Simultaneous Equations

• Two numbers are not an answer. A pair is an answer. If your final line reads “\(x=3\) or \(x=-1\)” you have written down half a solution and left the other half unfound.
• Zones 1 and 2 are about pairs. Zones 3 to 5 are about legality. Zone 6 is the one that catches all five of the others — which is why verification is not optional decoration at the end of a solution.
• One pair, both equations, at the same time. If a candidate satisfies only one of them, it is not a near miss — it is not a solution.
• Choose the method for a stated reason, and write that reason down. “Since \(xy=3\ne0\), division is valid” is one line that turns a step which might have destroyed a solution into one that provably cannot.
• Substitute the bracket, expand on its own line, collect to \(ax^2+bx+c=0\), keep every root, and send each root back through the equation you rearranged. That last step is what produces pairs.
• Elimination is fast but forgetful. It removes a term and, with it, the record of which root belongs to which. Rebuild that record before you write the answer.
• Constants may be cancelled. Variables must be factorised. “Since \(xy=3\ne0\)” is a licence; without it, division is a guess that sometimes deletes an answer.
• Symmetric in, symmetric out. If swapping \(x\) and \(y\) leaves the system unchanged, then every solution has a mirror image, and an answer with an odd number of pairs in it deserves a second look.

• Explain that a simultaneous solution is an ordered pair satisfying both original equations, and that graphically it is a point of intersection.
• Decide, and justify, whether substitution or elimination is the shorter route for a given system.
• Solve a system of one linear and one quadratic or nonlinear equation by substituting the full bracketed expression.
• Expand a squared bracket inside a substitution without losing a sign, and collect to a standard quadratic.
• Solve the reduced quadratic completely — by factorising, by completing the square or by formula — and keep every root.
• Recover the matching second coordinate for each root, and never cross-pair.
• Eliminate a shared term by adding or subtracting two nonlinear equations, then handle the \(\pm\) branches that result.
• Divide one equation by another only after proving the divisor is non-zero, and handle the zero case separately when it is not.
• Recognise when a factorisation such as \(x(y-2)=0\) creates two cases, and solve both.
• State excluded values before clearing denominators, and reject any candidate that violates them.
• Use \((x+y)^2=x^2+2xy+y^2\) and \((x-y)^2=x^2-2xy+y^2\) to solve symmetric systems, generating every sign combination.
• Leave exact answers in surd form, with matching signs written unambiguously.
• Interpret a repeated root as a single ordered pair, and a negative discriminant as no real solution.
• Verify every candidate pair by substitution into both original equations, not the simplified ones.
• Present a final answer as a complete, correctly matched set of ordered pairs.

Why this topic carries weight beyond itself. Every later chapter that asks where a line meets a curve, where two graphs cross, where a tangent touches, or where a stationary point sits ends in a simultaneous system. The discipline you build here — restrictions first, pairs not lists, verify in both originals — is reused for the rest of the course.

Substitution Method

A method of solving simultaneous equations in which one equation is rearranged to give one unknown explicitly in terms of the other, and that whole expression is then written in place of the unknown throughout the second equation. The result is a single equation in one unknown, and each solution of it generates its partner directly by returning to the rearranged equation, which is why the method never leaves ordered pairs unmatched.

Symmetric System

A pair of simultaneous equations that is unchanged when the two unknowns are interchanged, so that every expression in it is built from the sum x + y and the product xy. Such a system is solved most efficiently through the identities (x + y) squared equals x squared plus y squared plus 2xy and (x - y) squared equals x squared plus y squared minus 2xy, which convert the given information into the sum and the difference of the unknowns; because the solutions come in interchanged pairs, each sign combination produced must be tested rather than assumed.

Nonlinear Simultaneous System

A pair of equations in two unknowns in which at least one equation is not linear, so that it contains a squared term, a product of the unknowns, a reciprocal or a similar nonlinear expression. Reducing such a system normally produces a quadratic rather than a linear equation, so it usually has more than one solution, and the reduction step must be chosen to avoid dividing by a quantity that could be zero.

Elimination Method

A method of solving simultaneous equations in which the two equations are scaled so that one term appears identically in both, and are then added or subtracted so that the term cancels, leaving a single equation in one unknown. It applies to nonlinear systems as readily as to linear ones whenever the equations share a term such as x squared, y squared or xy, but because it does not automatically pair each root with its partner, the pairing must be established separately by returning to one of the original equations.

Ordered Pair

A solution of a system of two equations in two unknowns, written (x, y), in which the order of the two components carries meaning: the first is the value of x and the second the value of y that arose from it. The pairing is determined by the equation used to recover the second unknown, so the components cannot be recombined with those of another solution, and a set of x values listed separately from a set of y values is not an answer until the correct pairings are stated.

Division of Equations

A reduction step in which one equation of a system is divided by another, term for term, so that a common factor cancels and a simpler relation between the unknowns remains. The step is valid only when the dividing expression is known to be non-zero, which must be established from the equations themselves before the division is carried out; if the expression can be zero, the correct move is to factorise and treat the zero case as a separate branch, because dividing removes that branch permanently.

Simultaneous Solution

An ordered pair of values, written (x, y), that makes two equations in the same two unknowns true at the same time. Because both equations must hold together, the pair is a single indivisible answer rather than two separate results, and geometrically it is a point at which the graphs of the two equations intersect. A system may have no such pair, exactly one, or several.

Excluded Value

A value of an unknown for which one of the original equations is undefined, most commonly a value that would make a denominator zero. Excluded values must be written down before the denominators are cleared, because multiplying through produces a polynomial equation that is defined at those values even though the original system is not; any candidate solution that lands on an excluded value is therefore an artefact of the clearing step and must be rejected rather than reported.

• “\(x=3\) or \(x=-1\). Done.” WHY IT FAILS The question asked for solutions of a system in two unknowns, and you have answered a question about one unknown. The reduced quadratic was a tool for finding \(x\); it was never the problem. FIX After solving, return to the isolating equation and compute the partner of every root. Finish with \((x,y)=\ldots\).
• “\(x=3\) or \(-1\), and \(y=1\) or \(-3\), so there are four solutions.” WHY IT FAILS The \(y\) values are not an independent pool. Each was generated by one \(x\) through \(y=x-2\), so \((3,-3)\) corresponds to no equation at all. Substituting it gives \(x^2+y^2=18\), not \(10\). FIX Write each pair on the line where you compute it: “\(x=3\Rightarrow y=1\), so \((3,1)\)”. The pairing is then never separated from its source.
• “I’ll solve the first equation, then solve the second.” WHY IT FAILS Each equation alone has infinitely many solutions, so solving them separately produces two infinite families and no information about where they overlap. “Simultaneous” is the whole content of the question. FIX Combine the equations first — substitute or eliminate — so that one unknown disappears.
• “My answer is \(x=2,5\) and \(y=3,0\).” WHY IT FAILS Two coordinate lists do not state which value goes with which. The reader has to guess your matching, and an answer that has to be guessed at has not been stated — even when every number in it is correct. FIX Report ordered pairs: \((2,3)\) and \((5,0)\). Where the question says “coordinates”, use coordinate notation.
• “\(y=x-2\), so \(y^2=x^2-2\).” WHY IT FAILS Squaring is applied to the whole of \(y\), and \(y\) is the entire expression \(x-2\). Dropping the bracket squares only part of it and produces a different equation with different roots. FIX Write \((x-2)^2\) first, on its own, and expand it on the next line: \(x^2-4x+4\).
• “\((x-3)^2=x^2-9\).” WHY IT FAILS This confuses a square with a difference of two squares. Expanding properly, \((x-3)(x-3)=x^2-3x-3x+9=x^2-6x+9\); the middle term is not optional. FIX Expand squared brackets as two binomials until the pattern \(a^2-2ab+b^2\) is automatic. Test with \(x=1\): \((1-3)^2=4\), while \(1-9=-8\).
• “\(-3x(x-3)=-3x^2-9x\).” WHY IT FAILS The negative multiplies both terms in the bracket, and \(-3x\times-3=+9x\). Two negatives give a positive; keeping the sign negative changes the entire reduced quadratic. FIX Expand products involving a leading negative on a line of their own, before combining anything.
• “Subtracting \(5x-2y=11\) from \(4x+6y=24\) gives \(-x+4y=13\).” WHY IT FAILS The subtraction was applied to the first term only. Subtracting \(-2y\) adds \(2y\), so the \(y\) term is \(6y+2y=8y\). FIX Rewrite the subtraction as an addition of the negated equation: add \(-5x+2y=-11\). Every sign is then already changed on the page.
• “\(xy=2x\), so cancel \(x\): \(y=2\).” WHY IT FAILS Cancelling \(x\) assumes \(x\ne0\), and nothing here proves that. The value \(x=0\) satisfies \(xy=2x\) for every \(y\), so the cancellation deletes a whole branch of the solution — silently, with no sign of error in the remaining work. FIX Move everything to one side and factorise: \(x(y-2)=0\), giving the two cases \(x=0\) and \(y=2\). Solve both.
• “\(5y^2=20y\), so \(5y=20\) and \(y=4\).” WHY IT FAILS Same error in a different costume: dividing by \(y\) requires \(y\ne0\), and \(y=0\) is a root. In the circle-and-line example this discards the genuine intersection \((-5,0)\). FIX Never cancel a variable across an equals sign. Write \(5y(y-4)=0\) and take both roots.
• “Multiply through by \(xy\) and carry on.” WHY IT FAILS The cleared equation is equivalent to the original only where \(xy\ne0\). Everywhere else it is a different, larger equation, and it can hand you candidates the original never permitted. FIX Write \(x\ne0,\ y\ne0\) before multiplying, and test every candidate against those exclusions at the end.
• “Dividing \(xy^2\) by \(xy\) is obviously fine.” WHY IT FAILS It is fine here, but only because the companion equation \(xy=3\) proves \(xy\ne0\). Without that proof the same move on a different system loses solutions. An unstated justification is not a justification. FIX Write the licence explicitly: “since \(xy=3\ne0\), we may divide by \(xy\)”. One line, and the method is airtight.
• “\((x-y)^2=4\), so \(x-y=2\).” WHY IT FAILS Every positive number has two square roots. Taking only the positive branch halves the solution set, and in a symmetric system that means losing two of the four ordered pairs. FIX Write \(\pm\) at the moment you take the root, not afterwards: \(x-y=\pm2\).
• “\((x-2)(x+7)=0\), and \(x=2\) looks right, so \(x=2\).” WHY IT FAILS A root is discarded only when something in the question forbids it — a stated restriction, an excluded value, or a context such as a length. “It looks nicer” is not one of those. FIX Carry both roots to the end. If one must go, name the condition that removes it in writing.
• “\(x=1\pm\sqrt3\), so \(x=2.73\) or \(x=-0.73\).” WHY IT FAILS Unless the question asks for a decimal, rounding discards accuracy that was already in your hands, and the rounded values will not verify exactly. Exact form is also easier to check, because conjugate surds cancel. FIX Leave surds as surds, simplified: \(\sqrt{12}=2\sqrt3\), then cancel. Decimalise only on explicit instruction.
• “I checked it in \(x^2-2x-3=0\) and it worked.” WHY IT FAILS That equation is your own product. It carries any error you made while deriving it, and it has forgotten every restriction lost during clearing or dividing. It will confirm a wrong answer without hesitation. FIX Verify in the two equations the question printed. Both of them, for every pair.
• “It satisfies the first equation, so it is a solution.” WHY IT FAILS The first equation alone has infinitely many solutions. Satisfying one equation is the definition of “lies on one curve”, not of “lies on both”. A rejected candidate very often passes exactly one check. FIX Two substitutions per pair. Always both, even when the first is obviously satisfied by construction — then say so and spend the time on the other.
• “The graphs cross at about \((3,1)\), so the answer is \((3,1)\).” WHY IT FAILS A reading from a sketch is an estimate. It cannot distinguish \((3,1)\) from \((2.98,0.98)\), and it demonstrates no method at all. It also silently misses solutions outside the part of the plane you drew. FIX Use the graph to predict how many pairs to expect and to sanity-check your answers. Obtain the answers algebraically.

• On presentation. Finish with a single, unambiguous line such as \((x,y)=(3,1)\) or \((-1,-3)\). Two columns of loose numbers leave the matching for the reader to guess, and an answer that has to be guessed at has not really been given. Where a question says “coordinates”, write them as coordinates.
• What to leave on the page. The substitution line and the collected quadratic are the two steps that show how the system became a single equation. Jumping from the system straight to “\(x=3\) or \(x=-1\)” leaves that reasoning invisible, and the syllabus asks for all necessary working to be shown. It also removes the two lines you would need in order to find your own error if the roots turned out wrong.
• How much verification to write. Under time pressure, one line per pair is enough: “Check \((3,1)\): \(1=3-2\) ✓, \(9+1=10\) ✓.” That is a complete verification and takes about eight seconds. What must never happen is verifying only the pair you feel confident about.
• Self-marking. The tariffs above are this chapter’s own, set to give you a sense of proportion rather than to reproduce any official scheme. Give yourself the reduction credit only if the substituted line and the collected quadratic are both written down, and the answer credit only if the pairs are matched and complete. Solving every quadratic correctly and never matching a partner leaves every question on this set unfinished.

• Additional Mathematics 4037 is examined by two written papers of equal weight, each 2 hours and each carrying 80 marks. Paper 1 is a non-calculator paper; Paper 2 requires a scientific calculator. Both may draw on any part of the syllabus, so Topic 5 can appear in either. There is no multiple-choice component, and the syllabus states that candidates must show all necessary working.
• The split matters in this topic more than in most. On Paper 1 every expansion, every factorisation, every quadratic formula and every surd simplification in this chapter has to be done by hand — which is exactly why the chapter keeps answers exact and teaches you to verify a surd pair by cancelling conjugates rather than by reaching for a decimal.
• The syllabus requires all necessary working to be shown, so each of the following belongs on the page rather than in your head.
• Visible substitution. Write the line where the bracketed expression enters the second equation. Done mentally, the reduction leaves no evidence that it happened.
• A recognisable quadratic. Collect to the form \(ax^2+bx+c=0\) before solving, so that the equation you solved is the equation on the page.
• Both roots, then both partners. A list of roots answers a different question from the one that was asked.