Solving Equations from Real Problems Revision Notes | GCSE AQA Higher Maths
To form an equation, name the unknown, write each condition as algebra, then solve and check. Free notes with worked examples and 12 questions answered.
Updated 21 Mar 2026 · Revise Right Team
Solving Equations from Real Problems is part of Solving equations and inequalities in GCSE AQA Higher Maths.
5 min
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39
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GCSE AQA Higher
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Content review
What this lesson covers
To form an equation, name the unknown, write each condition as algebra, then solve and check. Free notes with worked examples and 12 questions answered.
Derive an equation or pair of simultaneous equations from a real-world or geometrical problem
Solve the derived equation(s) to find unknown values
Interpret the solution in context, including rejecting values that are not valid in the situation
Apply equation-building to geometrical problems involving perimeter, area, and angle relationships
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Use this lesson to stay inside Solving equations and inequalities while connecting the detail back to the wider Algebra topic.
Deriving an equation: Translating a written or geometrical problem into algebraic form by assigning a variable to an unknown quantity and expressing all given conditions as an equation or equations.
Interpreting a solution: Relating the algebraic answer back to the original context, checking whether it is valid (e.g. a length cannot be negative) and stating what it means in the problem.
Simultaneous equations: A pair of equations in two unknowns that must both be satisfied at the same time. Used when a problem introduces two separate unknown quantities.
Geometrical equation problem: A problem where an unknown angle, length, or area is expressed algebraically, and a known geometric property (e.g. angles in a triangle sum to 180°) provides the equation to solve.
Revision notes
Read the core explanations from Solving Equations from Real Problems before testing yourself from memory.
1. From Words to Algebra: Why This Skill Matters
Most real exam questions do not hand you an equation. They describe a situation — a garden, a triangle, two friends sharing money — and expect you to build the equation yourself. That process is called deriving an equation, and it is one of the highest-value skills on the Higher-tier paper.
The good news is that the method is always the same: name the unknown, write every condition as algebra, then solve. Once you see the pattern, you will spot it in every context the examiner throws at you.
There are three broad contexts you will meet: problems set in words (ages, money, lengths described in sentences), geometrical problems (angles, perimeters, areas expressed with algebra), and two-unknown problems that need simultaneous equations. Each section below covers one of these in depth.
2. Deriving and Solving from a Written Context
Start by reading the problem once for the story, then again to extract the maths. Ask: what is unknown? Call it $x$. Then ask: what relationships does the problem give me? Each relationship becomes an algebraic expression, and together they form your equation.
A common structure is: one quantity is described in terms of another. For example, "Aisha is three years older than Ben" immediately gives you $A = B + 3$. A second condition — say, their ages add to 41 — gives $A + B = 41$. You now have two equations and two unknowns.
Worked example: Deriving and Solving: Age Problem
Aisha is three years older than Ben. The sum of their ages is 41. Find both their ages. [4 marks]
Let Ben's age $= x$
Then Aisha's age $= x + 3$ — Choosing one variable and expressing the other in terms of it reduces the problem to one unknown. [M1 — correct setup]
$x + (x + 3) = 41$ — The sum of their ages equals 41. This is the derived equation. [M1 — correct equation]
$2x + 3 = 41$
$2x = 38$
$x = 19$ — Subtract 3 from both sides, then divide by 2. [M1 — correct algebraic manipulation]
Ben $= 19$
Aisha $= 19 + 3 = 22$ — Substitute back to find the second quantity. [A1 — both correct values]
$19 + 22 = 41$ ✓
Aisha is $22 - 19 = 3$ years older ✓ — Always verify both conditions from the original problem. Ages are positive integers — both values are valid in context.
Answer: Ben is 19 years old; Aisha is 22 years old.
Notice step 5: interpreting the solution means checking the answer makes sense in the real context. Ages must be positive. Lengths must be positive. If your algebra gives $x = -5$ for a length, you have either made an error or the problem has no valid solution — flag it.
3. Geometrical Problems: Algebra Meets Shape
Geometry gives you built-in equations for free. Angles in a triangle sum to $180^\circ$. Angles on a straight line sum to $180^\circ$. The perimeter of a rectangle is $2(l + w)$. Whenever a diagram labels sides or angles with algebraic expressions, those geometric facts become your equation.
The key move is to write the geometric rule first, then substitute the algebraic expressions in. Do not try to do both at once — it is a reliable source of sign errors.
Worked example: Deriving and Solving: Perimeter Problem
A rectangle has length $(3x - 1)$ cm and width $(x + 4)$ cm. Its perimeter is 46 cm. Find the value of $x$ and hence find the area of the rectangle. [5 marks]
Perimeter $= 2(\text{length} + \text{width})$ — State the geometric rule before substituting. [M1 — correct formula used]
$2\big((3x - 1) + (x + 4)\big) = 46$ — Replace length and width with the given algebraic expressions. [M1 — correct substitution]
$2(4x + 3) = 46$ — Collect like terms: $3x + x = 4x$ and $-1 + 4 = 3$.
$8x + 6 = 46$
$8x = 40$
$x = 5$ — Subtract 6, then divide by 8. [A1 — correct value of $x$]
Length $= 3(5) - 1 = 14$ cm
Width $= 5 + 4 = 9$ cm — Substitute $x = 5$ back into each expression.
Area $= 14 \times 9 = 126$ cm$^2$ — [A1 — correct area with units]. Check: perimeter $= 2(14 + 9) = 2 \times 23 = 46$ cm ✓
Answer: $x = 5$; area $= 126$ cm$^2$
Worked example: Deriving and Solving: Angles in a Triangle
A triangle has angles $(2x + 10)^\circ$, $(3x - 5)^\circ$, and $(x + 15)^\circ$. Find $x$ and state each angle. [4 marks]
Angles in a triangle sum to $180^\circ$ — Write the rule explicitly — examiners award a method mark for this.
Answer: $x = \frac{80}{3}$; angles are $\frac{190}{3}^\circ$, $75^\circ$, and $\frac{125}{3}^\circ$
The same approach works for any polygon. For a quadrilateral, angles sum to $360^\circ$. For angles on a straight line, the sum is $180^\circ$. Always write the geometric rule as your first line — it earns a method mark and keeps your working organised.
4. Two Unknowns: Deriving Simultaneous Equations
When a problem introduces two genuinely separate unknowns — say, the price of an adult ticket and the price of a child ticket — one equation is not enough. You need two conditions from the problem to form two equations, then solve them simultaneously.
The strategy is: read the problem and identify two separate statements of fact. Each statement becomes one equation. Label your unknowns clearly at the start — this prevents confusion when you substitute back at the end.
Worked example: Deriving and Solving Simultaneous Equations from Context
Two adult tickets and three child tickets cost £37. One adult ticket and two child tickets cost £21. Find the cost of each type of ticket. [5 marks]
Let adult ticket cost $= a$ (£)
Let child ticket cost $= c$ (£) — Two unknowns need two letters. State them clearly. [M1 — correct setup]
Equation 1: $2a + 3c = 37$
Equation 2: $a + 2c = 21$ — Each sentence in the problem gives one equation. [M1 — both equations correct]
5. A Harder Geometrical Problem: Quadratic Equations from Context
Some geometrical problems produce a quadratic equation when you substitute the algebraic expressions. The process is identical — derive, then solve — but the solving step now requires factorisation or the quadratic formula. After solving, you must reject any root that is not valid in context.
This is where interpretation becomes critical. A quadratic typically gives two roots. One will often be negative or produce a negative length, making it geometrically impossible. You must state explicitly why you reject it.
Worked example: Deriving a Quadratic from an Area Problem
A rectangular garden has length $(x + 5)$ m and width $(x - 2)$ m. Its area is 36 m². Find the value of $x$ and hence find the dimensions of the garden. [5 marks]
$(x + 5)(x - 2) = 36$ — Area of rectangle = length × width. [M1 — correct equation derived]
$x^2 + 3x - 10 - 36 = 0$
$x^2 + 3x - 46 = 0$ — Subtract 36 from both sides to get the quadratic equal to zero. [M1 — correct rearrangement]
$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-46)}}{2(1)}$
$= \frac{-3 \pm \sqrt{9 + 184}}{2}$
$= \frac{-3 \pm \sqrt{193}}{2}$ — $\sqrt{193} \approx 13.89$. Use the quadratic formula since this does not factorise neatly. [M1 — correct use of formula]
$x = \frac{-3 + 13.89}{2} \approx \frac{10.89}{2} \approx 5.45$
$x = \frac{-3 - 13.89}{2} \approx \frac{-16.89}{2} \approx -8.45$ — Two roots arise from the $\pm$.
Reject $x \approx -8.45$ because width $= x - 2 \approx -10.45$ m, which is negative.
Accept $x \approx 5.45$. — A length cannot be negative. Stating the reason for rejection earns the interpretation mark. [A1 — correct root selected with reason]
Length $= 5.45 + 5 \approx 10.45$ m
Width $= 5.45 - 2 \approx 3.45$ m
Check: $10.45 \times 3.45 \approx 36.1 \approx 36$ m² ✓ — Round to 3 significant figures unless the question specifies otherwise.
Answer: $x \approx 5.45$; length $\approx 10.45$ m, width $\approx 3.45$ m (to 3 s.f.)
6. Exam Technique: Deriving, Solving, and Interpreting
mark schemes for these questions typically award marks in three stages: one or two marks for the correct equation(s), one or two marks for the correct algebraic method, and one mark for the correct interpreted answer. You can lose the final mark even with correct algebra if you forget to state what the value means in context.
When the command word is derive, write your equation clearly on its own line before solving. When the command word is interpret, write a sentence: "$x = 5$, so the width of the garden is 5 cm." Never leave the examiner to infer what your number means.
Every problem in this topic — whether it involves ages, tickets, perimeters, areas, or angles — follows that same five-step chain. Practise naming it out loud as you work: define, express, derive, solve, interpret. That habit alone will keep your working organised under exam pressure.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-21.
Derive means building an equation from the information given: choose a letter for the unknown and turn every condition into algebra until the whole problem is captured. The derived equation should be written clearly on its own line before solving.
How do you form and solve an equation from a word problem?
Name the unknown, write each condition as algebra, then solve and check. If Aisha is three years older than Ben and their ages sum to 41, let Ben's age be x: x + (x + 3) = 41, so 2x + 3 = 41, x = 19 and Aisha is 22.
How do you find angles in a triangle using algebra?
Write the geometric rule first, then substitute the expressions. For angles (2x + 10)°, (3x - 5)° and (x + 15)°, the angles in a triangle sum to 180°, so 6x + 20 = 180 and x = 80/3, giving angles of 190/3°, 75° and 125/3°.
How do you solve perimeter problems with algebra?
Write the perimeter formula, substitute the expressions, then solve. For length (3x - 1) cm, width (x + 4) cm and perimeter 46 cm, 2(4x + 3) = 46 gives 8x + 6 = 46, so x = 5, the rectangle is 14 cm by 9 cm and its area is 126 cm².
Why do you reject a negative solution in a word problem?
A solution must make sense in context, and lengths and ages must be positive. For a garden (x + 5) m by (x - 2) m with an area of 36 m², the roots are x ≈ 5.45 and x ≈ -8.45, and the negative root is rejected because it gives a negative width.
How do you form simultaneous equations from a context?
Use one letter for each unknown and turn each separate statement into an equation. If two adult and three child tickets cost £37 and one adult and two child tickets cost £21, then 2a + 3c = 37 and a + 2c = 21, giving £11 per adult ticket and £5 per child ticket.
How do you solve an age problem with algebra?
An age problem is solved by letting one age be x and writing the other in terms of it. If Aisha is three years older than Ben and their ages add to 41, then x + (x + 3) = 41, so 2x = 38 and x = 19: Ben is 19 and Aisha is 22.
How do you solve adult and child ticket problems?
Ticket problems are solved by letting a and c be the two prices and writing one equation for each statement. If 2 adult and 3 child tickets cost £37 and 1 adult and 2 child tickets cost £21, then 2a + 3c = 37 and a + 2c = 21, giving c = 5 and a = 11.
How do you solve an area problem that gives a quadratic?
An area problem giving a quadratic is solved by writing length × width = area, expanding and rearranging to zero. For (x + 5)(x - 2) = 36, this gives x² + 3x - 46 = 0, and the quadratic formula gives x ≈ 5.45, with x ≈ -8.45 rejected because the width would be negative.
How do you interpret a solution in context?
A solution is interpreted in context by writing a sentence saying what the value means, such as 'x = 5, so the width of the garden is 5 cm'. Values that make no sense, such as a negative length, should be rejected with a reason.
What are the steps for solving a word problem with algebra?
Word problems are solved with the same five-step chain: define the unknown, express each quantity with algebra, derive the equation, solve it, then interpret the answer in context. Checking the answer against every condition in the original problem shows up errors.
How do you find angles in a triangle given as (2x + 10)°, (3x - 5)° and (x + 15)°?
Angles of (2x + 10)°, (3x - 5)° and (x + 15)° add to 180°, so 6x + 20 = 180, 6x = 160 and x = 80/3. The angles are then 190/3°, 75° and 125/3°, which add to 540/3 = 180°.