A Algebra · notation and manipulation — revision question pack

5 specification points · notes, questions, answers and worked methods

A1 · 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

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. Write $m\times m\times n\times n\times n$ using indices.[1 mark]

Tier 2 · Standard

1. Write $p\times p\times p\times q\div5$ in conventional algebraic notation, and state its coefficient.[2 marks]

Tier 3 · Hard

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

A2 · Substitute numerical values into formulae and expressions, including scientific formulae

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

1. Work out $2x^2-5$ when $x=-3$ .[2 marks]

Tier 2 · Standard

1. The kinetic energy of an object is given by $E=\frac12mv^2$ . Work out $E$ when $m=3.2\,\text{kg}$ and $v=5\,\text{m s}^{-1}$ .[3 marks]

Tier 3 · Hard

1. 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.[3 marks]

A3 · Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

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. State whether $5x+7$ is an expression, equation or inequality.[1 mark]

Tier 2 · Standard

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

Tier 3 · Hard

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

A4 · 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

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

1. Simplify $7a+3b-2a+5b$ .[2 marks]

Tier 2 · Standard

1. Factorise $6x^2+x-2$ .[3 marks]

Tier 3 · Hard

1. Simplify $\frac{x^2-9}{x^2+x-6}$ , stating every value of $x$ excluded from the original expression.[4 marks]

A5 · Understand and use standard mathematical formulae; rearrange formulae to change the subject

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. Make $w$ the subject of $A=lw$ .[1 mark]

Tier 2 · Standard

1. 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 $b=11\,\text{cm}$ .[3 marks]

Tier 3 · Hard

1. Make $a$ the subject of $P=\frac{a+b}{a-b}$ .[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

A1 · 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

Tier 1 · Easy

1. Answer
- $m^2n^3$
Method: There are two factors of $m$ and three factors of $n$ , so $m\times m=m^2$ and $n\times n\times n=n^3$ . Therefore the product is $m^2n^3$ .

Tier 2 · Standard

1. Answer
- $\frac{p^3q}{5}$
- Coefficient: $\frac15$
Method: The three factors of $p$ give $p^3$ , and division by $5$ is shown by a fraction bar. Hence the expression is $p^3q/5=\frac15p^3q$ , so its coefficient is $\frac15$ .

Tier 3 · Hard

1. Answer
- Area: $\frac{3xy}{2}$
- Perimeter: $3x+2y$
Method: Area is length multiplied by width, so $A=(3x/2)y=3xy/2$ . Perimeter is twice the length plus twice the width: $P=2(3x/2)+2y=3x+2y$ .

A2 · Substitute numerical values into formulae and expressions, including scientific formulae

Tier 1 · Easy

1. Answer
- $13$
Method: Substitute using brackets: $2(-3)^2-5=2\times9-5=18-5=13$ .

Tier 2 · Standard

1. Answer
- $40\,\text{J}$
Method: Substitute both values: $E=\frac12\times3.2\times5^2$ . Since $5^2=25$ , $E=0.5\times3.2\times25=40\,\text{J}$ .

Tier 3 · Hard

1. Answer
- $3.78\times10^9\,\text{J}$
Method: Substitute before evaluating: $E=(4.2\times10^{-8})(3\times10^8)^2$ . Squaring gives $9\times10^{16}$ , so $E=4.2\times9\times10^8=37.8\times10^8=3.78\times10^9\,\text{J}$ .

A3 · Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

Tier 1 · Easy

1. Answer
- Expression
Method: The algebra has no equality or inequality sign, so it is an expression.

Tier 2 · Standard

1. Answer
- Two terms
- Factors $4x$ and $(2x-3)$
Method: The addition or subtraction signs separate $8x^2$ and $-12x$ , so there are two terms. On the right, $4x$ is multiplied by the bracket $(2x-3)$ , so these are the two factors.

Tier 3 · Hard

1. Answer
- $3(2x-1)=9$ is an equation
- $3(2x-1)\equiv6x-3$ is an identity
- $3(2x-1)<9$ is an inequality
Method: The first statement is true only for a particular value of $x$ , so it is an equation. Expanding the second gives $6x-3$ for every $x$ , so it is an identity. The final statement compares two quantities using $<$ , so it is an inequality.

A4 · 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

Tier 1 · Easy

1. Answer
- $5a+8b$
Method: Collect the like $a$ -terms and the like $b$ -terms separately: $(7a-2a)+(3b+5b)=5a+8b$ .

Tier 2 · Standard

1. Answer
- $(3x+2)(2x-1)$
Method: The product of the leading and constant coefficients is $6\times(-2)=-12$ . Split the middle term using $4x-3x$ : $6x^2+4x-3x-2=2x(3x+2)-(3x+2)=(3x+2)(2x-1)$ .

Tier 3 · Hard

1. Answer
- $\frac{x-3}{x-2}$
- $x\ne-3$ and $x\ne2$
Method: Factorise both parts: $x^2-9=(x-3)(x+3)$ and $x^2+x-6=(x+3)(x-2)$ . Cancelling the common factor gives $(x-3)/(x-2)$ . The original denominator is zero at $x=-3$ or $x=2$ , so both values remain excluded.

A5 · Understand and use standard mathematical formulae; rearrange formulae to change the subject

Tier 1 · Easy

1. Answer
- $w=\frac{A}{l}$
Method: Divide both sides by $l$ to isolate $w$ : $A/l=w$ , so $w=A/l$ .

Tier 2 · Standard

1. Answer
- (a) $h=\dfrac{2A}{a+b}$
- (b) $h=5\,\text{cm}$
Method: (a) Multiply both sides by $2$ : $2A=(a+b)h$ . Divide by $(a+b)$ : $h=\dfrac{2A}{a+b}$ . (b) Substituting, $h=\dfrac{2\times45}{7+11}=\dfrac{90}{18}=5\,\text{cm}$ .

Tier 3 · Hard

1. Answer
- $a=\frac{b(P+1)}{P-1}$
Method: Multiply by $a-b$ : $P(a-b)=a+b$ . Expanding gives $Pa-Pb=a+b$ . Collect the $a$ -terms: $Pa-a=Pb+b$ , so $a(P-1)=b(P+1)$ . Dividing by $P-1$ gives $a=b(P+1)/(P-1)$ .