Deep Dive: Simultaneous Equations By Substitution

#CODEX Deep-Dive: Simultaneous Equations by Substitution (WMA11 §1.6)

Provenance — written 2026-09-29. Built from a direct read of research/veridian/WMA11-verified-facts.md (grepped for "simultaneous" and read in full at §1 spec table, §2 assessment structure, §3 mark-scheme conventions, §4.1's own simultaneous-equations entries, §6 and §7, plus the simultaneous-equations-by- substitution.ts lesson-audit log at the file's end); the live lesson file content/veridian/curriculum/wma11/simultaneous-equations-by-substitution.ts (read in full, 1,088 lines, including its own header, every teach block, the diagram, both marked-solutions, the trap-taxonomy and the closing flag); the three companion technique-hub documents in this folder — codex-formula-and-method- bank.md §1.1.4/§2.1.4, forge-exam-technique-guide.md §4.1, signal-frequency-and-boundaries.md §1 — and content/veridian/question-bank/wma11.ts (checked for overlap; its one seeded question, June 2023 Q2, is the same real anchor the lesson already uses). No new primary-source PDF was fetched for this document; every real-exam fact below is drawn from citations already verified in the sources just listed, and every arithmetic step shown (discriminants, factorisations, cross-multiplications) was independently recomputed by hand against those sources' own quoted numbers before being written in here, not copied blind.

#What this document is, and isn't

A single-topic reference for spec point 1.6 only. It does not repeat the lesson's full worked chains (the $y=x+3$/$y=x^2-2x-1$ chain, the chain-drill, the three-method rectangle comparison, either marked-solution, the diagram, or the five MCQs) stage-by-stage — those already exist, are pedagogically paced, and re-typing them here would be restatement, not addition. What follows instead is: the general theory stated as compact reference rather than teaching narrative; the real distinct sub-cases this exact spec point actually covers, several of which the lesson mentions only in passing (as a header disclosure or a trap-taxonomy quote) rather than working through mechanically; short, fresh micro-examples for each; and a single consolidated synthesis of every real WMA11 exam appearance of this topic this course's research has actually found — with the evidentiary gaps stated as plainly as the findings.

Two labels are used consistently below:

• [real] — a verified fact from a genuine Pearson question paper, mark scheme or examiner report, with series and question number given.
• [VERIDIAN, illustrative] — a fresh invented micro-example built to demonstrate a mechanism, with no claim to resemble any specific real Pearson question — the same disclosure convention the lesson file itself uses throughout.

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#1. What spec 1.6 actually asks for

The specification's own wording for this point is exactly nine words: "Solve simultaneous equations; analytical solution by substitution" (spec P1.3, item 1.6). Two things follow directly from that wording and matter for scoping this document: first, there is no restriction anywhere in the spec text to any particular kind of second equation — a line meeting a quadratic is the combination this paper's own lessons teach it through, but the spec point itself covers any pair of equations solvable by substitution; second, "analytical" rules out reading an approximate answer off a graph — the method has to be algebraic, which is why every sub-case below ends in an exact value, never a sketch-and-estimate.

#2. The general mechanism, stated as reference

A simultaneous solution to two equations in $x$ and $y$ is a pair $(x,y)$ satisfying both at once — geometrically, a point common to both graphs. The method is always the same four stages: (1) isolate one variable in whichever of the two equations makes that easiest (almost always a linear one, since it needs no root or square to rearrange); (2) substitute that expression into the other equation, wherever the isolated variable appears; (3) solve the single-variable equation that substitution just produced; (4) take every value found in stage 3 and substitute it back into one of the two original equations to recover the other variable — a coordinate needs both.

Why the method loses nothing and invents nothing. At any point lying on both graphs, the two equations' own expressions for the isolated variable must, at that shared point, be equal — they are two different descriptions of the same number. Substituting one directly for the other doesn't discard any genuine solution (every point on both graphs still satisfies the combined equation, since it satisfied both parts of it individually) and doesn't manufacture a spurious one (any $x$ solving the combined equation, once paired with its correct $y$, satisfies both original equations by construction). This is why the check step — verifying a claimed pair against both original equations, not just the one used for back-substitution — is a genuine safety net rather than a formality: it catches an algebra slip made during rearrangement, which wouldn't show up against the very equation the slip is hiding inside.

What determines the resultant equation's shape. Substituting a linear expression for $y$ into a quadratic curve's equation $y=ax^2+bx+c$ ($a\neq0$) always leaves the $ax^2$ term completely untouched, because substitution only ever replaces $y$ — so the resultant equation is always a genuine quadratic in that case, never something lower-degree by accident. Once it is in $\ldots=0$ form, its discriminant does the same job spec 1.4 already established for a curve meeting the $x$-axis: $b^2-4ac>0$ gives two distinct real $x$-values (two crossing points); $b^2-4ac=0$ gives one repeated value (a tangency — the line touches without crossing); $b^2-4ac<0$ gives no real value at all (the line and curve never meet, however far the picture is drawn).

The one genuine degenerate sub-case the general theory has to admit. If both equations are linear — $y=m_1x+c_1$ and $y=m_2x+c_2$ — substitution gives $m_1x+c_1=m_2x+c_2 \Rightarrow (m_1-m_2)x = c_2-c_1$. Here the "leading term" is $x$ itself, and it survives only when $m_1\neq m_2$. If the two lines share a gradient ($m_1=m_2$), the $x$-terms cancel completely: either $c_1\neq c_2$, and the equation reduces to a false statement with no solution at all (the lines are parallel and distinct — they never meet), or $c_1=c_2$, and the equation reduces to a true statement for every $x$ (the two equations describe the same line, so every point on it is a "solution"). This case never arises when one equation is a genuine quadratic or reciprocal curve, since a line's $y=mx+c$ substituted into $y=ax^2+bx+c'$ can never cancel the $ax^2$ term — only two linear equations can produce it. It is stated here as part of the complete theory rather than as a claim about this paper: no WMA11 anchor located in this course's research turns on this degenerate case, and none is claimed.

#3. The real distinct sub-cases, with the evidence for each

#3.1 Line meets quadratic curve — the shape every worked example on this paper's lessons teaches

Mechanically this is the case worked out in full in §2 above: a genuine quadratic resultant every time, classified directly by its discriminant. Evidentiary status, stated plainly: this is the one sub-case with no real WMA11 anchor at all. A targeted search across four series with cached primary-source text (Jan 2019, Oct 2021, Jan 2023, Jan 2025) for a question pairing a line with a quadratic curve by substitution found none — confirmed independently in both WMA11-verified-facts.md §4.1/§6 and the lesson's own header. Every worked example of this exact combination in the live lesson is therefore [VERIDIAN, illustrative], not a reproduction, and the same is true of anything built here — so this document adds no new worked example for this sub-case, to avoid manufacturing a second illustrative instance where the lesson already has several perfectly good ones. What the general theory in §2 adds beyond the lesson's own teaching prose is the specific reason this sub-case is mechanically the "clean" one: the substitution can never accidentally drop to a lower degree, unlike the two sub-cases below.

#3.2 Line meets reciprocal curve — genuine anchor: Jan 2023, Q7(c)–(d) [real]

This sub-case has a real primary-source anchor, and it differs mechanically from 3.1 in one specific way: because the curve has $x$ in a denominator, direct substitution does not immediately produce a polynomial equation — a clearing-the-fraction step (multiplying both sides by the denominator) has to happen first, and only then does a quadratic (and its discriminant) become available. The real question: line $y=kx+7$ meets curve $y=6/(x-2)$ at two points $P$ and $Q$, with $P$'s $x$-coordinate given as $-4$.

[real] micro-reconstruction, independently recomputed: substituting $x=-4$ into both equations directly (no need to eliminate $y$ at all, since a coordinate is already known) gives $y=-4k+7$ from the line and $y=6/(-4-2)=-1$ from the curve, so $-4k+7=-1 \Rightarrow k=2$ — matching the real mark scheme's own answer [M1 A1, 2 marks]. For $Q$, substitute $k=2$ and equate: $2x+7=6/(x-2)$. Clearing the fraction: $(2x+7)(x-2)=6 \Rightarrow 2x^2+3x-14=6 \Rightarrow 2x^2+3x-20=0$, which factorises as $(x+4)(2x-5)=0$ [check: $2x^2-5x+8x- 20=2x^2+3x-20$ ✓] — recovering $x=-4$ (point $P$, confirming the equation is right) and $x=5/2$ (point $Q$), matching the real mark scheme's $x=5/2$ exactly [dM1 A1]. Back-substituting, $y=2(5/2)+7=12$, giving $Q=(5/2,\,12)$ — the real mark scheme's own stated answer.

Two things this sub-case teaches that 3.1 cannot: first, when one variable's value is already known (as $P$'s $x$-coordinate was here), the fast route is direct substitution into either original equation, not equating the two expressions for $y$ first — the real examiner report records candidates who took the slower, equate-and-rearrange route making "more errors in the algebra" [real, Jan 2023 Q7(c)] than those who substituted the known value directly. Second, a reciprocal expression with a shifted denominator does not distribute over subtraction — the real, quoted algebra error is $6/(x-2)$ rewritten as $(6/x)-3$ [real, Jan 2023 examiner report, quoted in forge-exam-technique-guide.md §4.1], which is not a valid identity for any $x$; the only correct way to clear that denominator is multiplying through by the whole bracket, exactly as shown above, never splitting the fraction term-by-term.

#3.3 Curve meets curve, no line at all — genuine anchor: Jan 2019, Q11(b)–(c) [real]

The general method in §2 never actually required one of the two equations to be linear — "isolate one variable, substitute into the other" works for any pair of equations already (or easily) written as $y=\ldots$. A real WMA11 question confirms this sub-case directly: quadratic $y=x(3-x)$ meeting cubic $y=x(x-2)(5-x)$, asked as a "show that" leading into an exact-value part.

[real] micro-reconstruction, independently recomputed: equating the two expressions for $y$, $3x-x^2 = x(x-2)(5-x)$. Expanding the right side: $x(x-2)(5-x)=x(-x^2+7x-10)=-x^3+7x^2-10x$. Collecting everything to one side: $3x-x^2+x^3-7x^2+10x=0 \Rightarrow x^3-8x^2+13x=0 \Rightarrow x(x^2-8x+13)=0$ — exactly the printed target in the real question, "the solutions to the equation $x(x^2-8x+13)=0$" [real, Jan 2019 Q11(b), 3 marks]. The factor $x=0$ is one genuine intersection (the origin, common to both curves); the quadratic factor $x^2-8x+13=0$ has discriminant $(-8)^2-4(1)(13)=64-52=12>0$, giving $x=\dfrac{8\pm\sqrt{12}}{2}=4\pm\sqrt{3}$ — irrational, which is exactly why the real question's next part asks for an exact coordinate rather than a decimal one, and why it is worth 5 marks rather than being folded into part (b) [real, Jan 2019 Q11(c)].

What this sub-case adds to the general theory: eliminating $y$ between two non-linear curves can produce a resultant of higher degree than either curve alone (degree 2 meets degree 3 here, producing a degree-3 resultant) — the discriminant tool from §2 doesn't apply directly to a cubic, but factoring out a common root (here, $x=0$) can still reduce it to a quadratic the discriminant does apply to. WMA11-verified- facts.md is explicit that this is "genuine spec 1.6 content, but not a line-meets-quadratic pairing" — a different flavour of the same technique, not a counterexample to it.

#3.4 Two linear equations, a coordinate-geometry application — genuine anchor: Jan 2023, Q2(b) [real]

When both equations are linear (finding a rectangle's fourth vertex from its other three), the resultant from §2's degenerate-case analysis is itself linear — no discriminant, no quadratic formula, just one equation in one unknown. This makes the mechanical step for this sub-case genuinely the simplest of the four — the real interest here isn't the algebra's difficulty but its volume: two full line equations (each needing its own gradient calculation) have to be built correctly before substitution even starts, and a real examiner report records that cost showing up in practice, not as a wrong method but as candidates who "rarely produced enough work to score full marks" [real, Jan 2023 Q2(b)] against two other valid, shorter routes (a vector between two known vertices; the shared midpoint of the rectangle's diagonals) that the same report records performing more reliably. The lesson file already builds this into a full three-method, mark-coded comparison and is not repeated here; the point that belongs in this document's theory section is narrower: solving two linear equations by substitution is mechanically the easiest member of this topic's sub-case family, and that is precisely why it is worth checking, before starting it, whether a non-algebraic shortcut (a geometric fact licensing a vector or midpoint argument) makes the substitution step avoidable altogether.

#3.5 Modelling from prose, then a constraint-based choice between two solutions — genuine anchor: June 2023, Q2 [real]

This sub-case adds two skills the mechanism in §2 doesn't itself contain, because it needs the two equations built before substitution can start, and it can leave two algebraically valid answers where substitution alone can't choose between them. First: a real-world scenario in prose ("a rectangular sports pitch has length $x$ metres and width $y$ metres... the perimeter... is 350 m... the area... is 7350 m²") translates directly into $2x+2y=350$ and $xy=7350$ using nothing beyond the standard perimeter and area formulae — the translation itself is not new algebra, but recognising which named quantity maps to which standard formula is [real, June 2023 Q2(a)–(b), 1 mark each]. Second: solving those two equations by substitution ($y=175-x$ into $xy=7350$, giving $x^2-175x+7350=0$, which factorises as $(x-70)(x-105)=0$ [check: $x^2-105x-70x+7350=x^2-175x+7350$ ✓]) produces the two numbers $70$ and $105$ — both genuinely satisfy both equations, and the algebra alone gives no reason to prefer one pairing over the other. The question's own stated restriction, $x>y$, given nowhere in either equation, is what selects $x=105,\,y=70$ as the one labelled answer — and a real examiner report names dropping exactly that restriction as the commonest way to lose the question's final mark despite otherwise-perfect algebra: many candidates "did not realise the requirement that x > y" and left both root-pairings standing with no final selection made [real, June 2023 examiner report, Q2]. The lesson file already carries this question as a full mark-coded worked solution and it is not repeated here; what this document adds is the general framing — a word-problem variant of this topic is really two topics stacked on top of the substitution mechanism itself (translate, then select), and only the middle one is the mechanism §2 describes.

Scope boundary worth stating explicitly, since the vocabulary overlaps: the hidden-quadratic- substitution technique (an unfamiliar exponential like $3^x$ replaced by a new letter $u$ inside a single equation) shares the word "substitution" with this topic but is mechanically a different operation — it eliminates a type of expression within one equation, not a second equation. The live lesson's own header draws this line explicitly, and it is not re-argued here; readers wanting that topic's own real anchors and traps should go to hidden-quadratic-substitution.ts directly, not to this document.

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#4. Synthesis: every real exam angle this specific topic has actually been tested from

Two structurally distinct trap families recur across every real anchor above, independent of which sub-case is involved: stopping too early — reporting a value of the eliminated variable (or, in 3.4/3.5, an unassigned pair of numbers) as if it were the full answer the question asked for, confirmed on 3.2, 3.3 and 3.5 above — and doing more algebra than the question requires — equating and rearranging when a known value could be substituted directly (3.2), or building two full line equations when a geometric shortcut was available (3.4). Neither trap is specific to one sub-case; both recur because the underlying discipline (finish the four stages in §2, and check for a shortcut before starting stage 2) is the same whatever the second equation happens to be.

#5. Honest limits of the evidence

WMA11-verified-facts.md §6 — the section listing which topics have "enough verified material for a genuine standalone lesson" — does not include spec 1.6's archetypal line-meets-quadratic shape in that list, stating directly that the evidence for it is "not yet enough for a worked-example anchor of the exact combination a lesson on this spec point would most want." signal-frequency-and-boundaries.md §1 reports the same gap numerically: of the five WMA11 series with a fully-read examiner report (Jan 2019, Oct 2020, Oct 2021, Jan 2023, Jan 2025), a full line-meets-curve substitution question was seen in 1 of 5 series checked (Jan 2023, Q7) — and that one anchor pairs the line with a reciprocal curve, not a quadratic. Two further genuine anchors (Jan 2019 Q11's curve-meets-curve pairing; June 2023 Q2's word-problem variant) come from series read for one targeted question each, not the full paper, and sit outside that five-series count as corroboration rather than being folded into it. No likelihood beyond that plain count is claimed anywhere in this document, and no grade-boundary figure is used or implied — this topic's own evidence base does not support either.

#6. What this document deliberately does not repeat

The lesson file's diagram (three lines of gradient 2 against one fixed quadratic), its full worked-chain and chain-drill on [VERIDIAN, illustrative] line-meets-quadratic pairs, its three-method rectangle comparison with full mark coding, both of its marked-solutions, its five MCQs, its reference card, and its own trap-taxonomy prose all already exist in simultaneous-equations-by-substitution.ts and are not reconstructed here. This document's purpose was the theory those teaching blocks assume rather than state, and the sub-cases and citation synthesis that sit alongside — not beneath — that existing content.