Algebra Exam Practice

Having completed the final topic of the course, use the seven Edexcel-style questions below as a capstone check of every algebra skill covered in the nine previous lessons: simplifying, expanding and factorising, linear and quadratic equations, simultaneous equations, inequalities, sequences, straight-line and quadratic graphs, and functions. Each question is written in the style and mark allocation of an actual Edexcel GCSE Mathematics (1MA1) paper, with mark-scheme-style solutions and examiner notes so you can see exactly where marks are awarded and commonly lost.

Attempt these under timed conditions: target 35 minutes for the full set (the total is 35 marks, which at 1 minute per mark is the standard Edexcel pace). Then self-mark against the solutions and the examiner notes.

Question A1: Simplifying and Expanding (3 marks, Foundation)

Expand and simplify (2x + 5)(x − 3).

Mark-scheme solution

Expand term by term using the grid or FOIL method: $2x \times x + 2x \times (-3) + 5 \times x + 5 \times (-3)$2x×x+2x×(−3)+5×x+5×(−3). (M1)

$= 2x^{2} - 6x + 5x - 15$=2x2−6x+5x−15. (M1)

= $2x^{2} - x - 15$2x2−x−15. (A1)

Examiner note. The most common error is writing $2x^{2} - 6x + 5x - 15$2x2−6x+5x−15 and stopping there — candidates forget to collect the like terms −6x and 5x, losing the A1. Another trap is a sign slip on $2x \times (-3)$2x×(−3), written as +6x: if you produce $2x^{2} + 11x - 15$2x2+11x−15 you have made this mistake. Always double-check the middle term.

Question A2: Solving a Linear Equation with Brackets and Fractions (3 marks, Foundation)

Solve 3(2x − 1) = (x + 11) / 2.

Mark-scheme solution

Multiply both sides by 2: 6(2x − 1) = x + 11. (M1)

Expand: 12x − 6 = x + 11. (M1)

Collect: 11x = 17, so x = 17/11 (or 1.545 to 3 d.p.). (A1)

Examiner note. Clearing the fraction first (by multiplying both sides by 2) is the cleanest route. Candidates who try to leave the fraction in place typically lose track of signs after expanding. If you prefer to work in decimals, x = 1.5454… is a perfectly acceptable final answer, but give at least 3 significant figures. An improper fraction like 17/11 is marked correct — there is no need to convert to a mixed number.

Question A3: Factorising and Solving a Quadratic (4 marks, Foundation/Higher crossover)

Solve $x^{2} - 7x + 12 = 0$x2−7x+12=0 by factorising.

Mark-scheme solution

Factorise: (x − 3)(x − 4) = 0. (M1 for attempt at factorisation; A1 for correct factors)

Set each factor to zero: x − 3 = 0 or x − 4 = 0. (M1)

x = 3 or x = 4. (A1)

Examiner note. You need two numbers that multiply to +12 and add to −7 — those are −3 and −4. A common mistake is using +3 and +4, which give $x^{2} + 7x + 12 ($x2+7x+12(wrong sign on the middle term). Always verify by quickly expanding: $(x - 3)(x - 4) = x^{2} - 7x + 12$(x−3)(x−4)=x2−7x+12 ✓. The A1 for the final answers is only awarded if both solutions are stated; writing only x = 3 scores 3/4. On non-calculator papers, factorisation is the expected method — candidates who use the quadratic formula for an easily-factorisable quadratic are taking unnecessary risk.

Question A4: Simultaneous Equations (4 marks, Higher)

Solve the simultaneous equations

3x + 2y = 16

5x − 2y = 8

Mark-scheme solution

Notice the y coefficients are +2 and −2 — add the equations to eliminate y. (M1)

(3x + 5x) + (2y − 2y) = 16 + 8

8x = 24, so x = 3. (A1)

Substitute into 3x + 2y = 16: $9 + 2y = 16 \to 2y = 7 \to y = 3.5$9+2y=16→2y=7→y=3.5. (M1)(A1)

Solution: x = 3, y = 3.5.

Examiner note. Always check both pairs in the other equation: 5(3) − 2(3.5) = 15 − 7 = 8 ✓. Candidates routinely find x correctly but make an arithmetic slip when substituting back — the check takes ten seconds and catches this. If the y coefficients had matched signs (e.g. +2 and +2), you would subtract instead of adding. A reliable rule: Same Signs Subtract, different signs add.

Question A5: Inequality with a Negative Coefficient (3 marks, Higher)

Solve the inequality $5 - 3x \geq 14$5−3x≥14 and represent the solution on a number line.

Mark-scheme solution

Subtract 5: $-3x \geq 9$−3x≥9. (M1)

Divide both sides by −3, reversing the inequality sign: $x \leq -3$x≤−3. (M1)

Number line: closed circle at −3, arrow pointing left. (A1)

Examiner note. This is the headline inequality trap: dividing (or multiplying) by a negative number flips the inequality sign. Candidates who forget typically write $x \geq -3$x≥−3 and lose the method mark. A closed circle on the number line indicates that x = −3 is included (because of the $\geq$≥ becoming $\leq)$≤); an open circle would mean "strictly less than". Edexcel mark schemes are explicit about this — an open circle for $a \leq$a≤ solution loses the A1, even if the algebra is correct.

Question A6: Nth Term of a Quadratic Sequence [H] (4 marks, Higher)

Find the nth term of the sequence 3, 10, 21, 36, 55, …

Mark-scheme solution

First differences: 7, 11, 15, 19 (not constant).

Second differences: 4, 4, 4 (constant). (M1)

Since the second difference is 4, the coefficient of $n^{2}$n2 is $4 \div 2 =$4÷2= 2. (M1)

So the nth term starts with $2n^{2}$2n2. Check: $2(1)^{2} = 2$2(1)2=2, $2(2)^{2} = 8$2(2)2=8, $2(3)^{2} = 18$2(3)2=18, $2(4)^{2} = 32$2(4)2=32, $2(5)^{2} = 50$2(5)2=50.

Subtract these from the actual sequence: 3 − 2 = 1, 10 − 8 = 2, 21 − 18 = 3, 36 − 32 = 4, 55 − 50 = 5.

The residual sequence 1, 2, 3, 4, 5 has nth term n. (M1)

Nth $term = 2n^{2} + n$term=2n2+n. (A1)

Verify: $2(1)^{2} + 1 = 3$2(1)2+1=3 ✓; $2(5)^{2} + 5 = 55$2(5)2+5=55 ✓.

Examiner note. Quadratic nth term questions are Higher-only territory. The structured method — constant second difference tells you the $n^{2}$n2 coefficient, then find the residual linear part — is bullet-proof if you follow it step by step. Candidates who try to guess the formula often arrive at the wrong middle term. Always verify the first and last term of your formula against the original sequence.

Question A7: Functions — Composite and Inverse [H] (6 marks, Higher)

f(x) = 3x − 2 and $g(x) = x^{2} + 1$g(x)=x2+1.

(a) Find gf(x), simplifying your answer. (2)

(b) Find $f^{-1}(x)$f−1(x). (2)

(c) Solve gf(x) = 10. (2)

Mark-scheme solution

(a) gf(x) means apply f first, then g. $gf(x) = g(3x - 2) = (3x - 2)^{2} + 1$gf(x)=g(3x−2)=(3x−2)2+1. (M1)

Expand: $(3x - 2)^{2} = 9x^{2} - 12x + 4$(3x−2)2=9x2−12x+4. So gf(x) = $9x^{2} - 12x + 5$9x2−12x+5. (A1)

(b) Let y = 3x − 2. Swap: x = 3y − 2. Rearrange: 3y = x + 2, so y = (x + 2)/3. (M1)

$f^{-1}(x) = (x + 2)/3$f−1(x)=(x+2)/3. (A1)

(c) Set gf(x) = 10: $9x^{2} - 12x + 5 = 10 \to 9x^{2} - 12x - 5 = 0$9x2−12x+5=10→9x2−12x−5=0. (M1)

Factorise: (3x − 5)(3x + 1) = 0. So x = 5/3 or x = −1/3. (A1)

Examiner note. This question packs composite functions, inverse functions, and quadratic solving into a single 6-mark item — the kind of combined algebra question Edexcel uses to distinguish grade 7 from grade 9. Critical moves: (i) gf means g of f, not f of g — candidates who compute fg here get $(x^{2} + 1 \to 3(x^{2} + 1) - 2 = 3x^{2} + 1)$(x2+1→3(x2+1)−2=3x2+1) and lose four marks; (ii) expanding $(3x - 2)^{2}$(3x−2)2 correctly requires the middle term −12x, not −6x — if you write $(3x - 2)^{2} = 9x^{2} + 4$(3x−2)2=9x2+4 you have lost both (a) and (c). Part (c) offers a second chance to earn a method mark: any correct quadratic in x (even from an incorrect part a) can still earn the M1 in (c) if solved correctly.

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Edexcel Mark-Scheme Quick Reference

Mark type	What earns it	Key phrase in mark scheme
M1 (method)	Correct method shown, even if the arithmetic is wrong	"for a correct method"
A1 (accuracy)	Correct final value, usually dependent on the M1	"cao" (correct answer only)
B1 (independent)	A correct standalone result (e.g. units, a conversion, a statement)	"for [specific value]"
dep	Awarded only if the preceding mark has been earned	"dep on M1"

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Common Algebra Mistakes Across the Whole Course

The questions above cover most of the high-frequency errors Edexcel examiners see in the algebra strand. The table summarises the headline issues from every lesson, with the lesson where each one was first explained.