The shape of the lesson
By the end of the lesson, learners can
- Eliminate one letter: In the lab choose Two lines and set the sliders to the equations \(2x + y = 7\) and \(x - y = 2\).
- Setting them up: Choose a letter for each unknown, then turn each sentence of information into one equation.
- A line and a curve (Extended): When one equation is quadratic, substitution is the only way.
- Match: multiply so one unknown has the same coefficient in both.
- Signs: different signs add; same signs subtract.
- Substitution: best when one equation already says \(y = \dots\).
- Back-substitute: find the second unknown, then check both equations.
- Non-linear (Extended): substitute the linear into the other and solve the quadratic; give pairs.
One picture every idea returns to
Make the coefficients of one unknown the same size, then add the equations if the signs differ or subtract if they are the same. Solve for the other unknown, substitute back, and check both values in both equations. The solution is where the two graphs cross.
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, Two equations, one answer.. Two unknowns need two equations. 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: Eliminate one letter | In the lab choose Two lines and set the sliders to the equations \(2x + y = 7\) and \(x - y = 2\). Project tab 3 of the lab (Equations lab, simultaneous equations) and let a student drive it while the class predicts. Model the worked example Add to eliminate on the board, one line at a time. Your turn: Solve \(4x + 3y = 25\) and \(2x + 5y = 23\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Setting them up | Choose a letter for each unknown, then turn each sentence of information into one equation. Model the worked example Ticket prices on the board, one line at a time. Your turn: 5 pens and 2 rulers cost $3.70. 3 pens and 4 rulers cost $3.90. Find the cost of one pen. | Pairs |
| 21 to 30 min | Teach and check: A line and a curve (Extended) | When one equation is quadratic, substitution is the only way. Model the worked example Line and parabola (Extended) on the board, one line at a time. Your turn: (Extended) Solve \(y = 2x - 1\) and \(y = x^{2} - 4\). | Round Table |
| 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, 22 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 A line and a curve (Extended), 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Same Signs Subtract. | When you subtract \(2x - y = 5\) from \(2x + 3y = 17\), the \(y\) terms give \(3y - (-y) = 4y\), not \(2y\). |
| Write down what each letter stands for, with units. | Many students find the right numbers and then give the child price when the question asked for the adult price. |
| \((x + 1)^{2}\) is \(x^{2} + 2x + 1\), not \(x^{2} + 1\). | And give the answers as pairs: each \(x\) goes with its own \(y\). |
Simultaneous equations: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Solve \(x + y = 10\) and \(x - y = 4\).
Solve \(4x + 3y = 25\) and \(2x + 5y = 23\).
Solve \(y = 3x - 4\) and \(2x + 3y = 21\).
5 pens and 2 rulers cost $3.70. 3 pens and 4 rulers cost $3.90.
Find the cost of one pen and the cost of one ruler.
(Extended) Solve \(y = 2x - 1\) and \(y = x^{2} - 4\).
(Extended) The line \(y = x + 1\) meets the circle \(x^{2} + y^{2} = 25\) at the points A and B.
(a) Find the coordinates of A and B.
(b) Find the length of AB.
Answer key and marking notes
- Add: \(2x = 14\), \(x = 7\)
- \(y = 3\)
- Double the second: \(4x + 10y = 46\)
- Subtract the first: \(7y = 21\), \(y = 3\)
- \(2x + 15 = 23\), \(x = 4\)
- \(2x + 3(3x - 4) = 21\)
- \(11x - 12 = 21\), \(x = 3\)
- \(y = 9 - 4 = 5\)
- \(5p + 2r = 3.70\) and \(3p + 4r = 3.90\)
- Double the first: \(10p + 4r = 7.40\); subtract the second: \(7p = 3.50\)
- \(p = 0.50\)
- \(2r = 3.70 - 2.50 = 1.20\), \(r = 0.60\). Pen $0.50, ruler $0.60.
- \(x^{2} - 4 = 2x - 1\)
- \(x^{2} - 2x - 3 = 0\), \((x - 3)(x + 1) = 0\)
- \(x = 3,\ y = 5\)
- \(x = -1,\ y = -3\)
- (a) \(x^{2} + (x + 1)^{2} = 25\), so \(2x^{2} + 2x - 24 = 0\)
- \(x^{2} + x - 12 = 0\), \((x + 4)(x - 3) = 0\)
- A\((3, 4)\) and B\((-4, -3)\)
- (b) \(AB = \sqrt{(3 - (-4))^{2} + (4 - (-3))^{2}} = \sqrt{49 + 49}\)
- \(= \sqrt{98} = 9.90\) (3 s.f.)
6 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 simultaneous equations \(3x + y = 11\) and \(x - y = 1\).
- A\(x = 3,\ y = 2\)
- B\(x = 3,\ y = -2\)
- C\(x = 4,\ y = -1\)
- D\(x = 2,\ y = 1\)
Solve the simultaneous equations \(2x + 3y = 13\) and \(5x + 2y = 16\).
- A\(x = 2,\ y = 3\)
- B\(x = 3,\ y = 2\)
- C\(x = 2,\ y = 6\)
- D\(x = 6.5,\ y = 0\)
3 cakes and 2 coffees cost $13.50. 1 cake and 2 coffees cost $7.50. Find the cost of one coffee.
- A$2.25
- B$3.00
- C$4.50
- D$2.50
The lines \(y = 2x + 1\) and \(y = 7 - x\) are drawn on the same grid. What are the coordinates of the point where they cross?
- A\((2,\ 5)\)
- B\((5,\ 2)\)
- C\((6,\ 13)\)
- D\((0,\ 7)\)
Solve the simultaneous equations \(y = x + 1\) and \(y = x^{2} - 2x - 3\).
- A\(x = 4,\ y = 5\) and \(x = -1,\ y = 0\)
- B\(x = -4,\ y = -3\) and \(x = 1,\ y = 2\)
- C\(x = 3,\ y = 4\) and \(x = -1,\ y = 0\)
- D\(x = 4,\ y = 5\) only
Solve the simultaneous equations \(x^{2} + y^{2} = 20\) and \(y = 2x\).
- A\(x = 2,\ y = 4\) and \(x = -2,\ y = -4\)
- B\(x = 2,\ y = 4\) only
- C\(x = 2,\ y = -4\) and \(x = -2,\ y = 4\)
- D\(x = 4,\ y = 2\) and \(x = -4,\ y = -2\)
Quiz key
Adding: \(4x = 12\), so \(x = 3\). Then \(3 - y = 1\) gives \(y = 2\). Check in the first: \(9 + 2 = 11\). \(x = 4, y = -1\) fits the first equation only and \(x = 2, y = 1\) fits the second only: a solution must satisfy both. \(y = -2\) is a slip in the substitution.
\(4x + 6y = 26\) and \(15x + 6y = 48\). Same signs, so subtract: \(11x = 22\), \(x = 2\). Then \(4 + 3y = 13\), \(y = 3\). Check: \(5(2) + 2(3) = 16\). \(x = 3, y = 2\) swaps the values: \(2(3) + 3(2) = 12\), not 13. \(x = 6.5, y = 0\) fits only the first equation.
Subtract: \(2c = 6.00\), so one cake costs $3.00. Then \(3.00 + 2f = 7.50\), \(2f = 4.50\), \(f = 2.25\). One coffee costs $2.25. $3.00 is the cake, not the coffee. $4.50 is the cost of TWO coffees: divide by 2.
\(2x + 1 = 7 - x\), so \(3x = 6\), \(x = 2\), and \(y = 2(2) + 1 = 5\). The point is \((2, 5)\); check: \(7 - 2 = 5\). \((5, 2)\) writes \(y\) before \(x\). \((6, 13)\) comes from \(2x + 1 = 7 + x\), losing the minus sign. \((0, 7)\) is only where the second line meets the \(y\)-axis.
\(x^{2} - 2x - 3 = x + 1\) gives \(x^{2} - 3x - 4 = 0\), so \((x - 4)(x + 1) = 0\): \(x = 4\) or \(x = -1\). Using \(y = x + 1\): \(y = 5\) and \(y = 0\). \(x = -4, x = 1\) take the wrong signs from the brackets. \(x = 3\) solves \(x^{2} - 2x - 3 = 0\), ignoring the line. A quadratic usually gives TWO pairs, so 'only' is wrong.
\(x^{2} + (2x)^{2} = 20\), so \(5x^{2} = 20\), \(x^{2} = 4\), \(x = \pm 2\). With \(y = 2x\): \(x = 2, y = 4\) and \(x = -2, y = -4\). Forgetting \(x = -2\) loses a solution. Pairing \(x = 2\) with \(y = -4\) breaks \(y = 2x\). \(x = 4, y = 2\) has \(x\) and \(y\) swapped.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.