Term 1 · Week 7 · Inequalities in two variables and the region of solutions
NameDate
For the teacher. This is for a learner listing a few solutions and stopping, or shading by the direction of the sign. It is not a shortened copy of
worksheet.html — it is a different route to the same competency, starting further
back. Task 4 is the first region on worksheet.html. Show each worked example first and let the learner copy the
method, never the answer. If task 1 will not come: If task 1 will not come, stay on one variable: x < 5. Ask for solutions until the learner runs out of patience rather than numbers.
1Does this point work?
— targets deciding by the look of the sign rather than by substituting
Say: “Put the numbers in. Read what it says.”
Worked example. Is (1, 2) a solution of y > x? Put them in: 2 > 1. True, so yes.
Is (3, 1) a solution of y > x? Substitute: ______ > ______ → ______
Is (0, 5) a solution of y < 2x + 1? Substitute: ______ → ______
Is (2, 4) a solution of y ≤ 2x? Substitute: ______ → ______
2Test a point, pick a side
— targets "greater than means shade above"
Say: “Choose a point NOT on the line. Test it. Shade its side.”
Worked example. For y > x + 1, test (0, 0): 0 > 1 is false, so shade the side (0, 0) is not on.
y < x + 2, test (0, 0): ______ → true / false → shade the side (0,0) IS / IS NOT on
y > −x, test (0, 0): 0 > 0 → ______. Why can (0,0) not be used here? ______________
3Solid or broken?
— targets the boundary treated as decoration
Say: “Does a point exactly on the line count?”
Worked example. y ≤ x includes the line, so draw it solid. y < x excludes it, so broken.
y > 2x → ______________
y ≥ x − 3 → ______________
x + y ≤ 10 → ______________
4How many answers?
— targets a list of points offered as the answer
Say: “Give me one more. And one more.”
Worked example. x + y ≤ 4 has (0,0), (1,1), (2,2), (0,4), (1.5, 2)… and it does not stop.
Three solutions of x + y ≤ 6: ______ ______ ______
Now one more that nobody would guess: ______
Write in words why a list can never be the whole answer: ______________________________
Answer key — not printed on the learner's copy
1 > 3 false, no · 5 < 1 false, no · 4 ≤ 4 true, yes
0 < 2 true → shade the side (0,0) IS on · 0 > 0 is false, and the point sits ON the boundary, so it cannot decide a side
broken · solid · solid
any three that total 6 or less · any valid point, decimals allowed · because the solutions fill a region and between any two there is another