Forming & Solving Equations from Context
SPEC A21 • MODULE 2: ALGEBRA • ⭐ HIGH PRIORITY
Turning word problems into algebra — then answering the question that was actually asked.
BOTH TIERS • ALL PAPERS • LESSON 2.25
The Hardest Step Is Usually Step 1
Translating a word problem into algebra is often harder than the algebra itself. These questions are worth 4–5 marks each, and every paper has one. Follow a clean four-step procedure and the marks are yours.
STEP 1Define — write "Let x = …" in words.
STEP 2Translate — each condition becomes an expression or equation.
STEP 3Solve — standard algebra; find the unknown.
STEP 4Interpret — answer the question that was asked, with units.
Why all four steps matter
The marks split across all four stages: a B1 is often awarded just for defining the variable, and an A1 specifically for interpreting x back in context. Skip either and you leave marks on the table.
Step 1: Define Your Variable
Before you write a single equation, tell the examiner what your letter represents. One sentence — nothing fancy.
The opening sentence
Let x = [the thing you're trying to find]Be specific. "Let x = the width" is better than just "let x = 9". "Let x = the number of red counters in the bag" is better than "x = number".
Vague
"Let x = the number"
Which number? Examiner has to guess.
Specific
"Let x = Ella's age in years"
No ambiguity. Units included.
Pick the unknown to minimise work
You can usually choose which quantity gets the letter. Pick the one the others are defined in terms of — e.g. if Jack is "3 years older than Ella", let x = Ella's age. That makes Jack = x + 3, which is simpler than the other way round.
Step 2: Translate Words Into Algebra
Every piece of information in the question becomes a piece of algebra. Work through the question one phrase at a time.
The phrase book
English phraseAlgebra"5 more than x" / "5 greater than x"x + 5"5 less than x" / "x decreased by 5"x − 5"twice x" / "double x"2x"half of x"x ÷ 2 or x/2"3 less than twice x"2x − 3"three consecutive integers starting at n"n, n + 1, n + 2"the sum of a and b"a + b"the product of a and b"a × b or ab"the mean of three numbers is 5"(sum of three) ÷ 3 = 5, i.e. sum = 15
Watch the direction
"5 less than x" is x − 5, NOT 5 − x. Read each phrase slowly — the order of subtraction matters.
Steps 3 & 4: Solve and Interpret
Step 3 — Solve
Standard algebra. Combine like terms, expand brackets, rearrange, isolate the unknown. You've done this many times — nothing new here. Show every line.
Step 4 — Interpret
After you find x, go back to the question. Re-read it. Answer what was actually asked.
If the question asked "how old is Jack?" — don't write "x = 6". Write "Jack is 9 years old". If the question asked for "the length and the width" — state both, with units.
The interpretation mark
Mark schemes typically reserve the final A1 for a correctly interpreted answer. You've done the algebra perfectly — spending 10 extra seconds to re-read the question and write a proper sentence is a guaranteed mark.
Walkthrough: Rectangle Perimeter
A common geometry-context problem. Let's apply the four-step procedure from scratch.
Question (4 marks): The perimeter of a rectangle is 50 cm. The length is 7 cm longer than the width. Find the length and width.
Step 1 — Define
Let x = the width of the rectangle, in cm.
Why width? Because the length is defined in terms of it ("7 cm longer than the width").
Step 2 — Translate
Width = x | Length = x + 7
Perimeter = 2(length + width) = 50
2(x + (x + 7)) = 50
Walkthrough: Finish It
Step 3 — Solve
2(x + (x + 7)) = 50
2(2x + 7) = 50
4x + 14 = 50
4x = 36
x = 9
Step 4 — Interpret
x is the width. So width = 9 cm. Length = x + 7 = 16 cm.
Answer: Width = 9 cm, Length = 16 cm.
Verify in the original question
Check perimeter: 2(9 + 16) = 2(25) = 50 ✓. Check length − width = 16 − 9 = 7 ✓. Both conditions satisfied — answer is correct.
Don't stop at x = 9
Students who stop at "x = 9" typically lose the final mark — the question asked for the length and width, not "x". State both values with units.
Examiner Alert: Always Define x
Weak Student
Dives straight into algebra with no definition:
"4x + 14 = 50
x = 9"
The examiner has no idea what x represents. Method mark may be lost if the working is ambiguous.
Strong Student
Opens with a clear definition:
"Let width = x cm.
Then length = (x + 7) cm.
2(x + (x + 7)) = 50
..."
Examiner can follow the reasoning. Full marks.
Examiner alert
Skipping the variable definition is a very common and easily-avoided error in A21 questions. Spend two seconds writing the opening sentence — it's almost always worth a B1 mark of its own, and it clears up any ambiguity in your algebra.
Examiner Alert: Answer the Question
The question asked for the length and width, not "x". Re-reading the question before writing your final answer is worth a mark every time.
Weak Student
Solves the algebra and stops:
"x = 9"
Hasn't answered the question. Loses the interpretation mark (A1).
Strong Student
Translates x back to the original context:
"Width = 9 cm, Length = 16 cm"
Full sentence, both quantities, units. Full marks.
Pause before the final line
Get in the habit: once you've found x, re-read the question. Look at the final sentence — what specifically is being asked for? A number of apples, an age, a length? Answer that.
Worked Example: Age Problem
Age problems are a classic A21 context. They test whether you can translate "older", "younger", "twice as old" correctly.
Question (5 marks): Ella is x years old. Jack is 3 years older than Ella. Priya is twice Ella's age. The mean of their ages is 9. How old is Jack?
Method
Ella = x, Jack = x + 3, Priya = 2x
Mean of three = 9 → Sum of three = 9 × 3 = 27
x + (x + 3) + 2x = 27
4x + 3 = 27
4x = 24
x = 6
Interpret — "How old is Jack?"
Jack = x + 3 = 6 + 3 = 9 years old.
Check: Ella = 6, Jack = 9, Priya = 12. Sum = 27, mean = 27 ÷ 3 = 9 ✓
Examiner Alerts: Age Problems
Examiner reports on age-context questions consistently flag the same three errors. Knowing them means you won't make them.
Error 1 — "older/younger" the wrong way
"Jack is 3 years older than Ella" → Jack = x + 3. If someone were "3 years younger" it would be x − 3. Mixing the direction (e.g. writing x + 3 for "younger") is one of the most common errors on age questions. It can still earn method marks if the algebra that follows is internally consistent, but costs accuracy.
Error 2 — forgetting to include x itself
When the question defines x as one person's age and then asks about all the people, the original person must still appear in the sum. Writing "(x + 3) + 2x = 27" instead of "x + (x + 3) + 2x = 27" — forgetting Ella herself — typically caps you at 2 of 5 marks.
Error 3 — "mean = 9, so total = 9"
Mean of N numbers = (sum of N numbers) ÷ N. So mean = 9 with 3 numbers means sum = 9 × 3 = 27, not 9. Students who equate the sum directly to the mean lose every subsequent mark.
Takeaway
Read each phrase carefully: "younger" → minus, "older" → plus, "mean is 9" → sum is 9 × (count). Before equating — make sure every person/quantity is in the sum.
Worked Example: L-Shape Perimeter
A geometry-context problem. Here the setup is the hard part — once you have the perimeter expression, the algebra is quick.
Question (4 marks): The length of a rectangle is 6 times the width. Two identical rectangles are joined (no overlap) to form the L-shape shown. The perimeter of the L-shape is 52 cm. Work out the perimeter of one of the rectangles.
Two rectangles joined to form the L-shape. Diagram not drawn to scale.
L-Shape: Solution
Step 1 — Define
Let x = width of one rectangle, in cm. Then length = 6x.
Step 2 — Translate (walk the perimeter)
Walk around the L-shape clockwise from the top-left corner, labelling each of the six outer edges:
top of vertical = x | right of vertical (above horizontal) = 6x − x = 5x
top of horizontal = 6x | right of horizontal = x
bottom (full width) = 6x + x = 7x | left of vertical = 6x
Perimeter of L = x + 5x + 6x + x + 7x + 6x = 26x
Step 3 — Solve
26x = 52 → x = 52 ÷ 26 = 2
Step 4 — Interpret
Perimeter of one rectangle = 2(x + 6x) = 14x = 14 × 2 = 28 cm.
Watch the final question
The question asks for the perimeter of one rectangle, not x. Students who stop at x = 2 lose the interpretation mark. Always re-read.
Common Problem Types
Most A21 questions fall into one of a few standard shapes. Spot the pattern, apply the template.
ContextSet-upTip
AgeLet x = the age of the person others are compared to. "5 years older" → x + 5; "half the age" → x/2."Older" = add; "younger" = subtract. Re-read each phrase.
Consecutive integersn, n + 1, n + 2 (or n − 1, n, n + 1 for symmetric setups)."Even" → 2n, 2n + 2, 2n + 4. "Odd" → 2n + 1, 2n + 3, ...
PerimeterExpress each side in terms of x. Sum all sides, set equal to given perimeter.Walk around the shape labelling every edge before summing.
AreaExpress length and width in terms of x. Multiply, set equal to given area.May give a quadratic — solve and reject negative or non-physical values.
Money / countsSet up an equation for the total, and sometimes a second for the count.Units matter: pence vs pounds must match throughout.
Mean / totalMean × count = total. "Mean of 3 is 5" → sum = 15.The total is almost always easier to work with than the mean.
Your Turn
The sum of three consecutive integers is 48. Find the three integers.
[3 marks]
M
Method: let the smallest integer be n. Then the three consecutive integers are n, n + 1, n + 2. Set their sum equal to 48 and solve.
C
n + (n + 1) + (n + 2) = 48
C
3n + 3 = 48
C
3n = 45 → n = 15
E
The three integers are 15, 16 and 17. Check: 15 + 16 + 17 = 48 ✓
Common Errors To Avoid
Students often writeCorrect versionWhy it matters
"Let length = x, width = x"Use one letter and relate the other to it: width = x, length = x + 7Two unknowns, one equation = can't solve
No opening "Let x = …" sentenceAlways define the variable explicitlyTypically a B1 mark just for the definition
"x + 1" for a quantity described as "younger""x − 1" — "younger" means subtractVery common error on age problems
Forgetting one of the terms in a sum (e.g. x + (x−1) = 15 instead of x + (x−1) + 2x = 15)Add every quantity listed — re-read and countTypically caps you at 2/5 marks
"Mean of 3 numbers = 5, so sum = 5"Sum = mean × count = 5 × 3 = 15Loses every subsequent mark
Final answer: "x = 9""The width is 9 cm, the length is 16 cm"Interpretation mark (A1) lost
Quick Self-Check
Before you move on: Have I defined my variable in words? Have I answered the actual question (not just stated x)? Have I included units? Does my answer check out in the original wording?
Summary
The Four-Step Procedure
Define — open with "Let x = …" and state the unit.
Translate — turn each condition into an expression or equation.
Solve — standard algebra; show every line.
Interpret — re-read the question; answer with words and units.
Translation Shortcuts
"older / more than" → add | "younger / less than" → subtract
"twice / double" → × 2 | "half of" → ÷ 2
"consecutive integers from n" → n, n + 1, n + 2
"mean of N values is M" → sum of the values is M × N
Geometry: walk around the shape labelling every edge before summing
Spec Coverage
Spec refContent
A21Translate situations into algebraic expressions / formulae; derive and solve equations (or simultaneous equations); interpret the solution in context.
Notes: including the solution of geometrical problems and problems set in context.
Exam Tip
This topic is a ⭐ high priority — every paper, both tiers, typically 4–5 marks. Marks are distributed across all four steps, so skipping any one (especially the definition or the interpretation) leaves marks on the table. Get in the routine: define, translate, solve, interpret.
ExamRizz GCSE Maths (AQA 8300) • Lesson 2.25 • Module 2: Algebra • Spec A21 • Foundation + Higher • ⭐ High priority
Next: The Hardest Step Is Usually Step 1