Simultaneous Equations Linear:Quadratic & Numerical Iteration

SPEC A19 (HIGHER), A20 • MODULE 2: ALGEBRA

Linear/Quadratic Simultaneous & Iteration

Where a line meets a curve, and how to close in on a root one step at a time.

HIGHER TIER ONLY • LESSON 2.24

Two Skills in One Lesson

At Higher tier, simultaneous equations get harder in two ways. You need to handle a system where one equation is a quadratic (or a circle), and you need to find approximate roots using an iterative formula on a calculator.

Part 1 — Linear / Quadratic

A straight line and a curve can meet at up to two points. Solve by substituting the linear into the quadratic, then solving the resulting single-variable quadratic.

Part 2 — Numerical Iteration

An iterative formula xₙ₊₁ = g(xₙ) generates a sequence that closes in on a root. You use the calculator's Ans key to grind out each term.

Calculator Paper

Iteration is effectively calculator-only (Paper 2 or Paper 3) — impossible to grind out many iterates without a calculator. Linear/quadratic can appear on any paper, though it's most common on the calculator papers.

Part 1: Substitute Linear into Quadratic

Your linear equation is easy to rearrange — make y (or x) the subject in one step. Then you have a single expression to drop into the quadratic, eliminating one variable.

STEP 1 Rearrange linear for y (or x) — one variable, one equation.

STEP 2 Substitute into the quadratic — get one equation in one variable.

STEP 3 Solve the quadratic — usually two values of x.

Then Back-Substitute

For each x value, plug back into the linear (simpler) equation to find the matching y. You should end up with two (x, y) pairs — a line typically crosses a curve at two points.

Walkthrough: y = x + 1 and x² + y² = 25

The second equation is a circle — centre origin, radius 5. The first is a straight line with gradient 1 and y-intercept 1. Where do they meet?

Step 1 — substitute

y is already the subject of the linear. Drop y = x + 1 into the circle equation:x² + (x + 1)² = 25

Step 2 — expand and tidy

x² + (x² + 2x + 1) = 25

2x² + 2x + 1 = 25

2x² + 2x − 24 = 0

x² + x − 12 = 0 (divide by 2)

Watch the bracket

(x + 1)² is NOT x² + 1. It's x² + 2x + 1 — the middle term catches most students. Always expand the bracket fully before combining.

Walkthrough: Factorise, Back-Substitute

Step 3 — factorise the quadratic

x² + x − 12 = 0 (x + 4)(x − 3) = 0 x = −4 or x = 3

Step 4 — back-substitute into y = x + 1

x valuey = x + 1Solution pairx = −4y = −4 + 1 = −3(−4, −3)x = 3y = 3 + 1 = 4(3, 4)

Check in the OTHER equation

(−4)² + (−3)² = 16 + 9 = 25 ✓ | 3² + 4² = 9 + 16 = 25 ✓

Both pairs satisfy the circle equation — both are genuine intersection points.

Why Two Solutions? The Graph

The algebra gave two answer pairs because the line crosses the circle at two points. Seeing the geometry makes this obvious and tells you when to expect two solutions.

The red line meets the blue circle at exactly the two points (−4, −3) and (3, 4).

Geometric Intuition

A line can meet a circle at 0, 1, or 2 points (miss, tangent, or chord). For a parabola it's the same picture. Whenever you see a linear–quadratic system, expect two solution pairs unless the discriminant says otherwise.

Examiner Alert: Report BOTH Pairs

Weak Student

Solves the quadratic, gets x = 3, finds y = 4, writes:

"x = 3, y = 4"

Stops. Loses half the marks — there was a second solution pair they never reported.

Strong Student

Recognises the quadratic has two roots and checks both:

"(3, 4) and (−4, −3)"

Reports both pairs as solutions. Full marks.

Examiner alert

Reporting only one (x, y) pair when two exist is a standard AQA trap. If your quadratic has two distinct roots, there are two intersection points — and both must be stated. Write them clearly: "(3, 4) and (−4, −3)".

MCE Worked Example01

Solve the simultaneous equations: y = x − 2 and x² + y² = 10. [5 marks]

M

Method: substitute y = x − 2 into the quadratic; solve for x; back-substitute.

C

x² + (x − 2)² = 10

C

x² + x² − 4x + 4 = 10

C

2x² − 4x − 6 = 0 → x² − 2x − 3 = 0

C

(x − 3)(x + 1) = 0 → x = 3 or x = −1

C

Back-sub: when x = 3, y = 1. When x = −1, y = −3.

E

Solutions: (3, 1) and (−1, −3). Check (3,1): 9 + 1 = 10 ✓. Check (−1,−3): 1 + 9 = 10 ✓.

Part 2: Numerical Iteration

Some equations can't be solved neatly by factorising or the quadratic formula — for example, x³ − x − 1 = 0 has an irrational root. Instead, you can rearrange the equation into a recursive formula, plug in a starting value, and generate a sequence that closes in on the answer.

Iterative formula

xₙ₊₁ = g(xₙ)This says: to get the next term, apply the function g to the current term. The subscript n indexes the term number.

Reading the Suffix Notation

x₀ (or x₁) — the starting value, given in the question.

x₁ — the first iterate: what you get after applying g once.

x₂ — apply g to x₁.

x₃, x₄, ... — each is the previous one plugged into g.

Don't confuse n with x

If the question says "use x₁ = 1", it means start with the value 1, not "substitute x = 1 into the formula once and stop." The whole point of iteration is to keep feeding the output back in.

Worked: xₙ₊₁ = 1 + 1/xₙ, start 1.5

This iteration converges to a famous number — the golden ratio φ ≈ 1.618. Watch how each step gets closer.

The sequence

StepCalculationValuex₀(given)1.5x₁1 + 1 ÷ 1.51.6667x₂1 + 1 ÷ 1.66671.6000x₃1 + 1 ÷ 1.60001.6250x₄1 + 1 ÷ 1.62501.6154x₅1 + 1 ÷ 1.61541.6190...(continues)→ 1.6180...

Convergence

The values bounce slightly above and below but the swings get smaller — the sequence converges to about 1.618 (the golden ratio φ = (1 + √5)/2). That limit is a root of the equation x² − x − 1 = 0.

The Ans Key — Your Calculator Shortcut

Typing each value by hand is slow and error-prone. Your calculator's Ans key stores the last displayed number, so you can chain iterations with a single press of =.

The routine for xₙ₊₁ = 1 + 1/xₙ

TYPE 1 Type 1.5 then press =. Display shows 1.5 — that's x₀. →TYPE 2 Type 1 + 1 ÷ Ans, press =. Display shows x₁. →REPEAT Press = again. Calculator re-runs the formula: next press gives x₂, then x₃, ...

Record as you go

Write down each iterate with full accuracy shown on the screen (at least 4 decimal places). Round only at the very end to the accuracy the question asks for.

Past Paper: Jun 2024 Paper 2 Higher Q22

An authentic AQA question. The examiner report noted that "some did not understand the suffix notation" — so read the set-up carefully.

Question: Use x₁ = 1 and the iterative formula xₙ₊₁ = 5 − 1/xₙ to work out an approximate solution to x = 5 − 1/x. Give your answer to 4 significant figures. [3 marks]

Method

nxₙCalculation11given245 − 1/1 = 434.755 − 1/4 = 4.7544.78955 − 1/4.75 = 4.78947...54.79125 − 1/4.7895 = 4.79121...64.79135 − 1/4.7912 = 4.79128...

Answer

Consecutive iterates agree to 4 s.f.: x ≈ 4.791 (to 4 s.f.)

Examiner Alert: The Suffix Trap

Weak Student

Sees "xₙ₊₁ = 5 − 1/xₙ" and thinks n must be 1, 2, 3 as actual numerical values:

"Try x = 2, x = 3..."

Substitutes the number 2 into the formula at the outset. Doesn't iterate. Zero marks.

Strong Student

Recognises xₙ means "the nth value in the sequence", and uses the starting value x₁ = 1:

"x₂ = 5 − 1/1 = 4..."

Keeps feeding the previous answer back in. Converges. Full marks.

Examiner alert (Jun 2024 P2H Q22)

"Substituting x = 2 or x = 3 into the formula at the outset" was a common wrong response in the written examiner report. The subscript is a label for the term number. Feed each answer back into the formula — not the label number.

Examiner Alert: When to Stop Iterating

Iteration questions always ask for a specific accuracy — usually 3 or 4 decimal places, or a number of significant figures. You must keep going until consecutive terms agree to that accuracy.

Stopping rule

You can stop — and give the answer — once two consecutive iterates round to the same value at the required accuracy. Stopping one step early risks an answer that's 0.001 outside the mark-scheme range.

Weak

"Question asks for 4 s.f., I computed x₃ = 4.75 — done."

x₄ would give 4.79 — different answer. Not converged yet.

Strong

"x₅ = 4.7912, x₆ = 4.7913. Same to 4 s.f. — stop."

Answer: 4.791 to 4 s.f.

Keep the full display

Don't round mid-chain. If you punch 4.75 into the next iteration instead of the full calculator value, your later iterates drift off and the answer can miss the accepted range. Use the Ans key and only round at the end.

Variation: Given a Middle Term

Occasionally, rather than giving you x₁ and asking you to iterate forwards, AQA gives you a term in the middle (like u₂) and asks for the values on either side. This needs iterating both forwards and backwards.

Nov 2024 Paper 2 Higher Q20: Here is a formula for an iterative process: uₙ₊₁ = 24/uₙ + 4. Given u₂ = 8, work out the values of u₁ and u₃. [3 marks]

Forward — find u₃

u₃ = 24/u₂ + 4 u₃ = 24/8 + 4 u₃ = 3 + 4 = 7

Backward — find u₁

u₂ = 24/u₁ + 4 8 = 24/u₁ + 4 24/u₁ = 4 u₁ = 24/4 = 6

Common error (Nov 2024 mark scheme)

Mark scheme explicitly calls out "SC1 u₁ = 7 and u₃ = 7" — students who iterated forward twice rather than backwards once. To find u₁, you must solve the formula for the previous term — not plug u₂ back into it.

Your Turn

Solve the simultaneous equations

y = 2x + 1 and y = x² + 2x − 3

[5 marks]

M

Method: both sides equal y, so set the two right-hand sides equal. This eliminates y in one step.

C

2x + 1 = x² + 2x − 3

C

0 = x² − 4 → x² = 4

C

x = 2 or x = −2

C

Back-sub into y = 2x + 1: when x = 2, y = 5. When x = −2, y = −3.

E

Solutions: (2, 5) and (−2, −3). Check in the parabola: 2² + 2(2) − 3 = 5 ✓; (−2)² + 2(−2) − 3 = −3 ✓.

Common Errors To Avoid

Students often writeCorrect versionWhy it matters

(x + 1)² = x² + 1(x + 1)² = x² + 2x + 1Middle term dropped — whole quadratic wrong

Report only one (x, y) pairReport both pairs from the quadraticHalf the marks lost

Substitute x = 2 into an iterative formula from the startIterate from the given x₁, feeding each answer backNot iterating at all — zero marks

Iterate forward twice when asked for u₁ from u₂Solve the formula backwards for the previous termNov 2024 Q20 standard error

Stop iterating after two or three stepsContinue until consecutive iterates agree to the required accuracyFinal answer outside mark-scheme range

Round mid-chain (e.g. to 2 d.p.)Keep the full calculator display; round only at the endRounding error drift

Quick Self-Check

After any linear/quadratic solution: did the quadratic give two roots? If yes, have you reported both pairs? After any iteration: do the last two values agree to the accuracy asked for?

Summary

Linear/Quadratic Simultaneous — The MCE Flow

M — rearrange linear for y (or x); substitute into quadratic.

C — expand, tidy to 0 = ax² + bx + c, factorise or use the formula.

C — back-substitute each x into the linear equation for y.

E — report both (x, y) pairs; check each in the other equation.

Numerical Iteration — The Routine

Start with the given x₁ (or x₀).

Use the Ans key: type the formula with Ans, press = repeatedly.

Keep the full calculator display each time — no mid-chain rounding.

Stop when two consecutive iterates agree to the required accuracy .

Round the final value to the decimal places / significant figures requested.

Spec Coverage

Spec refContent

A19 HigherSimultaneous equations including linear/quadratic

A20Numerical iteration with suffix notation in recursive formulae

Exam Tip

Iteration is effectively calculator-only — always Paper 2 or Paper 3 (typically 3 marks). Linear/quadratic can appear on any paper (typically 5 marks) but is most common on the calculator papers. Neither is a "⭐ highest priority" topic, but both are reliable mark-winners if you follow the routine.

ExamRizz GCSE Maths (AQA 8300) • Lesson 2.24 • Module 2: Algebra • Spec A19 (Higher), A20 • Higher tier only