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.

• “\(x=3\) 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.

• Every candidate takes both Cambridge IGCSE Additional Mathematics 0606 papers. Each is 2 hours, each carries 80 marks and each is worth half the qualification. Paper 1 is a non-calculator paper; Paper 2 requires a scientific calculator. Both consist of structured and unstructured questions, all of them compulsory, and either may assess any part of the subject content — so Topic 5 can appear on either. AO1 and AO2 each account for 45–55% of every paper and of the qualification. The syllabus states plainly that candidates must show all necessary working. The examination is available in the June and November series, and in the March series in India, for this cycle, and the candidate grade range is A* to E.
• 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.