of paper, one inequality each:
| Group | Inequality | Boundary | Broken/solid | Origin usable? | Region |
|---|---|---|---|---|---|
| 1 | y > x² − 4 | y = x² − 4 | broken | yes | with the origin |
| 2 | y ≤ x² − 1 | y = x² − 1 | solid | yes → false | away from the origin |
| 3 | y < −x² + 9 | y = −x² + 9 | broken | yes → true | with the origin |
| 4 | y ≥ 2x − 6 | y = 2x − 6 | solid | yes → true | with the origin |
| 5 | x + y < 4 | y = −x + 4 | broken | yes → true | with the origin |
| 6 | y > x² | y = x² | broken | no | use (0, 1) → true |
| 7 | y ≤ −2x | y = −2x | solid | no | use (0, −1) → true |
| 8 | x² − y ≤ 0 | y = x² | solid | no | use (0, 1) → true |
Roles
| Member | Does only this |
|---|---|
| 1 | Writes the boundary equation and nothing else |
| 2 | Says broken or solid and writes why in four words |
| 3 | Chooses a test point and states whether it is legal |
| 4 | Performs the substitution and writes TRUE or FALSE |
| 5 | Shades and labels the region |
Pass the paper right. When it reaches member 5, it goes back to member 1, who checks the whole chain and signs it.
Answer key
| Group | Boundary | Line | Test | Result | Shade |
|---|---|---|---|---|---|
| 1 | y = x² − 4 | broken | (0, 0) | 0 > −4 T | with origin |
| 2 | y = x² − 1 | solid | (0, 0) | 0 ≤ −1 F | opposite side |
| 3 | y = −x² + 9 | broken | (0, 0) | 0 < 9 T | with origin |
| 4 | y = 2x − 6 | solid | (0, 0) | 0 ≥ −6 T | with origin |
| 5 | y = −x + 4 | broken | (0, 0) | 0 < 4 T | with origin |
| 6 | y = x² | broken | (0, 1) | 1 > 0 T | with (0, 1) |
| 7 | y = −2x | solid | (0, −1) | −1 ≤ 0 T | with (0, −1) |
| 8 | y = x² | solid | (0, 1) | 0 − 1 ≤ 0 T | with (0, 1) |
Groups 6, 7 and 8 cannot use the origin, and are not told so. Member 3's job is to notice. If member 3 hands (0, 0) to member 4, member 4 must be the one to catch it — and if neither does, the group discovers it when member 1 checks the chain. That failure is the activity working, not the activity going wrong.
Groups 6 and 8 are the same region written two ways — y > x² and x² − y ≤ 0 differ only in that one includes the boundary. Put those two papers side by side at the end.
Managing 45+ learners at fixed desks
Nobody moves. The paper moves along the row and back. With eight strips and groups of five, a class of 50 runs two full sets of eight with two learners floating as checkers — give the checkers the answer key and the job of saying "which step is wrong?" without saying which answer is wrong.
Assessment — group rubric, 12 points
| Criterion | 3 | 2 | 1 | 0 |
|---|---|---|---|---|
| Boundary | Correct, and correctly broken or solid | Correct, wrong line style | Rearranged wrongly | Absent |
| Test point legality | Legal point chosen, and illegality of the origin spotted where it applies | Legal point, no comment | Illegal point used | None chosen |
| Substitution | Written out and correct | Correct, not shown | Shown, arithmetic wrong | Absent |
| The chain held | Member 1's check caught any error, or there were none | Error caught late | Error passed unnoticed | No check made |
G-2 · Three Ways, One Region
Time 40 minutes · Teams of 5–6 · Use Day 5, application
Sixteen cards. Four regions, each represented four ways — as an inequality with y isolated, as the same inequality rearranged, as a described boundary with a shading instruction, and as a named test point with its result.
Preparation — 15 minutes the night before
Cut sixteen cards per team. Four sets of four:
Region P
| Card | Text |
|---|---|
| P1 | y > x² − 4 |
| P2 | x² − y < 4 |
| P3 | Parabola, vertex (0, −4), broken; shade the side with the origin |
| P4 | Test (0, 0): 0 > −4 is TRUE |
Region Q
| Card | Text |
|---|---|
| Q1 | y ≤ x² − 4 |
| Q2 | x² − y ≥ 4 |
| Q3 | Parabola, vertex (0, −4), solid; shade the side away from the origin |
| Q4 | Test (0, 0): 0 ≤ −4 is FALSE |
Region R
| Card | Text |
|---|---|
| R1 | y < 2x + 1 |
| R2 | 2x − y > −1 |
| R3 | Line through (0, 1) and (1, 3), broken; shade the side with the origin |
| R4 | Test (0, 0): 0 < 1 is TRUE |
Region S
| Card | Text |
|---|---|
| S1 | y ≥ 2x + 1 |
| S2 | 2x − y ≤ −1 |
| S3 | Line through (0, 1) and (1, 3), solid; shade the side away from the origin |
| S4 | Test (0, 0): 0 ≥ 1 is FALSE |
Instructions
Round 1 — 12 min. Match. Cards are shuffled and dealt. Teams sort all sixteen into four sets of four. No graphing is allowed in this round — matching must be done by testing.
Round 2 — 10 min. Draw. For each of the four regions, one sketch on squared paper: boundary with the correct line style, test point circled, region shaded.
Round 3 — 10 min. Make a fifth. Each team invents a fifth region and writes all four of its cards. The rearranged form must be genuinely rearranged, not the same expression with the terms in a different order.
Round 4 — 8 min. Swap and solve. Teams exchange their fifth set, shuffled, and match it.
Answer key — Round 1
| Region | Cards | The pair most often confused |
|---|---|---|
| P | P1 P2 P3 P4 | P2 with Q2 — both begin x² − y |
| Q | Q1 Q2 Q3 Q4 | Q1 with P1 — same curve, opposite side |
| R | R1 R2 R3 R4 | R2 with S2 — both begin 2x − y |
| S | S1 S2 S3 S4 | S3 with R3 — same line, different style and side |
P and Q share a boundary. R and S share a boundary. The four cards that actually separate them are P4, Q4, R4 and S4 — the test results. A team that starts from those finishes fastest, and noticing that is worth more than finishing.
Why Round 4 is worth the eight minutes
Writing a set is easy to fake — teams produce four cards that look right without checking that the rearranged form really is equivalent. Handing it to another team to match makes that impossible to hide. The most common flaw in an invented set is a rearrangement that flips the inequality without flipping the sign, and the receiving team finds it in under a minute.
Assessment — group rubric, 12 points
| Criterion | 3 | 2 | 1 | 0 |
|---|---|---|---|---|
| Round 1 matching | All 16 placed, by testing | 12–15 placed | 8–11 placed | Fewer than 8 |
| Round 2 sketches | Four correct, line styles right, test points circled | Three correct | Two correct | One or none |
| Round 3 invention | Fifth set valid, rearrangement genuine | Valid, rearrangement trivial | One card wrong | Not attempted |
| Round 4 exchange | Matched another team's set and found any flaw | Matched it | Partly matched | Not attempted |
Which activity to reach for
| If the class… | Run |
|---|---|
| Has no devices and it is Day 1 | I-1 — it replaces the game exactly |
| Shades correctly but writes no substitution | I-1, marked strictly. No substitution, no point |
| Says "> means above" | I-2 — the odd one out is the one that punishes it |
| Uses the origin every time without checking | I-3 — three of its six boundaries pass through it |
| Needs the routine to become automatic | G-1 — one step each, nobody does two |
| Is secure and needs the inverse task | G-2 Rounds 3 and 4 |
Part of the E-turo MATATAG artifact set. Companion files:
syllabus.md · lesson-plan.md ·
lesson-plan-ilaw.md · assessment.md ·
slides.html · game.html ·
worksheet.html