Mathematics • Year 7 • Unit 2 • Lesson 3
Substitution — Mixed Challenge
Pull everything from Lesson 3 together: substitute positives, negatives, multiple variables, powers and fractions, spot a BODMAS mistake, and finish with an open-ended target-number puzzle.
1. Mixed problems — substitute and evaluate
Each question uses a different part of Lesson 3. Always use brackets when substituting. Show your working. 2 marks each
1.1 If x = 5, find the value of 3x − 2.
1.2 If a = 4 and b = 3, find the value of 2a + 3b − 1.
1.3 If n = −2, find the value of 5n + 14.
1.4 If x = 3, find the value of x² + 2x.
1.5 If a = 8 and b = 4, find the value of (a + b) ⁄ (a − b).
1.6 If x = −3, find the value of x² − 4x + 1.
2. Find the mistake
Another Year 7 student has tried to evaluate 4x² − 3x + 2 when x = −2. Their working is shown below. Exactly one line contains the key mistake. Spot it, explain why, then redo the working correctly. 3 marks
Student's working — evaluate 4x² − 3x + 2 at x = −2:
Line 1: = 4x² − 3x + 2
Line 2: = 4(−2)² − 3(−2) + 2
Line 3: = 4(−4) − 3(−2) + 2
Line 4: = −16 + 6 + 2
Line 5: = −8
(a) Which line contains the key 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? (−2)² is NOT −4. It's (−2) × (−2). A negative number, squared, becomes positive.3. Open-ended challenge — hit the target
More than one answer may work. 4 marks
3.1 Consider the expression 2x + y. Find at least THREE different pairs of whole-number values for x and y (positive or negative) so that the expression evaluates to 10.
Show each substitution and check.
Bonus: Can you find a pair where x is negative? Can you find a pair where y is negative? Show both.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 3x − 2 at x = 5
= 3(5) − 2 = 15 − 2 = 13.
1.2 — 2a + 3b − 1 at a = 4, b = 3
= 2(4) + 3(3) − 1 = 8 + 9 − 1 = 16.
1.3 — 5n + 14 at n = −2
= 5(−2) + 14 = −10 + 14 = 4.
1.4 — x² + 2x at x = 3
= (3)² + 2(3) = 9 + 6 = 15.
1.5 — (a + b) ⁄ (a − b) at a = 8, b = 4
Top = 8 + 4 = 12. Bottom = 8 − 4 = 4. Result = 12 ÷ 4 = 3.
1.6 — x² − 4x + 1 at x = −3
= (−3)² − 4(−3) + 1 = 9 − (−12) + 1 = 9 + 12 + 1 = 22. (Brackets matter: (−3)² = +9, and −4 × (−3) = +12.)
2 — Find the mistake
(a) The mistake is on Line 3.
(b) (−2)² = (−2) × (−2) = +4, not −4. A negative number squared is positive. The student treated (−2)² as if it were −2².
(c) Corrected working:
Line 3 (fixed): = 4(4) − 3(−2) + 2.
Line 4: = 16 + 6 + 2.
Line 5 (fixed): = 24.
3 — Open-ended (sample solutions)
Any pair (x, y) with 2x + y = 10 works. Examples:
• x = 0, y = 10 → 2(0) + 10 = 10 ✓
• x = 3, y = 4 → 2(3) + 4 = 10 ✓
• x = 5, y = 0 → 2(5) + 0 = 10 ✓
Bonus (x negative): x = −1, y = 12 → 2(−1) + 12 = −2 + 12 = 10 ✓.
Bonus (y negative): x = 6, y = −2 → 2(6) + (−2) = 12 − 2 = 10 ✓.
Marking: 2 for three valid pairs with checks; 1 for a negative-x bonus; 1 for a negative-y bonus.