A1, A2, A3 - Algebraic notation, vocabulary and substitution

Algebra lets you describe a calculation before you know its inputs. Learn to read its shorthand, distinguish expressions from mathematical statements, and substitute numbers into expressions and formulae, including scientific formulae.

Expressions and mathematical statements

A ticket costs £3, with a £2 booking fee for the whole order: how can you describe the cost of any number of tickets in one line?

For four tickets the cost is $3\times4+2=14$ pounds. For $n$ tickets, $3n+2$ describes the same calculation without choosing $n$ . A letter might represent an unknown number, a changing quantity or a fixed parameter; the situation tells you which.

You already know

Multiplication happens before addition unless brackets change the order. In $3\times4+2$ , multiply first; in $3(4+2)$ , add inside the brackets first.

The parts of an expression have different jobs. Learning their names makes instructions such as “collect the terms” much easier to follow.

Expression, term and factor

An expression represents a value without asserting an equality or inequality. Its terms are the pieces added or subtracted, including their signs. A factor is a quantity being multiplied.

In $3x-2y+5$ , the terms are $3x$ , $-2y$ and $5$ . The quantities $3$ and $x$ are factors of $3x$ ; they are not two separate terms.

A coefficient is the numerical multiplier of a letter term. The coefficient of $-2y$ is $-2$ . In $x$ it is $1$ , and in $-x$ it is $-1$ . A constant term, such as $5$ , has no letter attached.

Mathematical statements

An equation states that two expressions are equal. An identity states an equality that holds for every permissible value of its letters. A formula is a rule connecting quantities. An inequality compares quantities using $<$ , $>$ , $\le$ or $\ge$ .

These descriptions can overlap: a formula is written as an equation, but its job is to describe a relationship. Compare the following examples.

Form	What it says
$3n+2$	An expression for the cost in pounds
$3n+2=14$	An equation true when $n=4$
$C=3n+2$	A formula connecting cost $C$ and ticket number $n$
$3(n+2)\equiv3n+6$	An identity, true for every real $n$
$3n+2\le14$	An inequality allowing costs up to and including £14

The identity symbol $\equiv$ emphasises “equal for all permissible inputs”. An identity may also use $=$ ; its meaning depends on the claim, not just the symbol. By contrast, $3n+2=14$ is false at $n=2$ . The symbol $\ne$ means “not equal to”. A strict inequality such as $n>4$ excludes $4$ , whereas $n\ge4$ includes it.

Key idea

Terms are added; factors are multiplied. An expression names a value, while a statement makes a claim about values.

Keep the sign with its term in the next check.

Reading compact algebraic notation

Algebra omits some multiplication signs, but it keeps every operation.

Write $ab$ for $a\times b$ , $3y$ for $3\times y=y+y+y$ , and $4(x+1)$ for four copies of the entire bracket. Do not write $3\times4$ as $34$ : two adjacent digits form a different number.

Power and fractional coefficient

A positive integer power records repeated multiplication: $a^2=a\times a$ , $a^3=a\times a\times a$ , and $a^2b=a\times a\times b$ . A fractional coefficient, such as $\frac34$ in $\frac34x$ , multiplies the letter by that fraction.

In $a^2b$ , only $a$ is squared. In $(ab)^2$ , the whole product is squared, giving $a^2b^2$ . Repeated addition is different: $a+a=2a$ , while $a\times a=a^2$ .

Division is usually a fraction: $\frac ab=a\div b$ , with $b\ne0$ . The bar groups the entire numerator and denominator. Thus $\frac{x+3}{2}$ halves the total, while $x+\frac32$ halves only $3$ . Exact coefficients matter: $0.75x=\frac34x$ , but $0.33x\ne\frac13x$ in general.

Worked example 1: recording a repeated product

Question. Multiply $p$ by itself, multiply by $q$ , then divide the result by five. Write one compact expression.

Plan. Record the operations, then replace the repeated factor by a power.

\begin{aligned} \text{expression}&=\frac{p\times p\times q}{5} &&\text{divide the whole product}\\ &=\frac{p^2q}{5} &&\text{two factors of }p \end{aligned}

Why this step? The fraction bar keeps every numerator factor inside the division.

Answer. $\frac{p^2q}{5}$ , or $\frac15p^2q$ .

Check. With $p=2$ , $q=5$ , both the instruction and expression give $2\times2\times5\div5=4$ .

Now change one operation: add the first two copies of $p$ , rather than multiplying them. This changes a power into a grouped sum.

Worked example 2: grouping before multiplication

Question. Add $p$ to itself, multiply the total by $q$ , then divide by five. Write one compact expression.

Plan. Put the addition in brackets before applying the later operations.

\begin{aligned} \text{expression}&=\frac{(p+p)q}{5} &&\text{group the total}\\ &=\frac{2pq}{5} &&\text{two copies of }p \end{aligned}

Why this step? Adding two copies makes $2p$ , not $p^2$ .

Answer. $\frac{2pq}{5}$ .

Check. With $p=3$ , $q=5$ , the total is $(3+3)\times5\div5=6$ ; the expression also gives $6$ .

The same grouping explains “add $p$ and $q$ , then multiply by five”: write $5(p+q)$ . Writing $5p+q$ would multiply only $p$ .

Watch out

Wrong: $a+a=a^2$ . Addition counts copies, so $a+a=2a$ . Multiplication counts factors, so $a\times a=a^2$ .

Change only the final divisor from Worked example 2 in your turn.

Substitution and the order of operations

Substitution replaces each letter with its given value while keeping the expression's operations.

Write the numerical calculation before evaluating it. Negative inputs need brackets: $(-3)^2=9$ , while $-3^2=-(3^2)=-9$ . Substituting $-3$ into $x^2$ means squaring the whole input.

Substitution

To substitute a value is to replace every occurrence of its letter by that value. The order of operations remains unchanged.

Powers come before multiplication, and a fraction bar groups its numerator and denominator. These two habits prevent many sign errors.

Worked example 3: substituting a negative input

Question. Find $2x^2-3y$ when $x=-3$ and $y=2$ .

Plan. Substitute with brackets, square, then multiply and subtract.

\begin{aligned} 2x^2-3y&=2(-3)^2-3(2) &&\text{replace both letters}\\ &=2(9)-6 &&\text{square before multiplying}\\ &=12 &&\text{subtract} \end{aligned}

Why this step? The square belongs to $x$ , so it includes the negative sign in the input.

Answer. $12$ .

Check. The squared term contributes $18$ , then $6$ is removed: $18-6=12$ .

Change only the sign of $y$ . The square stays the same; the subtraction now removes a negative value.

Worked example 4: subtracting a negative term

Question. Find $2x^2-3y$ when $x=-3$ and $y=-2$ .

Plan. Substitute both signed values and evaluate the two products separately.

\begin{aligned} 2x^2-3y&=2(-3)^2-3(-2) &&\text{include both signs}\\ &=18-(-6) &&\text{evaluate the products}\\ &=24 &&\text{subtracting a negative adds} \end{aligned}

Why this step? $3y$ is $-6$ , so the expression asks for $18-(-6)$ .

Answer. $24$ .

Check. Changing $y$ from $2$ to $-2$ increases $-3y$ by $12$ ; the answer rises from $12$ to $24$ .

For a fraction, work out the complete top and bottom separately. With $a=5$ , $b=-2$ ,

\frac{a-2b}{a+b}=\frac{5-2(-2)}{5+(-2)}=\frac93=3.

If $a=2$ , $b=-2$ , the denominator is zero and the expression is undefined. Fractional inputs can stay exact too: for $t=\frac12$ , $6t^2=6\times\frac14=\frac32$ .

Watch out

Wrong substitution into $2x^2$ : $2\times-3^2$ . Write $2(-3)^2$ , then evaluate the power first. Squaring $2(-3)$ instead would also change the expression.

Change only the value of $x$ from Worked example 4.

Using mathematical and scientific formulae

A formula needs compatible units as well as correct arithmetic.

For a triangle, $A=\frac12bh$ , where $h$ is perpendicular to the chosen base $b$ . With $b=8\,\mathrm{cm}$ , $h=5\,\mathrm{cm}$ , its area is $20\,\mathrm{cm}^2$ . Two lengths multiply to give square units; a sloping side cannot replace the perpendicular height.

Area of a triangle

A=\frac12bh

Here $b$ and perpendicular height $h$ use the same length unit.

Not on the formula sheet: learn it.

Scientific formulae work in the same way. The velocity rule below is provided in the question when needed; it is not on the formula sheet. If acceleration is constant, the supplied rule $v=u+at$ connects initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ . For $u=12\,\mathrm{m/s}$ , $a=-2\,\mathrm{m/s}^2$ , $t=3\,\mathrm{s}$ , substitution gives $v=12+(-2)(3)=6\,\mathrm{m/s}$ .

Kinetic energy

E=\frac12mv^2

In this version, mass $m$ is in kilograms, speed $v$ is in metres per second and energy $E$ is in joules.

Provided in the question when needed; it is not on the formula sheet.

The speed is squared before multiplying by half the mass. Here is the calculation with units already matching the formula.

Worked example 5: using a scientific formula

Question. Use $E=\frac12mv^2$ to find the kinetic energy of a $2\,\mathrm{kg}$ object moving at $3\,\mathrm{m/s}$ .

Plan. Substitute the mass and speed, then evaluate the square first.

\begin{aligned} E&=\frac12(2)(3^2) &&\text{substitute in the given units}\\ &=1\times9 &&\text{square the speed}\\ &=9\,\mathrm{J} &&\text{multiply} \end{aligned}

Why this step? Only $v$ is squared; the formula does not square the mass or the half.

Answer. $9\,\mathrm{J}$ .

Check. Half of $2$ is $1$ , so the numerical energy must equal $3^2=9$ .

Change only the way the same mass is given: grams now need converting to kilograms.

Worked example 6: matching the units first

Question. Use $E=\frac12mv^2$ , with mass in kilograms and speed in metres per second, to find the kinetic energy of a $2000\,\mathrm{g}$ object moving at $3\,\mathrm{m/s}$ .

Plan. Convert the mass, then use the same order of operations.

\begin{aligned} m&=2000\div1000=2\,\mathrm{kg} &&\text{convert grams to kilograms}\\ E&=\frac12(2)(3^2) &&\text{substitute after converting}\\ &=9\,\mathrm{J} &&\text{square and multiply} \end{aligned}

Why this step? Using $2000$ as the kilogram input would make the answer a thousand times too large.

Answer. $9\,\mathrm{J}$ .

Check. This is the same object and speed as Worked example 5, so changing the written unit must leave its energy unchanged.

There is a striking consequence of the square: doubling speed makes kinetic energy four times as large at fixed mass, since $(2v)^2=4v^2$ . Formulae describe patterns as well as individual answers.

Watch out

Wrong: using $500$ as the mass input when the formula requires kilograms. Convert $500\,\mathrm{g}=0.5\,\mathrm{kg}$ first. Also square only the speed: $\frac12mv^2$ is different from $(\frac12mv)^2$ .

Change the mass value in Worked example 6; the formula and speed stay the same.

Quick Recap

Expressions name values; equations, identities and inequalities make claims. Formulae connect quantities.
Terms are added or subtracted, factors are multiplied, and coefficients are numerical multipliers.
Brackets and fraction bars preserve grouping; powers record repeated multiplication.
Substitute every value with its sign, follow the order of operations and match a formula's units.

The ticket question has one compact answer: $C=3n+2$ pounds. Four tickets cost $3(4)+2=14$ pounds.

Last Accessed

Papers 1, 2 and 3

A1, A2, A3 - Algebraic notation, vocabulary and substitution

Expressions and mathematical statements

Reading compact algebraic notation

Substitution and the order of operations

Using mathematical and scientific formulae

Quick Recap

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