Starting A Level Maths: An Algebra Readiness Check

Before starting A Level Maths, it helps to know which algebra skills you can use without a prompt. A student may remember a formula but lose confidence when a question combines factorising, substitution and interpreting an answer. This readiness check makes those steps visible.

The eight original questions below are a starting point for practice, not an admissions test, a predicted grade or a complete assessment of readiness. Your school sets its own entry requirements and may provide transition work. Use that work alongside this guide.

Why begin with algebra?

Algebra lets you connect numbers, graphs and general relationships. For example, the current AQA A Level Mathematics 7357 algebra content includes indices, surds, quadratics, simultaneous equations, polynomial manipulation and functions. This check selects a few building blocks from those areas; it does not cover the entire course or every awarding body’s route.

The aim is to find the first uncertain step. Someone who expands brackets correctly but chooses the wrong equation needs a different practice task from someone who models the problem correctly and then makes an arithmetic slip.

Try the eight questions first

Use paper and work without a calculator initially. Show each step, including any restrictions on a variable. If a topic is new, mark it “not yet learned” rather than guessing. Set aside enough time to think; this is not a speed test.

1. Evaluate 27^2/3.
2. Factorise 2x² − 7x + 3, then solve 2x² − 7x + 3 = 0.
3. Simplify (x² − 4)/(x − 2), and state the excluded value.
4. Rearrange v² = u² + 2as to make a the subject, assuming s ≠ 0.
5. Solve y = 2x + 1 and y = x² − 2 simultaneously.
6. Write 1/(√5 − 2) with a rational denominator.
7. If f(x) = x² − 3x, expand and simplify f(a + 1).
8. Find the equation of the line through (4, 1) parallel to y = −2x + 7.

For each question, record one of three descriptions: “independent and explained”, “correct with help” or “not yet secure”. Keep that description after marking; an answer copied from a solution does not become an independent success.

Worked solutions and what each reveals

1. Fractional indices

27^2/3 = (∛27)² = 3² = 9. The denominator of the fractional index indicates the root; the numerator gives the power. Multiplying 27 by 2/3 is a different operation. Check your understanding with 16^3/4 = 8.

2. Factorising and solving

2x² − 7x + 3 = (2x − 1)(x − 3). Expanding gives 2x² − 6x − x + 3, which checks the middle term. A product is zero when at least one factor is zero, so x = 1/2 or x = 3. Factorising is rewriting an expression; solving also identifies values that make the equation true.

3. Algebraic fractions and restrictions

x² − 4 = (x − 2)(x + 2). For x ≠ 2, the common factor cancels, leaving x + 2. Keep the restriction x ≠ 2: simplifying does not make the original expression defined at a zero denominator. Test x = 5 to see that both forms give 7.

4. Rearranging a formula

Subtract u² to obtain v² − u² = 2as. Divide the whole left side by 2s: a = (v² − u²)/(2s). Brackets matter. As a check, set u = 2, v = 4 and s = 3. The result is a = 2, and the original equation becomes 16 = 4 + 12.

5. A line meeting a curve

Both expressions equal y, so x² − 2 = 2x + 1. Rearrange to x² − 2x − 3 = 0, then factorise: (x − 3)(x + 1) = 0. Thus x = 3 or −1. Substitute each value into y = 2x + 1 to obtain (3, 7) and (−1, −1). Both coordinates are needed for each intersection.

6. Rationalising a denominator

Multiply the numerator and denominator by √5 + 2. The denominator becomes (√5 − 2)(√5 + 2) = 5 − 4 = 1, so the answer is √5 + 2. You have multiplied by (√5 + 2)/(√5 + 2), which equals 1, preserving the value.

7. Function notation and substitution

Replace every x with the whole expression a + 1: f(a + 1) = (a + 1)² − 3(a + 1). Expanding gives a² + 2a + 1 − 3a − 3 = a² − a − 2. Writing f(a) + 1 answers a different question. Try a = 1: both the original substitution and the simplified expression give −2.

8. A parallel line

The required gradient is −2, so start with y = −2x + c. Substitute (4, 1): 1 = −8 + c, giving c = 9. Therefore y = −2x + 9. The point checks, and the coefficient of x matches the original line’s gradient.

Choose practice from the error, not the total score

• Questions 1 or 6 felt unfamiliar: practise the meaning of roots, indices and surds with simple exact values before longer expressions.
• Questions 2 or 3 broke down: expand to check factors, and distinguish cancelling factors from cancelling terms.
• Question 4 went wrong: write one inverse operation per line and keep the complete numerator in brackets.
• Questions 5 or 8 were difficult to start: practise translating a graph or relationship into equations before solving.
• Question 7 caused trouble: substitute into brackets first, then simplify. Separate the input from the function’s output.

Do not use these categories to decide whether someone is “an A Level student”. They identify manageable practice tasks. A teacher can judge which gaps need attention before the student’s particular course starts.

A manageable transition routine

Choose two weak skills for the first week. In each short session, explain a worked method, close it, attempt a fresh question and check the reasoning. At the end of the week, mix the skills with some questions you already know. Return to the weak skills the following week without the original solutions beside you.

For broader planning once the course is underway, read our A Level revision techniques guide. Keep your school’s transition tasks and specification as the main reference.

Explore A Level tutoring support or A Level Maths tutoring. If you would like to discuss specific gaps, request a free trial, sharing your board and a short description of where the working breaks down. Trial arrangements depend on confirmation.

Official course reference checked 14 October 2026. All eight questions and worked solutions are original learning examples, not official assessment items or evidence of exam-board endorsement.