Two equations, two unknowns — solved at the same time. One equation alone can’t give you a definite answer. But two equations together pin down both values exactly. Engineers use this every day: finding two unknown forces, two unknown currents, two unknown dimensions.
Two equations, two unknownsElimination methodSubstitution methodWord problemsEngineering applications
The big ideaWhy two equations are needed
If I tell you “x + y = 10”, that’s not enough — x could be 1 and y could be 9, or x=5 and y=5, or x=7 and y=3. Infinite possibilities!
But if I also tell you “x − y = 2”, now there’s only ONE pair that satisfies both at the same time: x=6, y=4.
💡 “Simultaneous” means “at the same time” — both equations must be true simultaneously for the same x and y values.
Setting-up checklistDo this before you write a single equation
Read the problem twice. Identify the two unknown quantities.
Choose simple letter names: x and y, or meaningful ones like c and t, a and b.
State what each letter represents including units (e.g. “let x = number of bolts”).
Go through each sentence of information — each piece of information gives one equation.
Number your equations (1) and (2) — you’ll refer to them by number when solving.
Before solving, check: do you have exactly 2 equations and 2 unknowns?
⚠️ Never mix up which letter is which! Clearly stating “let x = ...” at the top prevents the most common setup errors.
✖
The Elimination Method
Add or subtract the two equations to eliminate one variable completely. Best when coefficients of one variable are already equal — or can be made equal by multiplying.
Strategy: Make the coefficients of one variable the same in both equations. Then ADD (if signs differ) or SUBTRACT (if signs same) to eliminate it.
Continuing — substitute x = 4 into equation (2) to find y
Eq (2): x + 2y = 8
Sub x=4: 4 + 2y = 8
Subtract 4: 2y = 8 − 4 = 4
Divide by 2: y = 4 ÷ 2
Answer:y = 2 ✅
✔ Always check in BOTH originals: (1): 3(4)+2(2)=16 ✔ (2): 4+2(2)=8 ✔
Full worked example3x + 2y = 16 and x + 2y = 8 — complete solution
(1)3x + 2y = 16equation 1
(2)x + 2y = 8equation 2
Solve using elimination — every step shown
Notice: both equations have 2y → same sign → SUBTRACT to cancel
(1)−(2): (3x − x) + (2y − 2y) = (16 − 8)
Simplify: 2x + 0 = 8
So: 2x = 8
÷2 both: x = 8 ÷ 2
x found: x = 4
Sub x=4 into (2): 4 + 2y = 8
−4 both: 2y = 8 − 4 = 4
÷2 both: y = 4 ÷ 2
y found: y = 2
Check (1): 3(4) + 2(2) = 12 + 4 = 16 ✔
Check (2): 4 + 2(2) = 4 + 4 = 8 ✔
Answer:x = 4, y = 2 ✅
When to multiply firstCoefficients don’t match — scale up first
If neither variable has matching coefficients, multiply one or both equations to create a match.
(1)2x + 3y = 13
(2)5x − 2y = 1
Eliminate y: multiply to match y coefficients
(1)×2: 4x + 6y = 26 → call this (1a)
(2)×3: 15x − 6y = 3 → call this (2a)
6y diff sign: ADD (1a)+(2a): 19x = 29
x: x = 29/19 ≈ 1.526... Hmm! Let’s try eliminating x instead
(1)×5: 10x+15y=65
(2)×2: 10x−4y=2
Subtract: 19y=63 → y=63/19
Note:This one has non-integer answers — normal! Use the quadratic formula if needed.
Clean integer exampleMultiply first, then eliminate
(1)3x + y = 11
(2)x + 2y = 7
Eliminate y: multiply (1) by 2
(1)×2: 6x + 2y = 22 → call this (1a)
Now (1a)−(2): (6x − x) + (2y − 2y) = (22 − 7)
Simplify: 5x + 0 = 15
So: 5x = 15
÷5 both: x = 15 ÷ 5
x found: x = 3
Sub x=3 into (2): 3 + 2y = 7
−3 both: 2y = 7 − 3 = 4
÷2 both: y = 4 ÷ 2
Answer:x = 3, y = 2 ✅
💡 ADD when signs are different (one has +y and one has −y → they cancel when added). SUBTRACT when signs are the same (both have +2y → subtract to cancel).
🔄
The Substitution Method
Rearrange one equation to express one variable in terms of the other, then substitute that expression into the second equation. Best when one equation already has a variable on its own.
Full worked exampley = 2x − 1 and 3x + 2y = 12
(1)y = 2x − 1already has y alone!
(2)3x + 2y = 12
Solve by substitution — every step shown
y alone: equation (1) already gives y = 2x − 1 → substitute directly
Sub into (2): 3x + 2(2x − 1) = 12
Expand bracket: 3x + 4x − 2 = 12
Collect x: 7x − 2 = 12
+2 both: 7x = 12 + 2 = 14
÷7 both: x = 14 ÷ 7
x found: x = 2
Find y from (1): y = 2(2) − 1
Calculate: y = 4 − 1
y found: y = 3
Check (2): 3(2) + 2(3) = 6 + 6 = 12 ✔
Answer:x = 2, y = 3 ✅
When to use substitution
✅ Use substitution when: • One equation has a variable alone (x = ... or y = ...) • One equation is easy to rearrange to x = ... or y = ... • The coefficient of one variable is 1
📌 If both equations have large coefficients on every variable, elimination is usually faster. But both methods always give the same answer!
Another example4x + y = 14 and 2x − 3y = 2
(1)4x + y = 14
(2)2x − 3y = 2
Rearrange (1) for y, then substitute
From (1): y = 14 − 4x
Sub into (2): 2x − 3(14−4x) = 2
Expand: 2x − 42 + 12x = 2
Simplify: 14x = 44 → x = 44/14 ≈ 3.14
y: y = 14−4(44/14) = 14−176/14 = 20/14 ≈ 1.43
Exact:x = 22/7, y = 10/7 ✅
🛒
Word Problems — Everyday Contexts
Before engineering problems, practise with familiar everyday contexts. The maths is identical — the key skill is translating words into equations.
☕ Problem 1 — Cafe Order
At a work canteen, 2 coffees and 1 tea cost £5.50. 1 coffee and 2 teas cost £5.00.
Find the cost of one coffee and one tea.
Two engineers together complete 50 inspections per day.
Engineer A completes 8 more inspections per day than Engineer B.
Find how many inspections each engineer completes per day.
(1)A + B = 50total per day
(2)A − B = 8difference
Solution — add the equations (B cancels)
(1)+(2): 2A = 58 → A = 29
Sub in (1): 29 + B = 50 → B = 21
Answer:Engineer A = 29/day, Engineer B = 21/day ✅
💡 When one equation has +B and the other has −B, ADD to cancel B instantly.
🧱 Problem 3 — Material Costs
A workshop orders steel rod and aluminium sheet. 3 metres of rod and 2 sheets cost £46. 5 metres of rod and 1 sheet cost £43.
Find the cost of 1 metre of rod and 1 sheet.
(1)3r + 2s = 46r=rod, s=sheet
(2)5r + s = 43
Solution — substitution (s alone in (2))
From (2): s = 43 − 5r
Sub into (1): 3r + 2(43−5r) = 46
Expand: 3r + 86 − 10r = 46
Simplify: −7r = −40 → r = 40/7 ≈ £5.71
s: s = 43 − 5(40/7) = 43 − 200/7 = 101/7 ≈ £14.43
Answer:Rod = £5.71/m, Sheet = £14.43 ✅
🚚 Problem 4 — Delivery Distances
A depot makes two types of delivery run: short routes (x km) and long routes (y km).
On Monday: 3 short and 2 long runs totalled 260 km.
On Tuesday: 2 short and 5 long runs totalled 430 km.
Find the distance of each route type.
(1)3x + 2y = 260Monday
(2)2x + 5y = 430Tuesday
Solution — elimination (match x: ×2 and ×3)
(1)×2: 6x + 4y = 520
(2)×3: 6x + 15y = 1290
Subtract: −11y = −770 → y = 70 km
Sub in (1): 3x + 140 = 260 → 3x = 120 → x = 40 km
Check (2): 2(40)+5(70)=80+350=430 ✔
Answer:Short route = 40 km, Long route = 70 km ✅
⚙
Word Problems — Engineering Contexts
These are the exact types you’ll see in T Level / Level 3 Engineering examinations. Same method — two equations, two unknowns — but drawn from real engineering situations.
⚡ Engineering Problem 1 — Kirchhoff’s Voltage Law
A circuit has two loops. Applying Kirchhoff’s Voltage Law gives:
Loop 1: 4I₁ + 2I₂ = 12 (voltages sum to 12V)
Loop 2: 2I₁ + 6I₂ = 12 (voltages sum to 12V)
Find the two loop currents I₁ and I₂ in amperes.
(1)4I₁ + 2I₂ = 12loop 1
(2)2I₁ + 6I₂ = 12loop 2
Solution — eliminate I₁ (multiply (2)×2)
(1)×1: 4I₁ + 2I₂ = 12
(2)×2: 4I₁ + 12I₂ = 24
Subtract: −10I₂ = −12 → I₂ = 1.2 A
Sub in (1): 4I₁ + 2(1.2) = 12 → 4I₁ = 9.6 → I₁ = 2.4 A
Check (2): 2(2.4)+6(1.2) = 4.8+7.2 = 12 ✔
Answer:I₁ = 2.4 A, I₂ = 1.2 A ✅
🏗️ Engineering Problem 2 — Beam Reactions
A simply-supported beam carries a load. The two support reactions R₁ and R₂ satisfy: Vertical equilibrium: R₁ + R₂ = 800 N Moment equilibrium about left support: 3R₂ = 1800 N·m, so R₂ is already found directly.
But in a more complex case: R₁ + R₂ = 800 and 2R₁ − R₂ = 400
Find R₁ and R₂.
(1)R₁ + R₂ = 800vertical equilibrium
(2)2R₁ − R₂ = 400moment equation
Solution — add equations (R₂ has opposite signs)
(1)+(2): 3R₁ = 1200 → R₁ = 400 N
Sub in (1): 400 + R₂ = 800 → R₂ = 400 N
Answer:R₁ = 400 N, R₂ = 400 N ✅
🧪 Engineering Problem 3 — Alloy Mixing
An engineer needs to mix two alloys to produce 100 kg of a new alloy containing exactly 40% copper.
Alloy A contains 30% copper. Alloy B contains 60% copper.
How many kg of each alloy should be used?
(1)a + b = 100total mass (kg)
(2)0.3a + 0.6b = 40copper content (kg)
Solution — substitution from (1)
From (1): a = 100 − b
Sub into (2): 0.3(100−b) + 0.6b = 40
Expand: 30 − 0.3b + 0.6b = 40
Simplify: 0.3b = 10 → b = 33.3 kg
a: a = 100 − 33.3 = 66.7 kg
Answer:66.7 kg of Alloy A, 33.3 kg of Alloy B ✅
🚗 Engineering Problem 4 — Two Machines, Different Rates
Machine A and Machine B together produce 120 components per hour.
Machine A produces 20 more components per hour than Machine B.
Find the production rate of each machine.
(1)A + B = 120total output
(2)A − B = 20difference
Solution — add equations (B cancels)
(1)+(2): 2A = 140 → A = 70 components/hr
Sub in (1): 70 + B = 120 → B = 50 components/hr
Answer:Machine A = 70/hr, Machine B = 50/hr ✅
⚠
Common Mistakes & How to Avoid Them
These are the errors that appear most frequently in T Level exam scripts. Read each one carefully — recognising the mistake pattern is the first step to never making it yourself.
Mistake 1Subtracting when you should add (and vice versa)
The rule: same signs → subtract. Different signs → add.
❌ Wrong: 3x + 2y = 10 and x + 2y = 4 → student ADDs → 4x + 4y = 14 (doesn’t help!) ✅ Correct: same +2y → SUBTRACT → 2x = 6 → x = 3
Mistake 2Forgetting to multiply ALL terms when scaling
❌ Wrong: multiply 2x + 3y = 8 by 3 → 6x + 3y = 8 (only multiplied the x term!) ✅ Correct: EVERY term multiplied → 6x + 9y = 24
Mistake 3Finding x but forgetting to find y (or vice versa)
❌ Stops at: “x = 3” without finding y ✅ Always substitute back into one of the original equations to find the second variable. The question asks for BOTH!
💡 After finding x, always write: “Substitute x = 3 into equation (1)...”
Mistake 4Not checking the answer in BOTH equations
❌ Only checks in the equation they substituted into — but that’s the one they just used! ✅ Always check in BOTH original equations. If one fails, there’s an error somewhere.
📌 Checking takes 20 seconds and can recover 2–3 marks in an exam.
Mistake 5Sign error when substituting negative values
❌ y = −2, substitute into 3x + y = 7 → 3x + −2 = 7 → 3x = 9 → x = 3 This is actually correct — but students often write 3x − (−2) = 7 which gives x = 5/3!
✅ Replace “y” with “(−2)” in brackets: 3x + (−2) = 7 → no ambiguity
Mistake 6Setting up the word problem incorrectly
❌ “3 adult + 2 child = £32” written as: 3 + 2 = 32 (forgot the variables!) ✅ Must include the variable: 3a + 2c = 32
✅ Always define your letters first: “let a = price of adult ticket in £” before writing any equation.
Spot the ErrorsFind what’s wrong in each student’s working
ℹ️ How to use: Work through each question and write down your answer. When you are ready, click Show Answer to reveal the full worked solution and mark scheme.
Error: when multiplying (1) by 3, the student only multiplied 2x and the constant 9 — forgot to multiply y! Correct: (1)×3 should give 6x + 3y = 27 Then subtract (2): 5x + 0y = 20 → x = 4
Error B
Student finds x=5 from elimination. Then substitutes into eq (1): 2(5)+y=14, y=4. Writes final answer as “x=5” only.
Error: student found y=4 during working but forgot to include it in the final answer. Both values must be stated: x=5 AND y=4 Also should check in equation (2): substitute both values to verify.
🧮
Interactive Simultaneous Equation Solver
Enter your two equations in the form ax + by = c. Choose whether to see the Elimination or Substitution method, and every step is shown clearly.
🧮 Simultaneous Equation SolverStep-by-step working
ℹ️ How to use: Type coefficients and constants for each equation below. Use negative numbers for subtraction (e.g. −2). Pick a method tab and the full working appears instantly.
Equation 1: ax + by = c
x +
y =
Equation 2: dx + ey = f
x +
y =
①2x + 3y = 8
②3x + 1y = 7
Enter your equations above — the working will appear here.
✖ Elimination — best when one variable has the same (or easily matched) coefficient in both equations.
🔄 Substitution — best when one equation has a single variable on its own, or can be easily rearranged to give x = … or y = …
💡 Both methods always give the same answer — choose whichever feels clearest for your pair of equations.
Check your answerAlways substitute back in!
After solving, put your values of x and y back into both original equations. Both left-hand sides should equal their right-hand side.
If x = 1, y = 2: Eq ①: 2(1) + 3(2) = 8 ✔ Eq ②: 3(1) + 1(2) = 5 ≠ 7 ✘ → error somewhere!
✅ The solver above checks this automatically and shows you whether the answer verifies.
🎮
Quick Fire Quiz
Test your knowledge — 10 questions, instant feedback.
⚡ Simultaneous Blitz
ℹ️ How to play: A question appears on screen. Click the answer you think is correct from the four options. Your score and streak update after each question. Click Next to move on.
Score: 0Streak: 0 🔥Q 1/10
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Score: 0/10
⚙️
Applications
Where this topic is used in engineering, manufacturing, maintenance and daily life.
⚡Electrical:Kirchhoff's laws: two equations for two loop currents.
🔧Mechanical:ΣFx=0 and ΣFy=0. Find two reaction forces.
🏗️Construction:Two suppliers, two cost equations. Find equal-cost point.
🛒Daily Life:Two meal deals. One costs £1.50 more. Find each price.
✏
Practice Questions
15 questions: always define your unknowns, number your equations, show every step, and check BOTH equations at the end.
🟢 Tier 1 — Pure Algebra
Q1
Solve: x + y = 10 and x − y = 4
[3]
(1)+(2): 2x=14 → x=7 Sub in (1): 7+y=10 → y=3 x=7, y=3 Check: 7+3=10✔ 7−3=4✔
Q2
Solve: 3x + 2y = 16 and x + 2y = 8
[3]
(1)−(2): 2x=8 → x=4 Sub: 4+2y=8 → y=2 x=4, y=2
Q3
Solve by substitution: y = 3x − 1 and 2x + y = 9
[3]
Sub y=3x−1 into eq2: 2x+3x−1=9 → 5x=10 → x=2 y=3(2)−1=5 x=2, y=5
A school shop sells pens for x pence each and rulers for y pence each. 3 pens and 2 rulers cost 190p. 2 pens and 5 rulers cost 250p. Find the cost of each item.
The sum of two numbers is 31. Three times the larger minus twice the smaller equals 58. Find both numbers.
[4]
Let x=larger, y=smaller. (1): x+y=31 (2): 3x−2y=58 From (1): x=31−y. Sub: 3(31−y)−2y=58 → 93−5y=58 → y=7 x=31−7=24 Numbers are 24 and 7 Check: 3(24)−2(7)=72−14=58✔
Q8
Two resistors R₁ and R₂ in series give a total resistance of 18Ω. In parallel, their combined resistance is 4Ω. Find R₁ and R₂. (Hint: parallel formula: R₁R₂/(R₁+R₂) = 4)
[5]
(1): R₁+R₂=18 (2): R₁R₂=4×18=72 So R₁ and R₂ are roots of: t²−18t+72=0 Factorise: (t−6)(t−12)=0 R₁=6Ω, R₂=12Ω
Q9
A workshop produces x small brackets and y large brackets each day. Small brackets take 20 minutes each, large brackets take 45 minutes each. The total daily production time is 600 minutes. The workshop makes twice as many small brackets as large ones. Find x and y.
[4]
(1): 20x+45y=600 (2): x=2y Sub (2) into (1): 20(2y)+45y=600 → 85y=600 → y=600/85≈7.06 For exact: y=120/17, x=240/17. Practically: y=7, x=14 Check: 20(14)+45(7)=280+315=595 ≈600 (rounding) y≈7 large, x≈14 small brackets per day ✅
Q10
An engineer mixes solution A (20% acid) with solution B (50% acid) to make 60 litres of a 30% acid solution. How many litres of each are needed?
[4]
(1): a+b=60 (2): 0.2a+0.5b=18 (30% of 60=18 litres acid) From (1): a=60−b. Sub: 0.2(60−b)+0.5b=18 → 12−0.2b+0.5b=18 → 0.3b=6 → b=20 a=40 40 litres of Solution A, 20 litres of Solution B ✅
🔴 Tier 3 — T Level Challenge
Q11
Applying Kirchhoff’s laws to a circuit gives: 6I₁ − 2I₂ = 24 and −2I₁ + 8I₂ = 0. Find I₁ and I₂.
[4]
From (2): −2I₁+8I₂=0 → I₁=4I₂ Sub into (1): 6(4I₂)−2I₂=24 → 22I₂=24 → I₂=12/11≈1.09A I₁=4(12/11)=48/11≈4.36A I₁≈4.36A, I₂≈1.09A ✅
Q12
A component is made from two materials. Material X costs £8/kg and Material Y costs £15/kg. The component weighs 12 kg and costs £129 in materials. Find the mass of each material used.
[4]
(1): x+y=12 (2): 8x+15y=129 From (1): x=12−y. Sub: 8(12−y)+15y=129 → 96+7y=129 → y=33/7=4.71kg x=12−4.71=7.29kg Material X: 7.29kg, Material Y: 4.71kg ✅
Q13
Two forces F₁ and F₂ act on a bracket. For horizontal equilibrium: F₁cos30° + F₂cos45° = 100. For vertical equilibrium: F₁sin30° − F₂sin45° = 0. Find F₁ and F₂ to 3 s.f.
A CNC machine produces two types of part. Type A takes 8 minutes, Type B takes 12 minutes. In one shift (480 minutes), 50 parts are made. How many of each type were made?
[4]
(1): a+b=50 (2): 8a+12b=480 From (1): a=50−b. Sub: 8(50−b)+12b=480 → 400+4b=480 → b=20 a=30 30 Type A parts, 20 Type B parts ✅ Check: 8(30)+12(20)=240+240=480✔
Q15
An engineering company orders M metres of steel bar and N bolts. Order 1: 5M + 200N = £1850. Order 2: 3M + 150N = £1140. Find the unit cost of 1 metre of bar and 1 bolt. Then find the cost of an order for 8M + 500N.
[6]
(1): 5M+200N=1850 (2): 3M+150N=1140 (2)×(4/3): 4M+200N=1520 Subtract from (1): M=330 → 1m of bar = £330? That’s high. Try division: (1)÷50: 0.1M+4N=37 (2)÷30: 0.1M+5N=38 Subtract: −N=−1 → N=£1 per bolt Sub: 0.1M+4=37 → M=£330 per metre (or perhaps the numbers represent bundles) New order: 8(330)+500(1) = 2640+500 = £3140 ✅
🎯 Score guide: Q1–5 = algebra only (17 marks) · Q6–10 = word problems (21 marks) · Q11–15 = T Level challenge (23 marks)