DSE Maths | Inequalities and Linear Programming Lesson-Prep Guide (Tutor Edition) | Sign Flips, Feasible Regions and Vertex Testing

2026-11-02

Tutor lesson-prep guides · Secondary Maths

Students solve 2x + 7 <= 19 correctly and still divide by a negative number without flipping the sign. This lesson makes the flip a rule they can explain, then turns it into a feasible region they can shade and test.

Objectives — Students can: (1) solve a linear inequality and state when the inequality sign reverses, using a test value rather than memory; (2) draw the feasible region for a two-variable system, including the non-negative constraints; (3) find the maximum or minimum of an objective function by testing the vertices and explain why an interior point can never win.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Solving linear equations	Solve 3x + 2 = 11	x = 3
Plotting a straight line	Where does x + y = 10 cross the axes	(10, 0) and (0, 10)
Negative number arithmetic	What is 13 – 4 divided by -3	-3

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	One variable	Solve 2x + 7 <= 19, then -3x + 4 > 13	Get x <= 6 and x < -3
10-20 min	The flip	Multiply by a negative, test one value back	Predict, then verify
20-32 min	Two variables	Draw x + y = 10 and 2x + y = 16, test (0,0)	Shade the correct side
32-42 min	Vertices	Solve the two lines together	Get (6, 4)
42-54 min	Objective	Evaluate P = 3x + 2y at every corner	Maximum 26
54-60 min	Exit check	Three questions marked on the spot	All three right

03 Teaching script (read it out)

Scene 1: A flip is a fact about order, not a trick

Solve 2x + 7 <= 19 by subtracting 7 and dividing by 2: x <= 6. Reverse of the inequality is not needed because both operations preserve order.
Now solve -3x + 4 > 13. Subtract 4 to get -3x > 9, then divide by -3. The answer is x < -3, and the sign must reverse because dividing by a negative number reflects every number on the number line.
Do not let students memorise “flip when negative”. Make them check: take x = -4, which is less than -3. Then -3(-4) + 4 = 16, and 16 is greater than 13, so the inequality holds. Take x = 0: 4 is not greater than 13, so 0 must be excluded. That test value is the proof, and it is also one mark in the exam.

Ask students: substitute x = -4 and x = 0 into -3x + 4 > 13 and say which one satisfies it.

Scene 2: Shade by testing a point, never by guessing

For x + y <= 10 the boundary is the line through (10, 0) and (0, 10). Test the origin: 0 + 0 = 0, which is less than 10, so the side containing the origin is the wanted region. For 2x + y <= 16 the boundary passes through (8, 0) and (0, 16), and the origin again satisfies it.
Add x >= 0 and y >= 0 to keep the region in the first quadrant. The region is then a quadrilateral with corners (0, 0), (0, 10), (6, 4) and (8, 0).
The intersection is found algebraically: subtract x + y = 10 from 2x + y = 16 to get x = 6, so y = 4. Students who sketch by eye and read off (6, 4) by luck lose the method marks when the numbers are less friendly.

Ask students: why does testing the origin work for both lines (the origin is not on either line here, so it is a clean test point).

Scene 3: An optimum always sits on a corner

The objective is P = 3x + 2y. Evaluate it at the four corners: (0, 0) gives 0, (0, 10) gives 20, (6, 4) gives 26, and (8, 0) gives 24. The maximum is 26 at (6, 4).
The reason is that P is linear. Its level curves are parallel straight lines, and a family of parallel lines leaves a convex polygon either through a vertex or along a whole edge, never at a single interior point. So checking corners is a complete method, not a shortcut.
When two corners tie, the whole edge between them gives the same value, and that sentence is worth a mark: the maximum is not unique, and any point on that edge achieves it.

Ask students: if two corners give 26, what does the entire edge between them give (the same value, 26).

04 Question types and worked steps

Type	Example	Steps	Answer
Solve	Linear inequality with a negative coefficient	Isolate x, reverse the sign, verify with a test value	x < -3
Draw	Feasible region from two inequalities	Boundary, test point, shade, add non-negative axes	Quadrilateral
Compute	Intersection of two boundary lines	Solve the pair simultaneously	(6, 4)
Optimise	Maximum of an objective function	Evaluate at every vertex, compare, state the point	26 at (6, 4)

Sanity check (teach this): Divide or multiply by a negative number and the inequality sign must reverse; verify with one substituted value.
A strict inequality uses a dashed boundary and an open endpoint; a non-strict one uses a solid boundary.
Never evaluate the objective only at the corners you like: all corners, then compare.
If two constraints are parallel, they never intersect, so the corner count drops by one; a sketch that shows four corners from parallel lines is wrong.

05 Independent practice (3 questions)

Question	Answer	What to watch
Solve -3x + 4 > 13	x < -3	Flip the sign, then test x = -4
Find the intersection of x + y = 10 and 2x + y = 16	(6, 4)	Subtract the equations, then substitute back
Maximum of P = 3x + 2y on the region above	26 at (6, 4)	Test all four corners, then compare

06 Common mistakes and what to say

Mistake	Student writes	Say this
Forgetting to flip the sign	Writing x > -3 from -3x > 9	Substitute x = -4: the inequality holds, so the answer must be x < -3.
Shading the wrong side	Shading above x + y <= 10	Test the origin; if 0 <= 10 the origin side is the one you want.
Reading the corner by eye	Writing (6, 4) with no working	Always solve the two boundary equations; the working carries the marks.
Testing only one corner	Answering the first corner you compute	The optimum can be any corner: list all of them before you compare.

07 Homework (paper, section and question numbers)

Secondary maths chapter index — All chapters and practice
All lesson-prep guides — Compare other subjects

08 Exit check (last 5 minutes)

Question	Answer	Decision
Solve -2x + 1 <= 7	x >= -3	All three right: move to quadratic graphs
Corner shared by x + y = 10 and 2x + y = 16	(6, 4)	One wrong: reteach solving simultaneously
Why must the optimum be at a corner	Linear level lines leave a polygon at a vertex	Two wrong: return to act three

09 Tutor FAQ

Do students need to draw the region to full scale?

A clear sketch with correct intercepts is enough; the marks come from the tested side and the labelled vertices.

What if the question asks for integer values only?

Then test the integer points inside the region near the best corner, because the true optimum may not be a lattice point.

Is there a faster method than testing every corner?

Comparing gradients of the objective and the constraints works, but vertex testing is safer under exam pressure.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.