Algebra · notation and manipulation

Section A

5 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

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Use and interpret algebraic manipulation: ab for a × b, 3y for y + y + y and 3 × y, a² for a × a, a³ for a × a × a, a²b for a × a × b, a/b for a ÷ b, coefficients as fractions, brackets

Notes

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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus three correct distinct drills across at least two dates and two practice sources.

Explanation

Algebraic notation records operations compactly. Adjacent letters mean multiplication, so $ab=a\times b$ , while $3y$ means three lots of $y$ .
An index records repeated factors: $a^2=a\times a$ and $a^2b=a\times a\times b$ .
A fraction bar represents division and also groups its numerator and denominator.
Write numerical coefficients before variables, usually as exact fractions rather than decimals, and use brackets when an operation acts on a complete expression.
Examiners expect conventional notation and the operations to remain unambiguous when translating words or repeated products.

Worked example

A rectangle has length $\dfrac{3x}{2}$ and width $y$ . Write its area and perimeter in conventional algebraic notation.

1.Area $=\dfrac{3x}{2}\times y=\dfrac{3xy}{2}$ .
2.Perimeter $=2\left(\dfrac{3x}{2}\right)+2y$ .
3.Simplify to $3x+2y$ .

Answer: Area $=\dfrac{3xy}{2}$ and perimeter $=3x+2y$ .

Common mistakes

Don't read $a^2$ as $2a$ instead of $a\times a$ .
Don't interpret $3y$ as $3+y$ rather than $3\times y$ .
Don't drop brackets when a multiplier must act on a whole expression.

Exam tip

Translate one operation at a time and use brackets before simplifying the notation.

Tier 1 · Easy

1 mark

ORIGINAL

Write $m\times m\times n\times n\times n$ using indices.

Tier 2 · Standard

2 marks

ORIGINAL

Write $p\times p\times p\times q\div5$ in conventional algebraic notation, and state its coefficient.

Tier 3 · Hard

3 marks

ORIGINAL

A rectangle has length $\frac{3x}{2}$ and width $y$ . Write its area and its perimeter in conventional algebraic notation.

Substitute numerical values into formulae and expressions, including scientific formulae

Notes

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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus three correct distinct drills across at least two dates and two practice sources.

Explanation

Substitution replaces every occurrence of a variable with its given value while preserving the original operations. Put negative and fractional values in brackets so that powers and signs act on the whole value.
Follow the order of operations: evaluate powers before multiplication, division, addition and subtraction.
In a scientific formula, include the stated units and convert them first if necessary.
Keep full calculator precision until the requested rounding.
Examiners award method for a correct substituted expression, so write that line before evaluating rather than giving only a calculator answer.

Worked example

The kinetic energy formula is $E=\dfrac12mv^2$ . Find $E$ when $m=3.2\,\text{kg}$ and $v=5\,\text{m s}^{-1}$ .

1.Substitute both values: $E=\dfrac12\times3.2\times5^2$ .
2.Evaluate the power first: $5^2=25$ .
3. $E=0.5\times3.2\times25=40\,\text{J}$ .

Answer: $40\,\text{J}$ .

Common mistakes

Don't substitute $v=5$ into $v^2$ as $2\times5$ .
Don't write $(-3)^2$ as $-9$ after omitting the brackets.
Don't round an intermediate value and loses accuracy in the final answer.

Exam tip

Show the formula with every value substituted before entering it into the calculator.

Tier 1 · Easy

2 marks

ORIGINAL

Work out $2x^2-5$ when $x=-3$ .

Tier 2 · Standard

3 marks

ORIGINAL

The kinetic energy of an object is given by $E=\frac12mv^2$ . Work out $E$ when $m=3.2\,\text{kg}$ and $v=5\,\text{m s}^{-1}$ .

Tier 3 · Hard

3 marks

ORIGINAL

Use $E=mc^2$ to calculate $E$ when $m=4.2\times10^{-8}\,\text{kg}$ and $c=3\times10^8\,\text{m s}^{-1}$ . Give your answer in standard form.

Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

Notes

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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus three correct distinct drills across at least two dates and two practice sources.

Explanation

An expression has no equality or inequality sign. An equation is true only for particular values, whereas an identity is true for every permitted value and is written with $\equiv$ .
A formula links quantities, and an inequality compares a range of possible values.
Terms are separated by addition or subtraction; factors are quantities multiplied together.
For example, $6x^2-15x$ has two terms and factorises as $3x(2x-5)$ , whose factors are $3x$ and $2x-5$ .
Examiners expect the correct vocabulary and a reason based on the statement's structure or truth.

Worked example

Classify $3(2x-1)=9$ , $3(2x-1)\equiv6x-3$ , and $3(2x-1)<9$ .

1. $3(2x-1)=9$ is true only for a particular value, so it is an equation.
2.Expanding $3(2x-1)$ always gives $6x-3$ , so the second statement is an identity.
3.The symbol $<$ compares possible values, so the third statement is an inequality.

Answer: Equation, identity, inequality, in that order.

Common mistakes

Don't call every statement containing an equals sign an identity.
Don't count factors as terms even though terms are separated by addition or subtraction.
Don't use $=$ instead of $\equiv$ for a relationship true for all permitted values.

Exam tip

When asked to classify a statement, justify whether it is always true, sometimes true or a comparison.

Tier 1 · Easy

1 mark

ORIGINAL

State whether $5x+7$ is an expression, equation or inequality.

Tier 2 · Standard

3 marks

ORIGINAL

For $8x^2-12x=4x(2x-3)$ , state the number of terms on the left and name the two factors on the right.

Tier 3 · Hard

3 marks

ORIGINAL

Classify each statement as an equation, an identity or an inequality: $3(2x-1)=9$ , $3(2x-1)\equiv6x-3$ , and $3(2x-1)<9$ .

Simplify and manipulate algebraic expressions (incl. surds and algebraic fractions): like terms, common factors, expanding two or more binomials, factorising quadratics incl. ax² + bx + c, indices

Notes

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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus three correct distinct drills across at least two dates and two practice sources.

Explanation

Simplify by collecting only like terms and applying index laws only to matching bases. Expand brackets by multiplying every required pair of terms, then collect.
When factorising, first remove any common factor and check that re-expansion reproduces every term.
Foundation questions can include expanding two binomials and factorising $x^2+bx+c$ .
Higher tier: manipulation extends to surds, algebraic fractions, products of more binomials and quadratics $ax^2+bx+c$ .
Never cancel terms across addition; factorise complete numerators and denominators first, then retain values excluded by the original denominator.

Worked example

Factorise $x^2+7x+12$ fully.

1.Find two numbers with product $12$ and sum $7$ : $3$ and $4$ .
2.Write the factors $(x+3)(x+4)$ .
3.Check by expanding: $x^2+4x+3x+12=x^2+7x+12$ .

Answer: $(x+3)(x+4)$ .

Common mistakes

Don't collect unlike terms such as $3x+2x^2$ to make $5x^3$ .
Don't miss a cross-term when expanding two brackets.
Don't cancel terms across addition in an algebraic fraction.

Exam tip

After factorising, expand your answer mentally; it must reproduce the original expression exactly.

Tier 1 · Easy

2 marks

ORIGINAL

Simplify $7a+3b-2a+5b$ .

Tier 2 · Standard

3 marks

ORIGINAL

Factorise $6x^2+x-2$ .

Tier 3 · Hard

4 marks

ORIGINAL

Simplify $\frac{x^2-9}{x^2+x-6}$ , stating every value of $x$ excluded from the original expression.

Understand and use standard mathematical formulae; rearrange formulae to change the subject

Notes

Unseen

Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus three correct distinct drills across at least two dates and two practice sources.

Explanation

The subject of a formula is the variable isolated on one side. Changing the subject must preserve an equivalent relationship, so perform the same operation on both sides and undo operations in reverse order.
Clear fractions or brackets when this makes the structure easier to see.
Foundation questions usually isolate a subject that appears once.
Higher tier: the subject may appear more than once or inside a fraction, requiring expansion and collection of its terms.
Examiners expect each inverse operation to be visible and the final subject to appear alone.

Worked example

Make $t$ the subject of $v=u+at$ .

1.Subtract $u$ from both sides: $v-u=at$ .
2.Divide both sides by $a$ : $\dfrac{v-u}{a}=t$ .
3.Write the subject first: $t=\dfrac{v-u}{a}$ .

Answer: $t=\dfrac{v-u}{a}$ .

Common mistakes

Don't change a sign while moving a term without applying an operation to both sides.
Don't divide only one term of a sum instead of the entire side.
Don't stop with the requested subject still multiplied by another quantity.

Exam tip

State one balancing operation per line until the requested subject is alone.

Tier 1 · Easy

1 mark

ORIGINAL

Make $w$ the subject of $A=lw$ .

Tier 2 · Standard

3 marks

ORIGINAL

The area of a trapezium is given by $A=\dfrac12(a+b)h$ , where $A$ is measured in $\text{cm}^2$ and $a$ , $b$ and $h$ are measured in cm. (a) Make $h$ the subject of the formula. (b) Hence work out $h$ when $A=45\,\text{cm}^2$ , $a=7\,\text{cm}$ and $b=11\,\text{cm}$ .

Tier 3 · Hard

4 marks

ORIGINAL

Make $a$ the subject of $P=\frac{a+b}{a-b}$ .

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