Simultaneous Equations Linear:Linear — Algebraic & Graphical Methods

SPEC A19 (ADD) • ALGEBRA

Simultaneous Equations: Linear/Linear

⭐ High-priority. Foundation (Additional) + Higher. Two equations, two unknowns: solve algebraically by elimination or substitution, or read off the intersection on a graph.

Two Equations, Two Unknowns

A pair of simultaneous equations in x and y has one solution: the unique pair (x, y) that satisfies both equations at the same time. There are three standard methods: elimination, substitution, and graphical.

Authentic exam evidence

Jun 2023 P1H Q16: "Solve simultaneously: 2x − 5y = 13; 3x + 4y = 8" — 4 marks. Many students lost marks at the subtraction/addition stage.

Nov 2023 P3H Q11: "Solve: 7x + 2y = 100; 3x + 2y = 48" — 3 marks. Coefficients already match (subtract directly).

Jun 2024 P1H Q14: "4 chocolate bars and 3 packets of mints cost £4.70; 5 chocolate bars and 1 packet of mints cost £4.50. Find the cost of each." — 4 marks contextual.

Tier note

Foundation (Additional): linear/linear by algebra and from a graph. Linear/quadratic systems and iteration are Higher only — covered in Lesson 2.24.

Key Terms

Simultaneous equations

Two (or more) equations that must be true at the same time. The solution is the value(s) of the unknowns that satisfy every equation — not just one.

Solution as an ordered pair

For two equations in x and y, the answer is a pair (x, y) — geometrically, the point where the two lines meet. Both values are required; finding only one loses half the marks.

Coefficient

The number multiplying a variable. In 3x + 2y = 16, the coefficient of x is 3 and the coefficient of y is 2. Elimination works by making one set of coefficients match.

Three Methods Overview

MethodBest whenRisk

EliminationBoth equations are in standard form ( ax + by = c )Sign errors when subtracting

SubstitutionOne equation already has y = … (or x = …)Bracket / sign errors when expanding

GraphicalApproximate solution wanted, or graphs already drawnOnly as accurate as the grid

Pick the fastest route

Both algebraic methods give the same answer. Choose the one that matches the form of the equations: standard form → elimination; one variable already isolated → substitution.

Elimination — Procedure

1Match coefficients. Multiply one or both equations so that the coefficients of one variable are equal in size (the same number, possibly with different signs).

2Add or subtract. If signs are opposite, ADD the equations to eliminate. If signs are the same, SUBTRACT to eliminate. Either way, one variable disappears.

3Solve the resulting one-variable equation.

4Substitute back into either original equation to find the second variable.

5Check by substituting both values into the other original equation.

"Same Sign Subtract, Different Sign Add" (SSS / DSA)

When the matching coefficients have the same sign, subtract. When they have different signs, add. This rule eliminates the variable cleanly every time.

Different signs → ADD

3x + 2y = 16

5x − 2y = 8

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8x = 24 → x = 3

(2y and −2y cancel when added)

Same signs → SUBTRACT

7x + 2y = 100

3x + 2y = 48

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4x = 52 → x = 13

(2y − 2y = 0 when subtracted)

Subtraction sign trap

When subtracting, you must negate every term in the second equation, not just the matching one. Many students lose the A1 here. Safer alternative: multiply the second equation by −1 first, then ADD. Same result, fewer sign errors.

When Coefficients Don't Match

Most simultaneous equations need multiplication first to make a pair of coefficients match. Pick whichever variable gives the smallest multipliers.

Substitution Method

If one equation is already rearranged to y = … (or x = …), substitute that expression into the other equation. This usually beats elimination when one equation is already in this form.

Use brackets when substituting

When substituting an expression for y (or x), put it in brackets. x + (2x − 3) prevents sign errors during simplification — especially when the expression contains negatives.

Graphical Method

Plot both equations as straight lines on the same axes. The intersection point is the solution. Reading is approximate unless the intersection lands on a gridline.

Lines y = x + 1 (blue) and y = −x + 5 (green) intersect at (2, 3). This point is the simultaneous solution: x = 2, y = 3 satisfies both equations.

Why "approximate"

If the intersection lands between gridlines, the best you can do is read to the nearest 0.1 or 0.5. Algebra gives exact answers; graphs give visual confirmation or estimates when algebra is unavailable.

MCE Chain

Whichever method you choose, marks are awarded across these stages.

MMethod. Choose elimination, substitution, or graphical. For elimination, decide which variable to eliminate and what multipliers to use. For substitution, identify the rearranged equation.

CCalculate. Carry out the multiplication, addition/subtraction (or substitution + expansion), then solve the resulting one-variable equation. Substitute back to find the second variable.

EEvaluate. Report both values as a pair. Check by substituting into the equation you didn't use to back-substitute.

Worked Example: Setup

Ex

Higher — 4 marks (Jun 2023 P1H Q16 authentic)

Solve the simultaneous equations:

2x − 5y = 13 ...(1) 3x + 4y = 8 ...(2)

M — Method

Coefficients of y are −5 and +4. To eliminate y:

Multiply (1) by 4: 8x − 20y = 52

Multiply (2) by 5: 15x + 20y = 40

Now y coefficients are −20 and +20 (different signs) → ADD to eliminate.

Worked Example: Calculate & Evaluate

C — Calculate

8x − 20y = 52 ...(1')

15x + 20y = 40 ...(2')

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Add: 23x = 92 → x = 4

Substitute into (1): 2(4) − 5y = 13

8 − 5y = 13

−5y = 5 → y = −1

E — Evaluate (check)

Check in (2): 3(4) + 4(−1) = 12 − 4 = 8 ✓

Solution: x = 4, y = −1

Mark breakdown (Jun 2023 mark scheme)

M1 for equating coefficients (e.g. 8x − 20y = 52 and 15x + 20y = 40) — allow one term error. M1dep for correctly eliminating one variable (23x = 92 or similar). A2 for both x = 4 AND y = −1; A1 for one correct value from correct method.

Weak vs Strong: Sign Errors on Subtraction

The single most common loss of marks in elimination is forgetting to negate every term when subtracting equations. This is especially common when both equations have the same sign on the variable being eliminated (e.g. Nov23 P3H Q11: 7x + 2y = 100; 3x + 2y = 48).

Weak student

Subtracting 3x + 2y = 48 from 7x + 2y = 100:

Didn't eliminate y at all — no further progress possible.

Strong student

Writes the subtraction in full brackets:

Distributes the minus across every term. Eliminates y cleanly.

The safer alternative: ×−1 then ADD

If you find subtraction error-prone, multiply the second equation by −1 first, then ADD. For 7x + 2y = 100 and 3x + 2y = 48: multiplying (2) by −1 gives −3x − 2y = −48. Adding to (1) gives 4x = 52 → x = 13. Only one set of signs flips — much less to track.

Examiner Alerts

Sign error on subtraction

Jun 2023 P1H Q16 — many candidates lost marks at the subtraction stage. Distribute the minus across every term, or multiply the second equation by −1 and ADD instead.

Solving for one variable only

The answer is a pair (x, y). Solving for x and forgetting y loses half the A marks. Always back-substitute.

Trial and improvement

Jun 2024 P1H Q14 mark scheme: "Trial and improvement is 0, 3 (incorrect money notation) or 4 (fully correct)" — guessing without algebraic working scores zero unless the answer is fully correct. Always show your method.

Money / unit notation

For contextual questions, use correct units. Jun 2024 P1H Q14: "£0.8" (missing trailing zero) and "0.80p" (mixing pounds and pence) both score M3A0 — full method but wrong notation = no answer mark.

Define your variables (contextual)

For word problems, write "Let x = cost of one chocolate bar, y = cost of one packet of mints" before forming equations. The examiner needs to follow your variables.

Practice Question 1

Foundation-friendly elimination. Solve simultaneously:

x + y = 7 2x − y = 5

[3 marks]

M

Coefficients of y: +1 and −1. Different signs → ADD to eliminate y.

C

(1) + (2): 3x = 12 → x = 4. Substitute into (1): 4 + y = 7 → y = 3.

E

Check in (2): 2(4) − 3 = 5 ✓. x = 4, y = 3. Marks: M1 add, A1 x = 4, A1 y = 3.

Practice Question 2

Substitution method. Solve simultaneously:

y = x + 4 2x + 3y = 22

[3 marks]

M

Equation (1) is already y = …. Substitute (x + 4) for y in equation (2).

C

2x + 3(x + 4) = 22 → 2x + 3x + 12 = 22 → 5x = 10 → x = 2. Then y = 2 + 4 = 6.

E

Check in (2): 2(2) + 3(6) = 4 + 18 = 22 ✓. x = 2, y = 6. Marks: M1 substitution, A1 x, A1 y.

Practice Question 3 (Contextual)

Form & solve from context (Jun 2024 P1H Q14 authentic).

4 chocolate bars and 3 packets of mints cost £4.70.

5 chocolate bars and 1 packet of mints cost £4.50.

Work out the cost of a chocolate bar and the cost of a packet of mints. [4 marks]

M

Let x = cost of a chocolate bar (£), y = cost of a packet of mints (£). Form equations: 4x + 3y = 4.70 ...(1); 5x + y = 4.50 ...(2).

C

Multiply (2) by 3 to match the y-coefficients: 15x + 3y = 13.50 ...(2'). Subtract (1) from (2') (same signs on y): 11x = 8.80 → x = 0.80. Substitute into (2): 5(0.80) + y = 4.50 → 4 + y = 4.50 → y = 0.50.

E

Chocolate bar: £0.80; packet of mints: £0.50. Check in (1): 4(0.80) + 3(0.50) = 3.20 + 1.50 = £4.70 ✓. Mark scheme: M1 form one equation; M1dep equate coefficients; M1dep eliminate; A1 final answer with correct money notation. Note: "£0.8" or "0.80p" lose the A1 — must be £0.80 or 80p.

Summary

Three Methods

Elimination: match coefficients, then ADD (different signs) or SUBTRACT (same signs).

Substitution: use a " y = …" equation; substitute into the other.

Graphical: plot both lines; intersection point is the solution (approximate).

The SSS / DSA Rule

Same Signs → Subtract; Different Signs → Add

Always do

Number your equations (1) and (2)

State your operation ("× 3", "add", "subtract")

Substitute back to find the second variable

Check in the other original equation

Report both values as a pair

Never do

Subtract without distributing the minus

Stop after finding only one variable

Use trial and improvement (scores 0 unless fully correct)

Mix unit notations (£0.80p) in money problems

Skip defining variables in word problems

Top Exam Tip

Whenever you're about to subtract equations, write the second equation in brackets and distribute the minus across every term. Or — even safer — multiply by −1 first, then ADD. Either approach saves the M1dep mark.