Mathematics • Year 10 • Unit 2 • Lesson 10
Formulas and Rearrangement — Mixed Challenge
Mixed rearrangements drawing on every tool from Lesson 10: reverse BIDMAS, brackets-first, powers and roots. Catch a classmate's square-root error, then construct a formula of your own that fits a target output.
1. Mixed problems — choose the right tool
Each problem asks for a different rearrangement. Show every step and state restrictions. 3 marks each
1.1 Make x the subject of y = 2x − 7.
1.2 Make a the subject of v² = u² + 2as. (Hint: solve for a, not the square root.)
1.3 Make h the subject of V = πr²h. State the restriction.
1.4 Make c the subject of P = a + b + c.
1.5 Make L the subject of T = 2π√(L/g). (Pendulum period.) State the restriction.
1.6 Make x the subject of y = (3x + 1)/4. Verify by checking: if y = 7, your formula should give x = 9.
2. Find the mistake
A Year 10 student tries to make r the subject of A = πr² and writes the working below. One step contains a Year-10-realistic error. Find it. 3 marks
Student's working:
Line 1: A = πr²
Line 2: A − π = r²
Line 3: r = √(A − π)
(a) Which line is wrong?
(b) Explain the mistake in one or two sentences.
(c) Show the corrected working and verify at A = 16π.
Stuck? The lesson says: to undo multiplication by π, you must divide by π, not subtract π.3. Open-ended challenge — design a formula
Many valid answers. 4 marks
3.1 Design a real-world formula in three variables (say A, B, C) that uses (i) at least one multiplication, (ii) at least one addition or subtraction, (iii) a squared term. Then:
(a) Write the formula and a one-sentence context (e.g. cost, area, energy, distance).
(b) Make a non-subject variable the subject.
(c) Choose values for the other two so the rearranged formula gives a clean numerical answer.
(d) Substitute back into the original to verify.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — y = 2x − 7
Add 7: y + 7 = 2x. Divide by 2: x = (y + 7)/2.
1.2 — v² = u² + 2as
Subtract u²: v² − u² = 2as. Divide by 2s: a = (v² − u²)/(2s) (s ≠ 0).
1.3 — V = πr²h
Divide by πr²: h = V/(πr²), restriction r ≠ 0 (and the denominator πr² ≠ 0).
1.4 — P = a + b + c
Subtract a and b: c = P − a − b.
1.5 — T = 2π√(L/g)
Divide by 2π: T/(2π) = √(L/g). Square both sides: T²/(4π²) = L/g. Multiply by g: L = gT²/(4π²), restriction g ≠ 0 (and L ≥ 0).
1.6 — y = (3x + 1)/4
Multiply by 4: 4y = 3x + 1. Subtract 1: 4y − 1 = 3x. Divide by 3: x = (4y − 1)/3. Check y = 7: x = (28 − 1)/3 = 27/3 = 9 ✓.
2 — Find the mistake
(a) Line 2.
(b) The student subtracted π instead of dividing by π. The inverse operation for "× π" is "÷ π", not "− π". The − π step would only undo "+ π" (which isn't there).
(c) Corrected: A = πr² → divide by π → A/π = r² → take √ → r = √(A/π). At A = 16π: r = √(16π/π) = √16 = 4. Check: π(4)² = 16π ✓.
3 — Open-ended (sample solution)
Sample formula: K = (1/2)mv² + Pt (total energy = kinetic energy + power × time, in joules; m kg, v m/s, P watts, t s).
Make m the subject: K − Pt = (1/2)mv² → 2(K − Pt) = mv² → m = 2(K − Pt)/v² (v ≠ 0).
Choose K = 200 J, P = 10 W, t = 5 s, v = 6 m/s: m = 2(200 − 50)/36 = 300/36 = 25/3 ≈ 8.33 kg.
Verify: (1/2)(25/3)(36) + 10(5) = (25/3)(18) + 50 = 150 + 50 = 200 ✓.
Marking: 1 for a formula meeting all three criteria with a clear context; 1 for correctly making a new subject; 1 for a clean numerical substitution; 1 for the verification.