Mathematics • Year 10 • Unit 2 • Lesson 1
Algebraic Expressions — Mixed Challenge
Pull together every idea from Lesson 1: like terms (same letter and same power), substitution (brackets around negatives), and BIDMAS order. Choose the right tool for each problem, spot another student's mistake, then design an expression of your own that hits a target value.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 1. Decide which tool applies before you start writing. Show your working. 3 marks each
1.1 Simplify 8p − 3q + 2p + 5q − p.
1.2 If x = 3 and y = −2, evaluate 5x − 2y + xy.
1.3 Evaluate 18 − 4 × 3 + 6 ÷ 2 using BIDMAS.
1.4 Simplify 4a² + 6a − a² + 2a − 5.
1.5 If m = −5, evaluate m² + 4m − 3. Show the BIDMAS order.
1.6 A triangle has sides (2x + 1), (3x − 2), and (x + 5) cm. Write a simplified expression for the perimeter, then evaluate it when x = 6.
2. Find the mistake
Another Year 10 student has tried to evaluate 3x² − 2x + 1 when x = −2. 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 — evaluate 3x² − 2x + 1 when x = −2:
Line 1: = 3(−2)² − 2(−2) + 1
Line 2: = 3(−4) − 2(−2) + 1
Line 3: = −12 + 4 + 1
Line 4: = −7
(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.
Stuck? Revisit lesson § "Spot the Trap" — what is (−2)² actually equal to?3. Open-ended challenge — design an expression that hits a target
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design an algebraic expression in two variables (e.g. x and y) that uses all four of: a coefficient ≥ 3, a squared term, a subtraction, and a constant. Then choose values for x and y so that when you substitute them in, the expression evaluates to exactly 20.
In your submission, include:
(i) Your expression, written cleanly.
(ii) The values you chose for x and y.
(iii) The full BIDMAS working showing the expression equals 20 for those values.
Bonus: Use at least one negative value (for x or y) and show the bracket habit in the working.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Simplify 8p − 3q + 2p + 5q − p
p-group: 8p + 2p − p = 9p. q-group: −3q + 5q = 2q. Answer: 9p + 2q.
1.2 — 5x − 2y + xy when x = 3, y = −2
= 5(3) − 2(−2) + (3)(−2) = 15 − (−4) + (−6) = 15 + 4 − 6 = 13.
1.3 — 18 − 4 × 3 + 6 ÷ 2
BIDMAS: multiplication and division first (left to right): 4 × 3 = 12, 6 ÷ 2 = 3.
Then add/subtract left to right: 18 − 12 + 3 = 6 + 3 = 9.
1.4 — Simplify 4a² + 6a − a² + 2a − 5
a²-group: 4a² − a² = 3a². a-group: 6a + 2a = 8a. Constant: −5.
Answer: 3a² + 8a − 5.
1.5 — m² + 4m − 3 when m = −5
Substitute with brackets: (−5)² + 4(−5) − 3.
Indices: (−5)² = 25. So 25 + 4(−5) − 3.
Multiplication: 4 × (−5) = −20. So 25 + (−20) − 3 = 25 − 20 − 3 = 2.
1.6 — Triangle perimeter
P = (2x + 1) + (3x − 2) + (x + 5).
Collect like terms: (2x + 3x + x) + (1 − 2 + 5) = 6x + 4 cm.
When x = 6: 6(6) + 4 = 36 + 4 = 40 cm.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) (−2)² should equal +4, not −4. A negative squared is positive because (−2) × (−2) = +4. The student dropped the bracket and let only the 2 (not the −2) get squared.
(c) Corrected working:
= 3(−2)² − 2(−2) + 1
= 3(4) − 2(−2) + 1 (indices: (−2)² = +4)
= 12 − (−4) + 1
= 12 + 4 + 1
= 17.
Lesson § "Spot the Trap" flags this exact misconception.
3 — Open-ended challenge (sample solutions)
Many valid answers — here are two starter pairs that hit the target value 20.
Sample 1 — expression: 3x² + y − 7
Pick x = 2, y = 15. Working: 3(2)² + 15 − 7 = 3(4) + 15 − 7 = 12 + 15 − 7 = 20 ✓.
Coefficient ≥ 3 ✓ (the 3), squared term ✓ (x²), subtraction ✓ (− 7), constant ✓ (−7).
Sample 2 — expression with a negative substitution: 4y² + 2x − 8
Pick x = 10, y = −2. Working: 4(−2)² + 2(10) − 8 = 4(4) + 20 − 8 = 16 + 20 − 8 = 20 (wait, 16 + 20 − 8 = 28 — recheck). Adjust: pick y = −1, x = 12: 4(−1)² + 2(12) − 8 = 4(1) + 24 − 8 = 4 + 24 − 8 = 20 ✓.
Bonus achieved with the negative y inside brackets.
Marking: 1 for a valid expression meeting the four criteria; 1 for a sensible choice of x and y; 2 for working that uses BIDMAS, brackets around any negatives, and produces exactly 20. Full marks for any valid construction.