The shape of the lesson
By the end of the lesson, learners can
- The graph of \(y = |ax + b|\) is a V: Draw the straight line \(y = ax + b\) first, then reflect every part that lies below the \(x\)-axis in the \(x\)-axis.
- Equations: square, or split into cases: When BOTH sides are moduli, use \(|a| = |b| \iff a^2 = b^2\), or equivalently \(a = b\) or \(a = -b\).
- Inequalities: squeeze, square or sketch: Pick the method that fits the shape of the question.
- Sketch: vertex where \(ax + b = 0\), \(y\)-intercept \(|b|\), arms of gradient \(\pm a\).
- \(|f| = |g|\): square both sides, or solve \(f = g\) and \(f = -g\).
- \(|f| = \) line: split into cases and CHECK each root in the original equation.
- \(|x - a| < b\): \(a - b < x < a + b\); for \(>\) take the two outside pieces.
- \(|f| < |g|\): square, then solve the quadratic inequality.
- Modulus against a line: sketch, find valid crossings, read off the region.
One picture every idea returns to
\(|x|\) is the distance of \(x\) from 0. Two facts do most of the work: \(|a| = |b| \iff a^2 = b^2\), and \(|x - a| < b \iff a - b < x < a + b\). When one side is NOT a modulus, any answer found by splitting into cases must be checked in the original equation.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Distance, never negative.. Papers 2 and 3. \(|x|\) is the size of \(x\) with its sign thrown away. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The graph of \(y = |ax + b|\) is a V | Draw the straight line \(y = ax + b\) first, then reflect every part that lies below the \(x\)-axis in the \(x\)-axis. Model the worked example Sketching a modulus graph on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Equations: square, or split into cases | When BOTH sides are moduli, use \(|a| = |b| \iff a^2 = b^2\), or equivalently \(a = b\) or \(a = -b\). Model the worked example Both sides are moduli on the board, one line at a time. Your turn: Solve \(|x + 2| = |3x - 4|\). | Pairs |
| 21 to 30 min | Teach and check: Inequalities: squeeze, square or sketch | Pick the method that fits the shape of the question. Model the worked example Modulus less than a number on the board, one line at a time. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 21 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Inequalities: squeeze, square or sketch, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The vertex of \(y = |2x - 3|\) is at \(x = 1.5\), not \(x = 3\) or \(x = -1.5\). | Solve \(ax + b = 0\) properly; the most common slip is forgetting to divide by \(a\). |
| In Case 2, \(-(2x - 1)\) is \(-2x + 1\), not \(-2x - 1\). | Put brackets round the whole side before you change its sign. |
| Never square when one side might be negative: \(2x + 5 < |x + 1|\) squared gives a wrong answer, because squaring a negative left side flips which side is bigger. | Square only when both sides are moduli (or are known to be non-negative). |
The modulus function: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Sketch the graph of \(y = |3x + 6|\), stating the coordinates of the points where the graph meets the axes.
Solve the equation \(|x - 4| = |2x + 1|\).
Solve the inequality \(|2x - 3| \ge 5\).
Solve the inequality \(|x + 2| > |2x - 5|\).
Solve the inequality \(3x - 1 > |x + 5|\).
(a) On the same axes, sketch the graphs of \(y = |2x - 5|\) and \(y = x + 1\).
(b) Solve the equation \(|2x - 5| = x + 1\).
(c) Hence solve the inequality \(|2x - 5| < x + 1\).
Answer key and marking notes
- Vertex: \(3x + 6 = 0\) gives \(x = -2\), so \((-2, 0)\).
- \(y\)-intercept: \(y = |6| = 6\), so \((0, 6)\).
- A V shape with arms of gradient \(-3\) (left) and \(3\) (right).
- Square: \(x^2 - 8x + 16 = 4x^2 + 4x + 1\), so \(3x^2 + 12x - 15 = 0\).
- \(x^2 + 4x - 5 = 0\), so \((x + 5)(x - 1) = 0\).
- \(x = -5\) or \(x = 1\). (Check: \(|-9| = |-9|\), \(|-3| = |3|\).)
- \(2x - 3 \ge 5\) gives \(x \ge 4\).
- \(2x - 3 \le -5\) gives \(x \le -1\).
- So \(x \le -1\) or \(x \ge 4\).
- Both sides are non-negative, so square: \(x^2 + 4x + 4 > 4x^2 - 20x + 25\).
- \(0 > 3x^2 - 24x + 21\), so \(x^2 - 8x + 7 < 0\).
- \((x - 1)(x - 7) < 0\), so the quadratic is negative between its roots.
- \(1 < x < 7\). (Check \(x = 3\): \(5 > 1\).)
- For \(x \ge -5\): \(3x - 1 = x + 5\) gives \(x = 3\), which satisfies \(x \ge -5\). Valid crossing.
- For \(x < -5\): \(3x - 1 = -x - 5\) gives \(x = -1\), which does not satisfy \(x < -5\). Reject.
- The line \(y = 3x - 1\) is steeper than the right arm, so it is above the V to the right of \(x = 3\).
- \(x > 3\). (Check \(x = 0\): \(-1 > 5\) is false, as expected.)
- (a) V with vertex \((2.5, 0)\) and \(y\)-intercept \((0, 5)\); straight line through \((-1, 0)\) and \((0, 1)\), crossing both arms of the V.
- (b) Right arm: \(2x - 5 = x + 1\) gives \(x = 6\). Check: \(|7| = 7\). Valid.
- Left arm: \(5 - 2x = x + 1\) gives \(3x = 4\), \(x = \dfrac{4}{3}\). Check: \(\left|-\dfrac{7}{3}\right| = \dfrac{7}{3}\). Valid.
- So \(x = \dfrac{4}{3}\) or \(x = 6\).
- (c) The V is below the line between the crossings: \(\dfrac{4}{3} < x < 6\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Solve the equation \(|2x - 7| = 3\).
- A\(x = 2\) or \(x = 5\)
- B\(x = 5\) only
- C\(x = -5\) or \(x = 5\)
- D\(x = -2\) or \(x = -5\)
The graph of \(y = |3x + 6|\) meets the axes at two points. What are they?
- A\((-2, 0)\) and \((0, 6)\)
- B\((2, 0)\) and \((0, 6)\)
- C\((-2, 0)\) and \((0, -6)\)
- D\((-6, 0)\) and \((0, 6)\)
Solve the inequality \(|x - 4| < 3\).
- A\(1 < x < 7\)
- B\(x < 7\)
- C\(-7 < x < -1\)
- D\(x < 1\) or \(x > 7\)
Solve the equation \(|x + 1| = |x - 5|\).
- A\(x = 2\)
- B\(x = -3\)
- CThere is no solution.
- D\(x = -2\) or \(x = 2\)
Solve the equation \(|2x + 1| = 3x - 4\).
- A\(x = 5\) only
- B\(x = 5\) or \(x = \dfrac{3}{5}\)
- C\(x = \dfrac{3}{5}\) only
- D\(x = 5\) or \(x = 1\)
Solve the inequality \(|x - 2| > |2x + 5|\).
- A\(-7 < x < -1\)
- B\(x < -7\) or \(x > -1\)
- C\(1 < x < 7\)
- D\(-7 < x < 1\)
Solve the inequality \(2x - 1 < |x + 4|\).
- A\(x < 5\)
- B\(-1 < x < 5\)
- C\(x < -1\) or \(x > 5\)
- D\(x > 5\)
The diagram shows the graphs of \(y = |3x - 1|\) and \(y = x + 3\). Solve the inequality \(|3x - 1| \le x + 3\).
- A\(-\dfrac{1}{2} \le x \le 2\)
- B\(x \le -\dfrac{1}{2}\) or \(x \ge 2\)
- C\(\dfrac{1}{3} \le x \le 2\)
- D\(-2 \le x \le \dfrac{1}{2}\)
Quiz key
\(2x - 7 = 3\) gives \(x = 5\); \(2x - 7 = -3\) gives \(2x = 4\), \(x = 2\). Both satisfy the equation: \(|3| = 3\) and \(|-3| = 3\). The tempting "\(x = 5\) only" forgets the negative case. \(x = -5\) comes from writing \(-(2x - 7)\) as \(-2x - 7\), and \(x = -2\) or \(-5\) comes from solving \(2x + 7 = \pm 3\).
\(3x + 6 = 0\) at \(x = -2\), so the V touches the \(x\)-axis at \((-2, 0)\). At \(x = 0\), \(y = |6| = 6\). The tempting \((2, 0)\) solves \(3x - 6 = 0\) with the sign changed. \((0, -6)\) is impossible because \(|3x + 6| \ge 0\) for every \(x\), and \((-6, 0)\) sets \(x + 6 = 0\), forgetting the 3.
\(-3 < x - 4 < 3\), and adding 4 throughout gives \(1 < x < 7\). The tempting \(x < 7\) only uses \(x - 4 < 3\) and loses the lower bound. \(-7 < x < -1\) centres the interval on \(-4\) instead of 4, and "\(x < 1\) or \(x > 7\)" is the solution of \(|x - 4| > 3\).
Squaring: \(x^2 + 2x + 1 = x^2 - 10x + 25\), so \(12x = 24\) and \(x = 2\). Check: \(|3| = |-3|\). Geometrically, \(x = 2\) is the point midway between \(-1\) and 5. The case \(x + 1 = x - 5\) has no solution, which tempts some to answer "no solution", but the other case \(x + 1 = -(x - 5)\) gives \(x = 2\). \(x = -3\) comes from writing \(-(x - 5)\) as \(-x - 5\), and \(x = -2\) is not a solution: \(|-1| \ne |-7|\).
Case 1: \(2x + 1 = 3x - 4\) gives \(x = 5\); check: \(|11| = 11\), valid. Case 2: \(-(2x + 1) = 3x - 4\) gives \(-2x - 1 = 3x - 4\), \(5x = 3\), \(x = \dfrac{3}{5}\); check: the right-hand side is \(\dfrac{9}{5} - 4 = -\dfrac{11}{5} < 0\), but a modulus cannot equal a negative number, so reject it. The tempting "\(x = 5\) or \(x = \dfrac{3}{5}\)" skips the check. \(x = 1\) comes from writing \(-(2x + 1)\) as \(-2x + 1\) (and \(x = 1\) fails the check anyway).
\((x - 2)^2 > (2x + 5)^2\): \(x^2 - 4x + 4 > 4x^2 + 20x + 25\), so \(0 > 3x^2 + 24x + 21\), i.e. \(x^2 + 8x + 7 < 0\), \((x + 1)(x + 7) < 0\), giving \(-7 < x < -1\). Check \(x = -2\): \(|-4| = 4 > |1| = 1\). The tempting "\(x < -7\) or \(x > -1\)" takes the outside region, which is the answer to \(<\) instead of \(>\). \(1 < x < 7\) comes from factorising as \((x - 1)(x - 7)\), and \(-7 < x < 1\) from a sign slip in one factor.
For \(x \ge -4\): \(2x - 1 < x + 4\) gives \(x < 5\), so \(-4 \le x < 5\). For \(x < -4\): \(2x - 1 < -x - 4\) gives \(x < -1\), true for every \(x < -4\). Together: \(x < 5\). (A sketch shows the line is below the V everywhere to the left of their single meeting point at \(x = 5\).) The tempting \(-1 < x < 5\) comes from squaring both sides: \(3x^2 - 12x - 15 < 0\). Squaring fails because \(2x - 1\) is negative for small \(x\), e.g. \(x = -2\) gives \(-5 < 2\), which is true but is lost by squaring.
Right arm: \(3x - 1 = x + 3\) gives \(x = 2\). Left arm: \(-(3x - 1) = x + 3\) gives \(-3x + 1 = x + 3\), \(x = -\dfrac{1}{2}\). The V lies on or below the line between these, so \(-\dfrac{1}{2} \le x \le 2\). (Squaring gives the same: \(8x^2 - 12x - 8 \le 0\), \((2x + 1)(x - 2) \le 0\).) The tempting "\(x \le -\dfrac{1}{2}\) or \(x \ge 2\)" is where the V is ABOVE the line. \(\dfrac{1}{3} \le x \le 2\) starts at the vertex instead of the left intersection, and \(-2 \le x \le \dfrac{1}{2}\) has both signs wrong.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.