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This final lesson brings together all the algebra topics from this course and focuses on the types of questions you will encounter in the AQA GCSE Mathematics exam. We will look at common mistakes, effective problem-solving strategies, and a set of exam-style questions with full worked solutions. Use this lesson to consolidate your knowledge and practise under exam-like conditions.
The AQA examiners' reports consistently highlight the same mistakes year after year. Being aware of these will help you avoid them.
| Common Mistake | Correct Approach |
|---|---|
| Expanding (x + 3)^2 as x^2 + 9 | (x + 3)^2 = x^2 + 6x + 9 — always use FOIL |
| Combining unlike terms: 3x + 2x^2 = 5x^2 | These are NOT like terms — leave as 2x^2 + 3x |
| Forgetting both solutions to a quadratic | A quadratic has TWO solutions (or a repeated root) |
| Sign errors when expanding negative brackets | -2(x - 3) = -2x + 6, not -2x - 6 |
| Dropping the negative when squaring: (-3)^2 = -9 | (-3)^2 = 9 because negative times negative = positive |
| Not reversing inequality when dividing by negative | If -2x > 6, then x < -3 (sign flips) |
| Cancelling terms instead of factors in fractions | (x + 3)/(x + 1) cannot be simplified by "cancelling x" |
| Writing the nth term as "n + 3" when it should be "3n + ..." | The common difference goes BEFORE n |
Exam Tip: Before handing in your paper, go back and check for these specific errors. Most of them take only a few seconds to spot and correct, but each one could cost you 1-2 marks.
When you encounter a multi-step algebra problem in the exam, use the following approach:
flowchart TD
A[Read the question carefully] --> B[Identify what you are asked to find]
B --> C[Decide which algebra technique to use]
C --> D[Show each step clearly]
D --> E[Check: does the answer make sense?]
E --> F[Write a clear concluding statement]
Simplify: 3a + 5b - a + 2b
Solution:
Collect like terms:
Answer: 2a + 7b
Expand and simplify: 4(2x + 3) - 3(x - 2)
Solution:
Expand: 8x + 12 - 3x + 6
Collect like terms: 8x - 3x + 12 + 6
Answer: 5x + 18
Solve: 5(2x - 1) = 3(x + 4)
Solution:
Expand: 10x - 5 = 3x + 12
Subtract 3x: 7x - 5 = 12
Add 5: 7x = 17
Divide by 7: x = 17/7
Answer: x = 17/7 (or x = 2 and 3/7)
Factorise: x^2 - x - 20
Solution:
Find two numbers that multiply to -20 and add to -1: -5 and +4.
Answer: (x - 5)(x + 4)
Check: (x - 5)(x + 4) = x^2 + 4x - 5x - 20 = x^2 - x - 20. Correct.
Solve: x^2 + 2x - 15 = 0
Solution:
Find two numbers that multiply to -15 and add to 2: 5 and -3.
(x + 5)(x - 3) = 0
x + 5 = 0 gives x = -5
x - 3 = 0 gives x = 3
Answer: x = -5 or x = 3
Solve simultaneously:
3x + 2y = 16
x - y = 2
Solution:
From the second equation: x = y + 2
Substitute into the first: 3(y + 2) + 2y = 16
3y + 6 + 2y = 16
5y + 6 = 16
5y = 10
y = 2
Substitute back: x = 2 + 2 = 4
Answer: x = 4, y = 2
Check: 3(4) + 2(2) = 12 + 4 = 16. Correct. 4 - 2 = 2. Correct.
(a) Solve: 4x + 3 > 19
4x > 16
x > 4
(b) Solve: -3x <= 12
Divide by -3 and reverse the sign:
x >= -4
(c) List the integer values of x that satisfy both inequalities above.
x > 4 AND x >= -4
Combining: x > 4
The question asks for integer values. Since no upper limit is given, the first few integers are: 5, 6, 7, 8, ...
If the question specified a range, list accordingly.
The nth term of a sequence is 4n - 7.
(a) Find the first three terms.
n = 1: 4(1) - 7 = -3
n = 2: 4(2) - 7 = 1
n = 3: 4(3) - 7 = 5
First three terms: -3, 1, 5
(b) Is 101 a term in the sequence?
4n - 7 = 101
4n = 108
n = 27
Since 27 is a positive integer, yes, 101 is the 27th term.
Find the equation of the line perpendicular to y = 4x + 1 that passes through the point (8, 3).
Solution:
Gradient of given line = 4
Perpendicular gradient = -1/4
Substitute (8, 3) into y = (-1/4)x + c:
3 = (-1/4)(8) + c
3 = -2 + c
c = 5
Answer: y = (-1/4)x + 5 or equivalently y = -x/4 + 5
A farmer has 100 metres of fencing to enclose a rectangular pen against an existing wall. The wall forms one of the longer sides, so fencing is needed for only three sides. The width of the pen is x metres.
(a) Show that the length of the pen is (100 - 2x) metres.
The fencing covers two widths and one length: 2x + length = 100
Therefore length = 100 - 2x.
(b) The area of the pen is 1200 m^2. Form an equation and find the dimensions.
Area = x(100 - 2x) = 1200
100x - 2x^2 = 1200
2x^2 - 100x + 1200 = 0
x^2 - 50x + 600 = 0
(x - 20)(x - 30) = 0
x = 20 or x = 30
When x = 20: length = 100 - 40 = 60 m
When x = 30: length = 100 - 60 = 40 m
Both solutions are valid: the pen is either 20 m by 60 m or 30 m by 40 m.
Exam Tip: In "show that" questions, you must clearly demonstrate each step leading to the given result. Even if you can see the answer immediately, write out the algebraic working. Marks are awarded for the METHOD, not just the conclusion.
| Marks | What the Examiner Expects |
|---|---|
| 1 | A single correct step or recall of a fact |
| 2 | Two steps: usually form and solve, or identify and apply |
| 3 | A short chain of reasoning: typically 3 clear steps of working |
| 4 | Multiple techniques combined: e.g. form equation, expand, solve, interpret |
| 5-6 | Extended problem: set up, manipulate, solve, check, and give contextual answer |
Use this checklist to make sure you have covered every algebra topic:
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