Mathematics • Year 10 • Unit 2 • Lesson 7
Mixed Algebraic Techniques — Mixed Challenge
Pull together every idea from Lesson 7: expand brackets, distribute the minus across every term, collect like terms, and verify by substitution. Choose the right move for each problem, spot a Year 10 student's slip, then design an expression that simplifies to a target form.
1. Mixed problems — choose the right tool
Each question uses a different combination from Lesson 7. Decide which tools apply before you start writing. Show your working. 3 marks each
1.1 Expand and simplify (x + 7)(x + 3) − x².
1.2 Expand and simplify 5(x + 2) − 2(x − 3).
1.3 Simplify (x + 6)(x + 1) − (x + 2)(x + 3). (Watch the minus on EVERY term.)
1.4 Expand and simplify (x − 4)(x + 4). (This is a "difference of squares" — see what happens to the middle terms.)
1.5 Expand and simplify 3(x + 4) + (x − 1)(x + 2).
1.6 A rectangle has area (x + 5)(x + 4). A smaller rectangle of area (x + 3)(x + 2) is cut out. Write and simplify a single expression for the remaining area. Then verify your answer by substituting x = 1.
2. Find the mistake
Another Year 10 student has tried to simplify (x + 5)(x + 4) − (x + 2)(x + 3). Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — simplify (x + 5)(x + 4) − (x + 2)(x + 3):
Line 1: (x + 5)(x + 4) = x² + 9x + 20
Line 2: (x + 2)(x + 3) = x² + 5x + 6
Line 3: = x² + 9x + 20 − x² + 5x + 6
Line 4: = 14x + 26
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer. Then verify by substituting x = 1.
Stuck? Revisit lesson § "Spot the Trap" — what should happen to the +5x and +6 when distributing the minus?3. Open-ended challenge — design an expression that simplifies to a target
This question has many valid answers. Show every step. 4 marks
3.1 Design an algebraic expression that mixes a binomial product and a separate bracket, so that after expanding and collecting like terms the result is exactly 4x + 7. Your expression must include all four of: at least one product of two binomials (e.g. (x + a)(x + b)), at least one multiplier-bracket piece (e.g. k(x + d)), a subtraction somewhere, and a constant.
In your submission, include:
(i) Your expression, written cleanly.
(ii) The full expansion line by line.
(iii) The collected result, which must equal 4x + 7.
(iv) A check by substituting x = 1 into both the original expression and 4x + 7.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (x + 7)(x + 3) − x²
(x + 7)(x + 3) = x² + 10x + 21. Subtract x²: 10x + 21.
1.2 — 5(x + 2) − 2(x − 3)
= 5x + 10 − 2x + 6 = 3x + 16. The −2 distributes to both x and −3: −2 × (−3) = +6.
1.3 — (x + 6)(x + 1) − (x + 2)(x + 3)
(x + 6)(x + 1) = x² + 7x + 6. (x + 2)(x + 3) = x² + 5x + 6.
Difference: (x² + 7x + 6) − (x² + 5x + 6) = x² + 7x + 6 − x² − 5x − 6 = 2x.
1.4 — (x − 4)(x + 4)
FOIL: x² + 4x − 4x − 16. The middle terms cancel: x² − 16.
1.5 — 3(x + 4) + (x − 1)(x + 2)
3(x + 4) = 3x + 12. (x − 1)(x + 2) = x² + 2x − x − 2 = x² + x − 2.
Sum: 3x + 12 + x² + x − 2 = x² + 4x + 10.
1.6 — Remaining area (x + 5)(x + 4) − (x + 3)(x + 2)
(x + 5)(x + 4) = x² + 9x + 20. (x + 3)(x + 2) = x² + 5x + 6.
Difference: (x² + 9x + 20) − (x² + 5x + 6) = x² + 9x + 20 − x² − 5x − 6 = 4x + 14.
Check at x = 1: original = (6)(5) − (4)(3) = 30 − 12 = 18. Answer: 4(1) + 14 = 18. ✓
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The minus in front of (x² + 5x + 6) must distribute to every term, giving "− x² − 5x − 6". The student only flipped the x² term and left +5x and +6 unchanged.
(c) Corrected:
= (x² + 9x + 20) − (x² + 5x + 6)
= x² + 9x + 20 − x² − 5x − 6 (distribute the minus to every term)
= (x² − x²) + (9x − 5x) + (20 − 6)
= 4x + 14.
Check at x = 1: original = (6)(5) − (3)(4) = 30 − 12 = 18. Answer: 4(1) + 14 = 18. ✓
3 — Open-ended challenge (sample solutions)
Many valid answers — here are two starter constructions that simplify to 4x + 7.
Sample 1 — (x + 1)(x + 3) − x² + 4
(i) Has product of binomials ✓, subtraction ✓ (the − x²), constant ✓ (the +4). Add a small multiplier-bracket piece, e.g. (x + 1)(x + 3) − x² + 1(x + 4) for the fourth criterion.
(ii) (x + 1)(x + 3) = x² + 4x + 3. So expression = x² + 4x + 3 − x² + (x + 4) = x² + 4x + 3 − x² + x + 4.
(iii) Collect: x² − x² = 0; 4x + x = 5x; 3 + 4 = 7 → 5x + 7. Not 4x + 7 — adjust the multiplier from 1 to 0, i.e. drop that piece, and instead include 2(x + 2) − (x + 1): expression becomes (x + 1)(x + 3) − x² + 2(x + 2) − (x + 1). Expand: (x² + 4x + 3) − x² + (2x + 4) − (x + 1) = x² + 4x + 3 − x² + 2x + 4 − x − 1 = 5x + 6. Adjust constant: replace −1 with −0 (drop the +1) gives 5x + 7. To hit 4x + 7 use (x + 1)(x + 3) − x² + (x + 4) − x = 4x + 7. Working: (x² + 4x + 3) − x² + (x + 4) − x = x² + 4x + 3 − x² + x + 4 − x = 4x + 7. ✓
(iv) Check x = 1: original = (2)(4) − 1 + 5 − 1 = 8 − 1 + 5 − 1 = 11. Target: 4(1) + 7 = 11. ✓
Sample 2 — 2(x + 5) + (x + 1)(x + 2) − (x + 1)(x + 2) − 3 = 2(x + 5) − 3 = 2x + 7 (only 2x — not the target). Adjust to 2(x + 5) + (x + 1)(x + 2) − x² − 3x − 3: expand = 2x + 10 + x² + 3x + 2 − x² − 3x − 3 = 2x + 9 (not target). Better: (x + 3)(x + 4) − x² + (x − 5): expand = x² + 7x + 12 − x² + x − 5 = 8x + 7. Halve the coefficient by replacing (x + 3)(x + 4) with (x + 1)(x + 2): expand = x² + 3x + 2 − x² + x − 5 = 4x − 3 — wrong constant. Replace (x − 5) with (x + 5): (x + 1)(x + 2) − x² + (x + 5) → x² + 3x + 2 − x² + x + 5 = 4x + 7. ✓ Check at x = 1: original = (2)(3) − 1 + 6 = 6 − 1 + 6 = 11. Target: 4(1) + 7 = 11. ✓
Marking: 1 for a valid expression including all four required pieces; 1 for a clean expansion shown line by line; 1 for arriving at 4x + 7; 1 for the substitution check at x = 1. Full marks for any valid construction.