y = 2x + 1 is a line; y < 2x + 1 is every point below it. Draw the boundary (dashed if points on it are not included, solid if they are), then test a point to see which side to shade. Vertical lines are x = a; horizontal lines are y = a.
A delivery zone, a speed limit, a budget: often the answer is not one point but a whole region of points that work.
You can graph a line from its rule, such as y=2x+1. Now: what about y<2x+1? Every point whose y is less than 2x+1 makes it true, and those points fill a whole region of the graph.
You also meet lines with only one letter: x=3 (vertical) and y=−2 (horizontal).
Predict: Read the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
Try: Try it on the board and compare the result with your prediction.
Investigate: Work out why it happens, one step at a time.
Modify: Change the problem on purpose and see what happens.
Practise: A few on your own, to check you have it.
Make: Make your own problem using what you have learned. Expect a few tries; that is normal.
Make your prediction first. Help opens once you've locked it in.
PredictRead the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
In or out?
Put x=1 and y=4 into y<2x+1. Is it true?
TryTry it on the board and compare the result with your prediction.
Into the region
Try a point on the dashed line too. Does it make y<2x+1 true?
Try it yourselfYour edits are saved in this browser.
Drag P into the shaded region y<2x+1. Check its coordinates in the inequality. Was your prediction right about (1,4)?
Use the board. This ticks when you get there.
InvestigateWork out why it happens, one step at a time.
Boundary, then test
Here is how to graph an inequality.
Worked exampleBefore each step, say what you think comes next.
y<2x+1The boundary is the line y=2x+1.
dashed< does not include the line's points (for ≤ or ≥, draw it solid).
(0,0):0<2×0+1Test a point not on the line. 0<1 is true.
shade the side with (0,0)True: shade the test point's side. (False: shade the other side.)
x=3,y=−2One letter: x=3 is vertical (every point has x = 3); y=−2 is horizontal.
Solid or dashed for y≥x−2?
Read Ana's working. Which line does not match the sign?
Need a hint?
The orange region is x ≥ 3: on or right of x = 3.
Another hint
The other is y < 2: below the dashed line y = 2.
Try it yourselfYour edits are saved in this browser.
Ana graphs y>x+1. Click the line where her working goes wrong.
Two regions: x≥3 and y<2. Drag P to a point that is in both.
Use the board. This ticks when you get there.
ModifyChange the problem on purpose and see what happens.
One change at a time
Each task changes one thing.
Need a hint?
x = −2 is the vertical line; y = 3 the horizontal one.
A shaded region with a solid boundary.
Need a hint?
On the line: y = 4 − x. The boundary is solid, so those points count.
The region y<3.
Try it yourselfYour edits are saved in this browser.
Drag P to where x=−2 meets y=3.
Use the board. This ticks when you get there.
The region is y≤−x+4. Drag P to a point on its boundary line.
Use the board. This ticks when you get there.
PractiseA few on your own, to check you have it.
Regions
MakeMake your own problem using what you have learned. Expect a few tries; that is normal.
Fit the region
Find two different answers. What do they have in common?
Try it yourselfYour edits are saved in this browser.
Need a hint?
(0, 0) in y ≤ mx + c means 0 ≤ c.
Another hint
(0, 3) outside means c is less than 3. Then check B (2, 1): 1 ≤ 2m + c.
Set m and c so the region y≤mx+c contains A(0,0) and B(2,1) but not C(0,3).
Use the board. This ticks when you get there.
For teachers
Students graph vertical and horizontal lines and linear inequalities as regions, choosing a dashed or solid boundary and testing a point to decide which side to shade (AC9M8A02, node 1). About 45 minutes.
I can graph an inequality such as y < 2x + 1 as a region, with the right boundary (dashed or solid).
I can test a point to decide whether it is in a region.
Before class
Do the Run board and the Make task yourself first. Students need Lines and their rules (gradient and intercept).
The one idea: an inequality is every point that makes it true; the line is its boundary, dashed when the line's points are left out (< and >) and solid when they are in (≤ and ≥). The common mistake is the wrong boundary style; it is planted in Ana's working.
Running the lesson
Predict: take a show of hands. Students who chose yes compared 4 with 1, not with 2 × 1 + 1.
Run: ask students to read out a point in the region and check it aloud.
Investigate: model the boundary-then-test routine with (0, 0); ask when (0, 0) would be a bad test point (when it is on the line).
Modify: before the vertical line task, ask what every point on x = −2 has in common.
Make: collect students' m and c and test C in each.
When students get stuck
A student draws the boundary solid for < (Ana's mistake). Ask: Does a point on the line make y < 2x + 1 true? Fix: No (it makes the two sides equal), so the line is not included: dashed.
A student shades the wrong side. Ask: Which point did you test? Was the result true or false? Fix: True: shade the test point's side. False: the other side.
A student draws x = 3 horizontally. Ask: Name three points with x = 3. Fix: (3, 0), (3, 1), (3, −2): they line up vertically.
Stretch and support
Stretch: Shade the region where y ≥ x and y < 4 and x > 0 (a triangle). What are its corners?
Support: Use the Run board: drag P and read whether the coordinates make the inequality true.
Where it leads
Next: Solving with graphs, then (in Algebra) Inequalities, which solves one-variable inequalities and checks them graphically.
Where it leads: linear programming and simultaneous inequalities (Years 10 and up).
What each step is for
in or out?: Committing to whether (1, 4) satisfies y < 2x + 1 shows whether they substitute both coordinates, before the board shows the region.
into the region: Dragging P into the shaded region and reading its coordinates shows that every point there makes the inequality true, and the boundary is not included for <.
boundary, then test: Draw the boundary (dashed for < and >, solid for ≤ and ≥), test a point not on it, and shade the side that makes it true. Ana's mistake draws a strict inequality's boundary solid.
one change at a time: One change at a time: vertical and horizontal lines, reading an inequality from its region, a point on a solid boundary, a point on a dashed one, then a region with no x in it.
regions: Testing points, choosing boundaries and reading lines, mixed; the solid-dashed slip planted again; the review keeps gradient and intercept fresh.
fit the region: Choosing m and c so a region contains some points and not others reverses the lesson: from points to the inequality.
Now you can
Graph lines such as x = 3 and y = −2.
Graph an inequality such as y < 2x + 1 as a region, with the right boundary (dashed or solid).