What is substitution in algebra and how is it used in solving equations?

Substitution in algebra is a method used to solve equations by replacing a variable with a given value or another expression. This technique is particularly useful in solving systems of equations where one equation is solved for one variable, and that expression is substituted into another equation. This method simplifies the problem and helps find the values of the variables involved.

How do you perform substitution in a system of linear equations?

To perform substitution in a system of linear equations, first solve one of the equations for one of the variables. Then, substitute this expression into the other equation(s). This will result in an equation with one variable, which can be solved using standard algebraic methods. Once you find the value of one variable, substitute it back into the original equation to find the value of the other variable.

Can substitution be used to solve quadratic equations in the Cambridge IGCSE Mathematics curriculum?

Yes, substitution can be used to solve quadratic equations, especially when dealing with equations that can be simplified by substituting a variable with another expression. For example, if you have a quadratic equation in the form of ax^2 + bx + c = 0, and you can express x in terms of another variable, substitution can simplify the equation, making it easier to solve.

What are some common mistakes to avoid when using substitution in algebra?

Common mistakes when using substitution include not correctly solving for a variable before substituting, forgetting to substitute into all parts of the equation, and arithmetic errors during the substitution process. It's important to carefully follow each step, double-check calculations, and ensure that the substituted expression is correctly simplified.

Why is substitution an important skill in the Cambridge IGCSE Mathematics curriculum?

Substitution is an important skill in the Cambridge IGCSE Mathematics curriculum because it is a fundamental technique for solving various types of equations, including linear and quadratic equations. Mastery of substitution helps students develop problem-solving skills and understand the relationships between variables, which are essential for more advanced topics in mathematics.

Why is "Substituting a negative value without brackets" a common mistake in Substitution?

Why it happens: Students write -5^2 instead of (-5)^2 and end up with -25 instead of 25. How to avoid it: Always wrap the substituted value in brackets — especially when it's negative. Source: 0580 examiner reports — recurring across multiple sittings

Why is "Adding before squaring or multiplying" a common mistake in Substitution?

Why it happens: Reading left-to-right, students add adjacent terms before applying the indices/multiplications. How to avoid it: Underline the powers and multiplications in the substituted expression first; do those before any addition or subtraction.

Why is "Mishandling double negatives" a common mistake in Substitution?

Why it happens: - (-6) is written as -6 instead of +6. How to avoid it: Rewrite double signs explicitly: -(-6) = +6. Then continue.

Why is "Converting $\pi$ to a decimal when the question asks for "in terms of $\pi$"" a common mistake in Substitution?

Why it happens: Calculator habit — students hit the button to get a number. How to avoid it: Read the instruction. "In terms of " means leave the answer as k for some number k. Source: 0580/22 — geometry questions

Why is "Inverting the fraction in $\dfrac{a}{b/c}$" a common mistake in Substitution?

Why it happens: When the denominator is itself a fraction, students forget to flip-and-multiply. How to avoid it: Rewrite a/b/c as a · c/b before substituting numerical values.

Cambridge IGCSE Mathematics (0580)

Substitution

0% Complete

3 tasks remaining

Short Notes - Substitution

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Substitution into Algebraic Expressions — Cambridge IGCSE 0580 Maths Extended (2026)

Replace each variable in an expression with a numerical value, then evaluate. The arithmetic is the easy part — most lost marks come from sign mistakes and misreading BIDMAS.

At a glance

Substitution = replace every variable with the value, then evaluate.
Use brackets when substituting negative or fractional values: $a = -3 \Rightarrow a^{2} = (-3)^{2} = 9$ .
Apply BIDMAS strictly: brackets, indices, multiply/divide, add/subtract.
$2x$ means $2 \times x$ . Implicit multiplication is invisible — don't forget it.
$ab$ means $a \times b$ . $abc$ means $a \times b \times c$ .
Always show working: $(-3)^{2} = 9$ , not just '9'.
Final answers should be simplified: leave as fraction or surd if exact, otherwise round as instructed.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

C2.2 / E2.2 — Substitute numbers into expressions and formulae.

The substitution method

Replace each variable, wrap in brackets, then evaluate using BIDMAS.

Method.

Write down the expression.
For each variable, replace it with its value — wrapped in brackets.
Apply BIDMAS to evaluate.

Wrapping each value in brackets keeps the signs correct through BIDMAS.

Why brackets matter. Without them, a negative substitution can flip signs in unexpected ways:

$a = -3 \Rightarrow a^{2}$ . Without brackets: $-3^{2} = -(3^{2}) = -9$ . With brackets: $(-3)^{2} = 9$ . Different answers.

Always wrap a substituted negative in brackets — without them the sign is lost.

Worked. Find the value of $5x - 2y$ when $x = 3$ and $y = -4$ .

Substitute with brackets: $5(3) - 2(-4)$ .
Multiply: $15 - (-8) = 15 + 8$ .
$= 23$ .

Worked. Find the value of $2a^{2} - 3b$ when $a = -2$ and $b = 5$ .

Substitute: $2(-2)^{2} - 3(5)$ .
Indices first: $(-2)^{2} = 4$ , so $2(4) - 3(5) = 8 - 15$ .
$= -7$ .

Wrap every substitution in brackets, especially negatives.
Indices BEFORE multiplication.
$-3^{2} = -9$ , but $(-3)^{2} = 9$ .
Show every step — examiners credit method marks.

Substituting into formulae

Cambridge gives you a formula and a set of values; you compute. Common contexts: speed, area, volume, interest.

Many Paper 4 questions hand you a formula and a set of values. The skill is identifying which symbol gets which value, then evaluating.

Worked. The formula for the volume of a cone is $V = \dfrac{1}{3}\pi r^{2} h$ . Find $V$ when $r = 4$ cm and $h = 9$ cm. Take $\pi = 3.14$ .

$V = \dfrac{1}{3} \times 3.14 \times (4)^{2} \times 9$ .
Indices: $(4)^{2} = 16$ .
$V = \dfrac{1}{3} \times 3.14 \times 16 \times 9 = \dfrac{452.16}{3} = 150.72$ cm $^{3}$ .

Square the radius first (4² = 16), then multiply through and divide by 3.

Worked. Find $A$ when $P = 250$ , $r = 5$ , $n = 4$ in $A = P\left(1 + \dfrac{r}{100}\right)^{n}$ .

$A = 250 \times \left(1 + \dfrac{5}{100}\right)^{4} = 250 \times (1.05)^{4}$ .
$(1.05)^{4} \approx 1.21551$ .
$A \approx 250 \times 1.21551 \approx 303.88$ .

Substitute the three values, work out the bracket, raise to the power, then multiply.

Read the formula carefully — match each letter to its value.
If the question says 'take $\pi = 3.14$ ', use that, not the calculator's $\pi$ .
State units in the final answer.
Round only at the end of a multi-step calculation.

Substituting fractions and surds

Same method, brackets even more important. Keep exact form unless asked otherwise.

When the value to substitute is a fraction or a surd, brackets are non-negotiable.

Worked. Find $3x^{2} + 2x$ when $x = \dfrac{1}{2}$ .

$3\left(\dfrac{1}{2}\right)^{2} + 2\left(\dfrac{1}{2}\right)$ .
$3 \times \dfrac{1}{4} + 1 = \dfrac{3}{4} + 1 = \dfrac{7}{4}$ .

Wrap ½ in brackets so the square applies correctly: (½)² = ¼.

Worked. Find $a^{2} - b$ when $a = \sqrt{5}$ and $b = 2$ .

$(\sqrt{5})^{2} - 2$ .
$5 - 2 = 3$ .

The square root squared cancels neatly. Cambridge loves this.

Fractions and surds: always wrap in brackets.
$(\sqrt{n})^{2} = n$ — useful simplification.
Keep exact form for final answers when the question allows it.

Quick recap

Substitute = replace each variable with its value, then evaluate.
Always wrap negatives, fractions and surds in brackets.
Apply BIDMAS strictly: indices BEFORE multiplication.
$-3^{2} = -9$ vs $(-3)^{2} = 9$ — brackets change the answer.
Show every step for method marks.
Round only at the end of multi-step calculations.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Variable — a letter that stands for a number whose value can change.
Substitution — replacing a variable with its given value.
BIDMAS — Brackets, Indices, Division/Multiplication (left to right), Addition/Subtraction (left to right).

How it’s examined

Substitution appears on every paper, mostly as 1-2 mark steps inside a larger problem. Paper 2 gives a formula plus values and asks for a numerical answer. Paper 4 hides substitution inside formulae for area, volume, interest, motion, etc. Examiner reports flag negative-substitution sign errors and missed brackets as the recurring failures.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.2); 0580/22 May/Jun 2024 — Q15 (substitution into formula); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Substitution

Step-by-step solutions to past-paper-style questions on substitution, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Substitution

Almost every substitution exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the value, evaluate, work out — with the variable values listed and the expression or formula given.

How to approach it

Replace every variable with its value wrapped in brackets, then apply BIDMAS: indices first, then multiplications, then add and subtract. Show the powers step on its own line.

Common trap

Substituting a negative without brackets — $(-5)^2$ written as $-5^2 = -25$ . Examiner reports flag this every series; brackets protect the sign on a square.

Multi-step problem

Recognise it by

Composite-function notation such as $f(g(-2))$ — one substitution feeds its output into a second function.

How to approach it

Work from the inside out: evaluate the inner function at the given value first, then substitute that result into the outer function.

Common trap

Applying the functions in the wrong order. Examiner reports note candidates evaluate the outer function first — $f(g(x))$ always means do $g$ first, then $f$ .

1Substitute integers into a linear expression
CoreDirect calculation• substitution, linear
Question
Find the value of $5x - 3y$ when $x = 4$ and $y = -2$ .
Step-by-step solution
1. Step 1
  Replace each variable with its value, keeping the operation signs.
  $5x - 3y = 5(4) - 3(-2)$
2. Step 2
  Multiply each term, careful with the negative.
  $= 20 - (-6) = 20 + 6 = 26$
Answer
$26$
Examiner tip
Always wrap the substituted value in brackets, especially for negatives. Writing $5 \times 4 - 3 \times -2$ without brackets is the most frequent slip in the examiner reports.
2Substitute into expressions with squares and cubes
CoreDirect calculation• Adapted from 0580/22 May/Jun 2023 Q4• substitution, powers
Question
Evaluate $a^{2} - 2ab + b^{2}$ when $a = 3$ and $b = -5$ .
Step-by-step solution
1. Step 1
  Substitute, using brackets for the negative.
  $(3)^{2} - 2(3)(-5) + (-5)^{2}$
2. Step 2
  Apply BIDMAS: powers first, then multiplications.
  $= 9 - (-30) + 25$
3. Step 3
  Simplify the double negative.
  $= 9 + 30 + 25 = 64$
Answer
$64$
Examiner tip
$(-5)^{2}$ is $25$ , not $-25$ . Bracketing the substitution is what protects the sign on the square.
3Substitute into a physics-style formula
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q3• formula, substitution
Question
The formula $s = ut + \dfrac{1}{2}at^{2}$ gives the distance travelled. Find $s$ when $u = 5$ , $a = 2$ and $t = 4$ .
Step-by-step solution
1. Step 1
  Substitute carefully.
  $s = (5)(4) + \dfrac{1}{2}(2)(4)^{2}$
2. Step 2
  Powers first, then multiplications.
  $= 20 + \dfrac{1}{2}(2)(16)$
3. Step 3
  Simplify.
  $= 20 + 16 = 36$
Answer
$s = 36$
Examiner tip
Mark schemes credit the BIDMAS layout. Doing $5 \times 4 + \frac{1}{2} \times 2 \times 4 \times 4$ in one line is more error-prone — write the powers step out.
4Substitute fractions into a rational expression
ExtendedDirect calculation• substitution, fractions
Question
Find the value of $\dfrac{2x + y}{x - y}$ when $x = \dfrac{1}{2}$ and $y = -\dfrac{1}{4}$ .
Step-by-step solution
1. Step 1
  Substitute, again with brackets.
  $\dfrac{2(\tfrac{1}{2}) + (-\tfrac{1}{4})}{(\tfrac{1}{2}) - (-\tfrac{1}{4})}$
2. Step 2
  Numerator: $2 \times \tfrac{1}{2} = 1$ , then $1 - \tfrac{1}{4} = \tfrac{3}{4}$ .
3. Step 3
  Denominator: $\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}$ .
4. Step 4
  Divide.
  $\dfrac{3/4}{3/4} = 1$
Answer
$1$
5Substitute into an expression involving $\pi$
CoreDirect calculation• substitution, pi
Question
The volume of a cylinder is $V = \pi r^{2} h$ . Find $V$ when $r = 3$ and $h = 5$ , leaving your answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Substitute.
  $V = \pi (3)^{2}(5)$
2. Step 2
  Power first.
  $= \pi (9)(5) = 45\pi$
Answer
$V = 45\pi$
Examiner tip
When the question says "in terms of $\pi$ ", do not convert to a decimal — $45\pi$ is the final answer.
6Substitute into a quadratic expression
ExtendedDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q9• substitution, quadratic
Question
Given $f(x) = 3x^{2} - 4x + 1$ , find $f(-2)$ .
Step-by-step solution
1. Step 1
  Replace $x$ with $-2$ , using brackets to protect the sign.
  $f(-2) = 3(-2)^{2} - 4(-2) + 1$
2. Step 2
  Indices first: $(-2)^{2} = 4$ .
  $= 3(4) - 4(-2) + 1$
3. Step 3
  Then multiplications.
  $= 12 + 8 + 1$
4. Step 4
  Finally add.
  $= 21$
Answer
$f(-2) = 21$
Examiner tip
The examiner report flags candidates often write $-2^{2} = -4$ instead of $(-2)^{2} = 4$ . Bracket the negative every time you substitute it.
7Substitute into a kinematics formula with a negative acceleration
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q6• substitution, formula, negatives
Question
The formula $v = u + at$ gives the velocity after time $t$ . Find $v$ when $u = 12$ , $a = -3$ and $t = 5$ .
Step-by-step solution
1. Step 1
  Substitute, keeping the sign of $a$ in brackets.
  $v = 12 + (-3)(5)$
2. Step 2
  Multiplication before addition.
  $= 12 + (-15)$
3. Step 3
  Simplify.
  $= -3$
Answer
$v = -3$
Examiner tip
The 2024 mark scheme awards method marks for showing $(-3)(5) = -15$ separately before combining. Doing it in one step often introduces sign slips.
8Substitute into a surface-area formula
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q4• substitution, formula, geometry
Question
The total surface area of a closed cylinder is $A = 2\pi r^{2} + 2\pi r h$ . Find $A$ when $r = 4$ and $h = 7$ , giving your answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Substitute the values into each term.
  $A = 2\pi (4)^{2} + 2\pi (4)(7)$
2. Step 2
  Apply powers, then multiplications.
  $= 2\pi (16) + 2\pi (28)$
3. Step 3
  Compute each term.
  $= 32\pi + 56\pi$
4. Step 4
  Combine like terms in $\pi$ .
  $= 88\pi$
Answer
$A = 88\pi$
Examiner tip
Examiner reports note candidates often square only $r$ in the first term ( $2\pi \times 4^{2}$ ) and forget to keep the coefficient $2$ . Write out the full substitution before evaluating.
9BIDMAS trap with a negative substitution
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q5• substitution, BIDMAS, negatives
Question
Find the value of $p - q^{2} + 2pq$ when $p = -3$ and $q = -1$ .
Step-by-step solution
1. Step 1
  Substitute with brackets.
  $(-3) - (-1)^{2} + 2(-3)(-1)$
2. Step 2
  Indices first: $(-1)^{2} = 1$ .
  $= -3 - 1 + 2(-3)(-1)$
3. Step 3
  Multiplications next: $2 \times (-3) \times (-1) = 6$ .
  $= -3 - 1 + 6$
4. Step 4
  Add and subtract from left to right.
  $= 2$
Answer
$2$
Examiner tip
The 2024 mark scheme awards method marks for clearly separating the indices and multiplication steps. Doing $(-1)^{2}$ as $-1$ is the dominant slip — bracket the negative to protect the sign on the square.
10Substitute one function into another at a point
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2023 Q19• substitution, functions, composition
Question
Given $f(x) = 2x - 1$ and $g(x) = x^{2} + 3$ , find the value of $f(g(-2))$ .
Step-by-step solution
1. Step 1
  Work from the inside out — first evaluate $g(-2)$ .
  $g(-2) = (-2)^{2} + 3 = 4 + 3 = 7$
2. Step 2
  Now substitute this output into $f$ .
  $f(7) = 2(7) - 1$
3. Step 3
  Simplify.
  $= 14 - 1 = 13$
Answer
$f(g(-2)) = 13$
Examiner tip
The examiner report flags candidates often evaluate $f$ first then substitute into $g$ — the reverse order. The notation $f(g(x))$ means: apply $g$ first, then $f$ to the result.

Key Formulae — Substitution

The formulae you need to memorise for substitution on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Order of operations (BIDMAS / PEMDAS)
$B \to I \to DM \to AS$
$B$
Brackets first
$I$
Indices (powers and roots) next
$DM$
Division and Multiplication, left to right
$AS$
Addition and Subtraction, left to right
When to use
Always apply BIDMAS when evaluating substituted expressions — it determines what you simplify first.
Bracket-substitution rule
$\text{If } x = a, \text{ then } f(x) \to f((a))$
When to use
Always wrap the value you substitute in brackets, especially when the value is negative or a fraction. Brackets protect signs and indices.

Key Definitions and Keywords — Substitution

Definitions to memorise and the exact keywords mark schemes credit for substitution answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Variable
Examiner keyword
A symbol (typically a letter) that represents an unknown number.
Example
In $3x + 2$ , the variable is $x$ .
Coefficient
Examiner keyword
The numerical factor in front of a variable.
Example
In $5x$ , the coefficient is $5$ .
Constant
A term in an expression that has no variable — just a number.
Example
In $3x + 7$ , the constant is $7$ .
Term
Examiner keyword
Each part of an algebraic expression separated by a $+$ or $-$ sign.
Example
In $4x - 3y + 5$ , the terms are $4x$ , $-3y$ and $5$ .
Substitute
Examiner keyword
Replace a variable with a given numerical value before evaluating the expression.
Evaluate
Examiner keyword
Find the numerical value of an expression after substitution.

Common Mistakes and Misconceptions — Substitution

The traps other students keep falling into on substitution questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Substituting a negative value without brackets
0580 examiner reports — recurring across multiple sittings
Why it happens
Students write $-5^{2}$ instead of $(-5)^{2}$ and end up with $-25$ instead of $25$ .
How to avoid it
Always wrap the substituted value in brackets — especially when it's negative.
✕Adding before squaring or multiplying
Why it happens
Reading left-to-right, students add adjacent terms before applying the indices/multiplications.
How to avoid it
Underline the powers and multiplications in the substituted expression first; do those before any addition or subtraction.
✕Mishandling double negatives
Why it happens
$- (-6)$ is written as $-6$ instead of $+6$ .
How to avoid it
Rewrite double signs explicitly: $-(-6) = +6$ . Then continue.
✕Converting $\pi$ to a decimal when the question asks for "in terms of $\pi$ "
0580/22 — geometry questions
Why it happens
Calculator habit — students hit the $\pi$ button to get a number.
How to avoid it
Read the instruction. "In terms of $\pi$ " means leave the answer as $k\pi$ for some number $k$ .
✕Inverting the fraction in $\dfrac{a}{b/c}$
Why it happens
When the denominator is itself a fraction, students forget to flip-and-multiply.
How to avoid it
Rewrite $\dfrac{a}{b/c}$ as $\dfrac{a \cdot c}{b}$ before substituting numerical values.

Substitution — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Substitution

0% Complete

3 tasks remaining

Short Notes - Substitution

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Substitution into Algebraic Expressions — Cambridge IGCSE 0580 Maths Extended (2026)

Replace each variable in an expression with a numerical value, then evaluate. The arithmetic is the easy part — most lost marks come from sign mistakes and misreading BIDMAS.

At a glance

Substitution = replace every variable with the value, then evaluate.
Use brackets when substituting negative or fractional values: $a = -3 \Rightarrow a^{2} = (-3)^{2} = 9$ .
Apply BIDMAS strictly: brackets, indices, multiply/divide, add/subtract.
$2x$ means $2 \times x$ . Implicit multiplication is invisible — don't forget it.
$ab$ means $a \times b$ . $abc$ means $a \times b \times c$ .
Always show working: $(-3)^{2} = 9$ , not just '9'.
Final answers should be simplified: leave as fraction or surd if exact, otherwise round as instructed.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

C2.2 / E2.2 — Substitute numbers into expressions and formulae.

The substitution method

Replace each variable, wrap in brackets, then evaluate using BIDMAS.

Method.

Write down the expression.
For each variable, replace it with its value — wrapped in brackets.
Apply BIDMAS to evaluate.

Wrapping each value in brackets keeps the signs correct through BIDMAS.

Why brackets matter. Without them, a negative substitution can flip signs in unexpected ways:

$a = -3 \Rightarrow a^{2}$ . Without brackets: $-3^{2} = -(3^{2}) = -9$ . With brackets: $(-3)^{2} = 9$ . Different answers.

Always wrap a substituted negative in brackets — without them the sign is lost.

Worked. Find the value of $5x - 2y$ when $x = 3$ and $y = -4$ .

Substitute with brackets: $5(3) - 2(-4)$ .
Multiply: $15 - (-8) = 15 + 8$ .
$= 23$ .

Worked. Find the value of $2a^{2} - 3b$ when $a = -2$ and $b = 5$ .

Substitute: $2(-2)^{2} - 3(5)$ .
Indices first: $(-2)^{2} = 4$ , so $2(4) - 3(5) = 8 - 15$ .
$= -7$ .

Wrap every substitution in brackets, especially negatives.
Indices BEFORE multiplication.
$-3^{2} = -9$ , but $(-3)^{2} = 9$ .
Show every step — examiners credit method marks.

Substituting into formulae

Cambridge gives you a formula and a set of values; you compute. Common contexts: speed, area, volume, interest.

Many Paper 4 questions hand you a formula and a set of values. The skill is identifying which symbol gets which value, then evaluating.

Worked. The formula for the volume of a cone is $V = \dfrac{1}{3}\pi r^{2} h$ . Find $V$ when $r = 4$ cm and $h = 9$ cm. Take $\pi = 3.14$ .

$V = \dfrac{1}{3} \times 3.14 \times (4)^{2} \times 9$ .
Indices: $(4)^{2} = 16$ .
$V = \dfrac{1}{3} \times 3.14 \times 16 \times 9 = \dfrac{452.16}{3} = 150.72$ cm $^{3}$ .

Square the radius first (4² = 16), then multiply through and divide by 3.

Worked. Find $A$ when $P = 250$ , $r = 5$ , $n = 4$ in $A = P\left(1 + \dfrac{r}{100}\right)^{n}$ .

$A = 250 \times \left(1 + \dfrac{5}{100}\right)^{4} = 250 \times (1.05)^{4}$ .
$(1.05)^{4} \approx 1.21551$ .
$A \approx 250 \times 1.21551 \approx 303.88$ .

Substitute the three values, work out the bracket, raise to the power, then multiply.

Read the formula carefully — match each letter to its value.
If the question says 'take $\pi = 3.14$ ', use that, not the calculator's $\pi$ .
State units in the final answer.
Round only at the end of a multi-step calculation.

Substituting fractions and surds

Same method, brackets even more important. Keep exact form unless asked otherwise.

When the value to substitute is a fraction or a surd, brackets are non-negotiable.

Worked. Find $3x^{2} + 2x$ when $x = \dfrac{1}{2}$ .

$3\left(\dfrac{1}{2}\right)^{2} + 2\left(\dfrac{1}{2}\right)$ .
$3 \times \dfrac{1}{4} + 1 = \dfrac{3}{4} + 1 = \dfrac{7}{4}$ .

Wrap ½ in brackets so the square applies correctly: (½)² = ¼.

Worked. Find $a^{2} - b$ when $a = \sqrt{5}$ and $b = 2$ .

$(\sqrt{5})^{2} - 2$ .
$5 - 2 = 3$ .

The square root squared cancels neatly. Cambridge loves this.

Fractions and surds: always wrap in brackets.
$(\sqrt{n})^{2} = n$ — useful simplification.
Keep exact form for final answers when the question allows it.

Quick recap

Substitute = replace each variable with its value, then evaluate.
Always wrap negatives, fractions and surds in brackets.
Apply BIDMAS strictly: indices BEFORE multiplication.
$-3^{2} = -9$ vs $(-3)^{2} = 9$ — brackets change the answer.
Show every step for method marks.
Round only at the end of multi-step calculations.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Variable — a letter that stands for a number whose value can change.
Substitution — replacing a variable with its given value.
BIDMAS — Brackets, Indices, Division/Multiplication (left to right), Addition/Subtraction (left to right).

How it’s examined

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.2); 0580/22 May/Jun 2024 — Q15 (substitution into formula); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Substitution

Step-by-step solutions to past-paper-style questions on substitution, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Substitution

Almost every substitution exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the value, evaluate, work out — with the variable values listed and the expression or formula given.

How to approach it

Replace every variable with its value wrapped in brackets, then apply BIDMAS: indices first, then multiplications, then add and subtract. Show the powers step on its own line.

Common trap

Substituting a negative without brackets — $(-5)^2$ written as $-5^2 = -25$ . Examiner reports flag this every series; brackets protect the sign on a square.

Multi-step problem

Recognise it by

Composite-function notation such as $f(g(-2))$ — one substitution feeds its output into a second function.

How to approach it

Work from the inside out: evaluate the inner function at the given value first, then substitute that result into the outer function.

Common trap

Applying the functions in the wrong order. Examiner reports note candidates evaluate the outer function first — $f(g(x))$ always means do $g$ first, then $f$ .

1Substitute integers into a linear expression
CoreDirect calculation• substitution, linear
Question
Find the value of $5x - 3y$ when $x = 4$ and $y = -2$ .
Step-by-step solution
1. Step 1
  Replace each variable with its value, keeping the operation signs.
  $5x - 3y = 5(4) - 3(-2)$
2. Step 2
  Multiply each term, careful with the negative.
  $= 20 - (-6) = 20 + 6 = 26$
Answer
$26$
Examiner tip
Always wrap the substituted value in brackets, especially for negatives. Writing $5 \times 4 - 3 \times -2$ without brackets is the most frequent slip in the examiner reports.
2Substitute into expressions with squares and cubes
CoreDirect calculation• Adapted from 0580/22 May/Jun 2023 Q4• substitution, powers
Question
Evaluate $a^{2} - 2ab + b^{2}$ when $a = 3$ and $b = -5$ .
Step-by-step solution
1. Step 1
  Substitute, using brackets for the negative.
  $(3)^{2} - 2(3)(-5) + (-5)^{2}$
2. Step 2
  Apply BIDMAS: powers first, then multiplications.
  $= 9 - (-30) + 25$
3. Step 3
  Simplify the double negative.
  $= 9 + 30 + 25 = 64$
Answer
$64$
Examiner tip
$(-5)^{2}$ is $25$ , not $-25$ . Bracketing the substitution is what protects the sign on the square.
3Substitute into a physics-style formula
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q3• formula, substitution
Question
The formula $s = ut + \dfrac{1}{2}at^{2}$ gives the distance travelled. Find $s$ when $u = 5$ , $a = 2$ and $t = 4$ .
Step-by-step solution
1. Step 1
  Substitute carefully.
  $s = (5)(4) + \dfrac{1}{2}(2)(4)^{2}$
2. Step 2
  Powers first, then multiplications.
  $= 20 + \dfrac{1}{2}(2)(16)$
3. Step 3
  Simplify.
  $= 20 + 16 = 36$
Answer
$s = 36$
Examiner tip
Mark schemes credit the BIDMAS layout. Doing $5 \times 4 + \frac{1}{2} \times 2 \times 4 \times 4$ in one line is more error-prone — write the powers step out.
4Substitute fractions into a rational expression
ExtendedDirect calculation• substitution, fractions
Question
Find the value of $\dfrac{2x + y}{x - y}$ when $x = \dfrac{1}{2}$ and $y = -\dfrac{1}{4}$ .
Step-by-step solution
1. Step 1
  Substitute, again with brackets.
  $\dfrac{2(\tfrac{1}{2}) + (-\tfrac{1}{4})}{(\tfrac{1}{2}) - (-\tfrac{1}{4})}$
2. Step 2
  Numerator: $2 \times \tfrac{1}{2} = 1$ , then $1 - \tfrac{1}{4} = \tfrac{3}{4}$ .
3. Step 3
  Denominator: $\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}$ .
4. Step 4
  Divide.
  $\dfrac{3/4}{3/4} = 1$
Answer
$1$
5Substitute into an expression involving $\pi$
CoreDirect calculation• substitution, pi
Question
The volume of a cylinder is $V = \pi r^{2} h$ . Find $V$ when $r = 3$ and $h = 5$ , leaving your answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Substitute.
  $V = \pi (3)^{2}(5)$
2. Step 2
  Power first.
  $= \pi (9)(5) = 45\pi$
Answer
$V = 45\pi$
Examiner tip
When the question says "in terms of $\pi$ ", do not convert to a decimal — $45\pi$ is the final answer.
6Substitute into a quadratic expression
ExtendedDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q9• substitution, quadratic
Question
Given $f(x) = 3x^{2} - 4x + 1$ , find $f(-2)$ .
Step-by-step solution
1. Step 1
  Replace $x$ with $-2$ , using brackets to protect the sign.
  $f(-2) = 3(-2)^{2} - 4(-2) + 1$
2. Step 2
  Indices first: $(-2)^{2} = 4$ .
  $= 3(4) - 4(-2) + 1$
3. Step 3
  Then multiplications.
  $= 12 + 8 + 1$
4. Step 4
  Finally add.
  $= 21$
Answer
$f(-2) = 21$
Examiner tip
The examiner report flags candidates often write $-2^{2} = -4$ instead of $(-2)^{2} = 4$ . Bracket the negative every time you substitute it.
7Substitute into a kinematics formula with a negative acceleration
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q6• substitution, formula, negatives
Question
The formula $v = u + at$ gives the velocity after time $t$ . Find $v$ when $u = 12$ , $a = -3$ and $t = 5$ .
Step-by-step solution
1. Step 1
  Substitute, keeping the sign of $a$ in brackets.
  $v = 12 + (-3)(5)$
2. Step 2
  Multiplication before addition.
  $= 12 + (-15)$
3. Step 3
  Simplify.
  $= -3$
Answer
$v = -3$
Examiner tip
The 2024 mark scheme awards method marks for showing $(-3)(5) = -15$ separately before combining. Doing it in one step often introduces sign slips.
8Substitute into a surface-area formula
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q4• substitution, formula, geometry
Question
The total surface area of a closed cylinder is $A = 2\pi r^{2} + 2\pi r h$ . Find $A$ when $r = 4$ and $h = 7$ , giving your answer in terms of $\pi$ .
Step-by-step solution
1. Step 1
  Substitute the values into each term.
  $A = 2\pi (4)^{2} + 2\pi (4)(7)$
2. Step 2
  Apply powers, then multiplications.
  $= 2\pi (16) + 2\pi (28)$
3. Step 3
  Compute each term.
  $= 32\pi + 56\pi$
4. Step 4
  Combine like terms in $\pi$ .
  $= 88\pi$
Answer
$A = 88\pi$
Examiner tip
Examiner reports note candidates often square only $r$ in the first term ( $2\pi \times 4^{2}$ ) and forget to keep the coefficient $2$ . Write out the full substitution before evaluating.
9BIDMAS trap with a negative substitution
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q5• substitution, BIDMAS, negatives
Question
Find the value of $p - q^{2} + 2pq$ when $p = -3$ and $q = -1$ .
Step-by-step solution
1. Step 1
  Substitute with brackets.
  $(-3) - (-1)^{2} + 2(-3)(-1)$
2. Step 2
  Indices first: $(-1)^{2} = 1$ .
  $= -3 - 1 + 2(-3)(-1)$
3. Step 3
  Multiplications next: $2 \times (-3) \times (-1) = 6$ .
  $= -3 - 1 + 6$
4. Step 4
  Add and subtract from left to right.
  $= 2$
Answer
$2$
Examiner tip
The 2024 mark scheme awards method marks for clearly separating the indices and multiplication steps. Doing $(-1)^{2}$ as $-1$ is the dominant slip — bracket the negative to protect the sign on the square.
10Substitute one function into another at a point
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2023 Q19• substitution, functions, composition
Question
Given $f(x) = 2x - 1$ and $g(x) = x^{2} + 3$ , find the value of $f(g(-2))$ .
Step-by-step solution
1. Step 1
  Work from the inside out — first evaluate $g(-2)$ .
  $g(-2) = (-2)^{2} + 3 = 4 + 3 = 7$
2. Step 2
  Now substitute this output into $f$ .
  $f(7) = 2(7) - 1$
3. Step 3
  Simplify.
  $= 14 - 1 = 13$
Answer
$f(g(-2)) = 13$
Examiner tip
The examiner report flags candidates often evaluate $f$ first then substitute into $g$ — the reverse order. The notation $f(g(x))$ means: apply $g$ first, then $f$ to the result.

Key Formulae — Substitution

The formulae you need to memorise for substitution on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Order of operations (BIDMAS / PEMDAS)
$B \to I \to DM \to AS$
$B$
Brackets first
$I$
Indices (powers and roots) next
$DM$
Division and Multiplication, left to right
$AS$
Addition and Subtraction, left to right
When to use
Always apply BIDMAS when evaluating substituted expressions — it determines what you simplify first.
Bracket-substitution rule
$\text{If } x = a, \text{ then } f(x) \to f((a))$
When to use
Always wrap the value you substitute in brackets, especially when the value is negative or a fraction. Brackets protect signs and indices.

Key Definitions and Keywords — Substitution

Definitions to memorise and the exact keywords mark schemes credit for substitution answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Variable
Examiner keyword
A symbol (typically a letter) that represents an unknown number.
Example
In $3x + 2$ , the variable is $x$ .
Coefficient
Examiner keyword
The numerical factor in front of a variable.
Example
In $5x$ , the coefficient is $5$ .
Constant
A term in an expression that has no variable — just a number.
Example
In $3x + 7$ , the constant is $7$ .
Term
Examiner keyword
Each part of an algebraic expression separated by a $+$ or $-$ sign.
Example
In $4x - 3y + 5$ , the terms are $4x$ , $-3y$ and $5$ .
Substitute
Examiner keyword
Replace a variable with a given numerical value before evaluating the expression.
Evaluate
Examiner keyword
Find the numerical value of an expression after substitution.

Common Mistakes and Misconceptions — Substitution

The traps other students keep falling into on substitution questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Substituting a negative value without brackets
0580 examiner reports — recurring across multiple sittings
Why it happens
Students write $-5^{2}$ instead of $(-5)^{2}$ and end up with $-25$ instead of $25$ .
How to avoid it
Always wrap the substituted value in brackets — especially when it's negative.
✕Adding before squaring or multiplying
Why it happens
Reading left-to-right, students add adjacent terms before applying the indices/multiplications.
How to avoid it
Underline the powers and multiplications in the substituted expression first; do those before any addition or subtraction.
✕Mishandling double negatives
Why it happens
$- (-6)$ is written as $-6$ instead of $+6$ .
How to avoid it
Rewrite double signs explicitly: $-(-6) = +6$ . Then continue.
✕Converting $\pi$ to a decimal when the question asks for "in terms of $\pi$ "
0580/22 — geometry questions
Why it happens
Calculator habit — students hit the $\pi$ button to get a number.
How to avoid it
Read the instruction. "In terms of $\pi$ " means leave the answer as $k\pi$ for some number $k$ .
✕Inverting the fraction in $\dfrac{a}{b/c}$
Why it happens
When the denominator is itself a fraction, students forget to flip-and-multiply.
How to avoid it
Rewrite $\dfrac{a}{b/c}$ as $\dfrac{a \cdot c}{b}$ before substituting numerical values.

Substitution — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Substitution

Short Notes - Substitution

Detailed Study Notes

At a glance

What you’ll learn

The substitution method

Substituting into formulae

Substituting fractions and surds

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Substitution

1Substitute integers into a linear expression

2Substitute into expressions with squares and cubes

3Substitute into a physics-style formula

4Substitute fractions into a rational expression

5Substitute into an expression involving π\piπ

6Substitute into a quadratic expression

7Substitute into a kinematics formula with a negative acceleration

8Substitute into a surface-area formula

9BIDMAS trap with a negative substitution

10Substitute one function into another at a point

Order of operations (BIDMAS / PEMDAS)

Bracket-substitution rule

Variable

Coefficient

Constant

Term

Substitute

Evaluate

✕Substituting a negative value without brackets

✕Adding before squaring or multiplying

✕Mishandling double negatives

✕Converting π\piπ to a decimal when the question asks for "in terms of π\piπ"

✕Inverting the fraction in ab/c\dfrac{a}{b/c}b/ca​

Substitution — frequently asked questions

Substitution

Short Notes - Substitution

Detailed Study Notes

At a glance

What you’ll learn

The substitution method

Substituting into formulae

Substituting fractions and surds

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Substitution

1Substitute integers into a linear expression

2Substitute into expressions with squares and cubes

3Substitute into a physics-style formula

4Substitute fractions into a rational expression

5Substitute into an expression involving π\piπ

6Substitute into a quadratic expression

7Substitute into a kinematics formula with a negative acceleration

8Substitute into a surface-area formula

9BIDMAS trap with a negative substitution

10Substitute one function into another at a point

Order of operations (BIDMAS / PEMDAS)

Bracket-substitution rule

Variable

Coefficient

Constant

Term

Substitute

Evaluate

✕Substituting a negative value without brackets

✕Adding before squaring or multiplying

✕Mishandling double negatives

✕Converting π\piπ to a decimal when the question asks for "in terms of π\piπ"

✕Inverting the fraction in ab/c\dfrac{a}{b/c}b/ca​

Substitution — frequently asked questions

5Substitute into an expression involving $\pi$

✕Converting $\pi$ to a decimal when the question asks for "in terms of $\pi$ "

✕Inverting the fraction in $\dfrac{a}{b/c}$

5Substitute into an expression involving $\pi$

✕Converting $\pi$ to a decimal when the question asks for "in terms of $\pi$ "

✕Inverting the fraction in $\dfrac{a}{b/c}$