HookThe word 'algebra' is an instruction: restore the balance
Around the year 820, in the House of Wisdom in Baghdad, the mathematician Muhammad ibn Musa al-Khwarizmi wrote a book whose Arabic title contained the phrase al-jabr — 'the restoring' or 'the reunion of broken parts'. Latin scholars mangled al-jabr into 'algebra', and mangled al-Khwarizmi's own name into 'algorithm'. But the word he chose was a precise description of a method, not a subject: al-jabr meant taking a term away from one side of an equation and restoring it on the other to keep the two sides balanced. Twelve hundred years later, that is still exactly what you do.
An equation is a set of weighing scales. The equals sign is the pivot, and whatever sits on the left weighs precisely the same as whatever sits on the right. Solving it means peeling away everything around the unknown until it stands alone — and the one unbreakable rule is that anything you do to one side you must do to the other, or the scales tip and the equation stops being true. Add 3 to the left, add 3 to the right. Divide the left by 5, divide the right by 5. Everything in A2.3 — linear equations, equations you build yourself from a worded situation, pairs of simultaneous equations, quadratics, and inequalities — is that single balancing act applied to harder and harder arrangements. It is examined on both the calculator and non-calculator papers, and it is the machinery every science and finance question secretly runs on.
ModelLinear equations — undo the arithmetic in reverse
To solve 3x + 7 = 22, ask what has been done to x and undo it in reverse order. x was multiplied by 3, then 7 was added. So peel from the outside in: subtract 7 from both sides to get 3x = 15, then divide both sides by 3 to get x = 5. Each step is an inverse operation — the opposite that cancels what was done — and doing it to both sides keeps the scales level. This is why 'moving a number across the equals sign changes its sign': you are not really moving it, you are subtracting it from both sides, and the version left behind is negative.
The harder case AQA tests is the unknown on both sides, like 5x − 3 = 2x + 12. Here you must first gather the x-terms on one side and the numbers on the other. Subtract 2x from both sides so the x's collect on the left: 3x − 3 = 12. Now it is an ordinary equation — add 3 to both sides for 3x = 15, divide by 3 for x = 5. A reliable tactic is to move the x-terms to whichever side keeps them positive, because negative-x equations are where sign errors breed.
The single most valuable habit is checking by substitution, which turns a guess into a certainty for free. The classic Foundation mistake is treating 3x = 15 as though it says 'subtract': writing x = 15 − 3 = 12. It does not — 3x means 3 times x, so the inverse is to divide, giving x = 5.
Solve 5x − 3 = 2x + 12, showing every line. Subtract 2x from both sides to collect the unknowns: 5x − 2x − 3 = 12, which is 3x − 3 = 12. Add 3 to both sides: 3x = 15. Divide both sides by 3: x = 5. Now check by substituting back into the original: left-hand side = 5 × 5 − 3 = 22; right-hand side = 2 × 5 + 12 = 22. Both sides equal 22, so x = 5 is correct. On the non-calculator paper every one of those lines is a potential method mark, and the substitution check is what stops a careless slip becoming a wrong answer.
MechanismTurning words into equations — the skill exams reward most
Half the algebra marks at Foundation are hidden inside worded problems, because before you can solve an equation you have to build it. The method is to name the unknown with a letter, translate each phrase into algebra, and set the two sides equal. 'I think of a number, multiply it by 4 and subtract 5, and the answer is 23' becomes 4n − 5 = 23 — then solve it exactly as before: 4n = 28, n = 7.
Real situations follow the same route. A charge that has a fixed part plus a per-unit part is always fixed + rate × quantity, the same y = mx + c structure from the graphs section. A plumber's £40 call-out plus £30 an hour for h hours gives a bill of 40 + 30h. If the bill came to £160, you form 40 + 30h = 160 and solve for the hours. The power of this is that once the equation is built, the arithmetic is routine — so examiners load the marks onto the forming, and a correctly set-up equation scores even if you never finish solving it.
The interpretation step is where careless candidates lose the final mark. After solving, read the answer back into the situation: 'h = 4, so the plumber was there 4 hours.' A bare '4' can be marked down if the question asked for a number of hours, a length, or a price, and if a context makes a negative or fractional answer impossible, say so.
A taxi charges a fixed £3 booking fee plus £1.50 for each mile. Priya's journey cost £16.50. How many miles was it? Let the number of miles be m. The cost is the fixed fee plus the per-mile charge: 3 + 1.5m = 16.50. Subtract the fixed fee from both sides: 1.5m = 13.50. Divide both sides by 1.5: m = 9. So the journey was 9 miles. Check it against the story: 3 + 1.5 × 9 = 3 + 13.50 = £16.50 ✓. The two marks live in writing 3 + 1.5m = 16.50 correctly; the solving is the easy part, and interpreting 'm = 9' as '9 miles' secures the last one.
MechanismSimultaneous equations — two unknowns need two facts
One equation with two unknowns, like x + y = 10, has endless solutions — (1, 9), (2, 8), (3, 7) all fit. To pin down a unique pair you need a second, different equation linking the same two letters. Solving the pair together is called solving simultaneous equations, and the Foundation method is elimination: combine the two equations so that one letter cancels, leaving a single ordinary equation you already know how to solve.
The trick is that adding or subtracting two true equations gives another true equation — the balance is preserved on both. If the two equations have a matching term, subtracting eliminates it. Take 3x + 2y = 16 and x + 2y = 8: both contain +2y, so subtracting the second from the first removes the y entirely — (3x − x) + (2y − 2y) = 16 − 8 — leaving 2x = 8, so x = 4. Then substitute x = 4 back into either original equation to find y. When the matching terms have the same sign you subtract; when they have opposite signs you add.
A solution is a pair, not a single number — the point (x, y) where the two lines would cross if you drew them. That geometric picture is also how you solve them approximately from a graph: plot both straight lines and read off the coordinates of the intersection. The commonest error is stopping at x = 4 and forgetting y, or finding both but never checking them in the second equation.
Solve 3x + 2y = 16 and x + 2y = 8 simultaneously. Both equations contain +2y, so subtract the second from the first to eliminate y: (3x − x) + (2y − 2y) = 16 − 8, giving 2x = 8, so x = 4. Substitute x = 4 into the simpler equation x + 2y = 8: 4 + 2y = 8, so 2y = 4 and y = 2. The solution is x = 4, y = 2 — the point (4, 2). Always check in the equation you did not use for substitution: 3 × 4 + 2 × 2 = 12 + 4 = 16 ✓. Both numbers are needed for full marks; an answer of 'x = 4' alone is only half the job.
MechanismQuadratic equations — the trick of the zero
A quadratic equation contains an x² term, like x² + 3x − 10 = 0. The Foundation method is to factorise and then use a fact so simple it is easy to miss: if two things multiply to give zero, at least one of them must be zero. Nothing else can produce a product of zero. So once you write the quadratic as two brackets multiplied together, equal to zero, you can set each bracket to zero separately.
Factorise x² + 3x − 10 by finding two numbers that multiply to −10 and add to +3. Test the factor pairs of 10: +5 and −2 multiply to −10 and add to +3. So x² + 3x − 10 = (x + 5)(x − 2), and the equation becomes (x + 5)(x − 2) = 0. Now apply the zero fact: either x + 5 = 0, giving x = −5, or x − 2 = 0, giving x = 2. A quadratic typically has two solutions, and dropping one is the classic error — students find x = 2, feel finished, and never notice x = −5.
The non-negotiable first move is that the equation must equal zero before you factorise, because the whole method depends on the product being zero. If a question gives x² + 3x = 10, rearrange to x² + 3x − 10 = 0 first. (At Foundation you only ever factorise; the quadratic formula, completing the square and rearranging harder quadratics are Higher-tier tools.) You can also read the two solutions approximately off a graph — they are exactly where the parabola crosses the x-axis.
Solve x² + 3x − 10 = 0. It already equals zero, so factorise: find two numbers multiplying to −10 and adding to +3 — those are +5 and −2 — giving (x + 5)(x − 2) = 0. Set each bracket to zero: x + 5 = 0 gives x = −5; x − 2 = 0 gives x = 2. The two solutions are x = −5 and x = 2. Check both in the original: for x = 2, (2)² + 3(2) − 10 = 4 + 6 − 10 = 0 ✓; for x = −5, (−5)² + 3(−5) − 10 = 25 − 15 − 10 = 0 ✓. Both check out, so both are genuine roots — and writing both is what earns the second mark.
ModelInequalities — solve like an equation, but watch the flip
An inequality such as 3x − 5 ≤ 7 is solved with the same balancing moves as an equation: add 5 to both sides for 3x ≤ 12, divide both sides by 3 for x ≤ 4. The answer is not a single number but a range — every value of x up to and including 4 satisfies it. Writing the answer as '4' instead of 'x ≤ 4' throws away the mark, because the question asked which values work, and the range is the answer.
There is exactly one extra rule, and it is the one AQA probes: multiplying or dividing both sides by a negative number reverses the inequality sign. To solve 4 − 2x < 10, subtract 4 to get −2x < 6, then divide by −2 — and flip the < to > — giving x > −3. The flip is not a quirk to memorise blindly; it is forced by the number line. Multiplying by a negative reflects every value to the other side of zero, so the one that was smaller is now larger, and the sign has to turn to stay true. Test it: −5 < 3 is true, but multiply both sides by −1 and 5 < −3 is false; you must flip to 5 > −3.
To represent the solution on a number line, draw the boundary value and shade the range. Use an open (unfilled) circle for < or > because the endpoint is not included, and a closed (filled) circle for ≤ or ≥ because it is. The open-versus-closed distinction is marked in its own right, and it is the same ≤/< difference that error intervals were built on earlier in the course.
Solve 3x − 5 ≤ 7, then 4 − 2x < 10, and show each on a number line. First: add 5 to both sides, 3x ≤ 12; divide by 3, x ≤ 4. On the number line, a closed circle at 4 with shading to the left, because 4 is included. Second: subtract 4 from both sides, −2x < 6; divide by −2 and flip the sign, x > −3. On the number line, an open circle at −3 with shading to the right, because −3 is not included. Check a value in the range of the second, say x = 0: 4 − 2 × 0 = 4, and 4 < 10 is true, and 0 > −3 is true — consistent. The flip on the second inequality is the mark most candidates lose.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
On the non-calculator paper this section is marked line by line, so write every balancing step down: the marks are M marks for correct method, and a complete solution with one slipped digit keeps most of them, while a bare wrong answer keeps nothing. Show the operation you are doing to both sides, and check your final value by substituting it back into the original equation — that check is free insurance and catches the sign errors examiners deliberately set up with unknown-on-both-sides questions.
Read the command word precisely. 'Solve' means find the value(s) algebraically and show working; 'solve graphically' or 'use the graph' means read the answer off the intersection or the x-axis crossings, and an algebraic answer there can miss the point. For quadratics, always rearrange to equal zero before factorising and always give both solutions. For worded problems the biggest marks are in forming the equation — define your letter, translate each phrase, and set the sides equal — because a correct set-up scores even if the solving goes wrong, and interpreting the final value back into the context ('so it was 9 miles') secures the last mark.
Inequalities have two examiner traps worth rehearsing: the answer must be an inequality such as x > −3, not a lone number, and the sign must flip whenever you divide by a negative. On the number line, the open-versus-closed circle is a separate mark — open for strict < or >, closed for ≤ or ≥ — so match the circle to the symbol every time. Across both papers, label simultaneous solutions clearly as a pair, and never leave a quadratic with only one of its two roots.