Substituting and expanding · Expressions, formulae & change of subject

Algebra is arithmetic with some numbers written as letters. Every rule you know for numbers still holds. This lesson covers the three things you do with expressions all the time: substituting, collecting like terms and expanding brackets.

Substituting: put the numbers in

To find the value of an expression, replace each letter with its number, in brackets, then work it out in the usual order: brackets, powers, multiply and divide, then add and subtract.

For $a = 4$ and $b = -6$ :

\begin{aligned} (a - b)^2 &= (4 - (-6))^2 \\ &= 10^2 = 100 \end{aligned}

Given a ratio instead of the numbers

Sometimes you’re told only $m : n = 3 : 2$ (or $\frac mn = \frac32$ ), and asked for the value of an expression. You don’t know $m$ and $n$ , and you don’t need to. Write both in terms of one letter: $m = 3k$ and $n = 2k$ . When every term has the same power of the letters, $k$ cancels:

One block sizem : n = 3 : 2 means m = 3k and n = 2k

\frac{m^2 + n^2}{m^2 - n^2} = \frac{9k^2 + 4k^2}{9k^2 - 4k^2} = \frac{13k^2}{5k^2} = \frac{13}{5}

If two ratios share a letter, such as $\frac pq$ and $\frac qr$ , write all three letters in terms of the shared one.

More: substituting numbers and ratios

Like terms

Like terms have exactly the same letters to the same powers: $3x$ and $-5x$ are like terms; $3x$ and $3x^2$ are not; $2ab$ and $5ba$ are (the order of multiplying doesn’t matter). You can add or take away like terms, and only like terms:

4x + 3y - x + 2y = 3x + 5y

Writing an expression from words

Word problems often ask you to write the expression yourself. Name each quantity with its letter, then say the sentence in symbols one piece at a time. A worker paid $r$ naira an hour for 40 hours, and then $1\frac12$ times that rate for $h$ more hours, earns

40r + 1\tfrac12 rh = r\left(40 + \tfrac32 h\right) \text{ naira}

Test your expression with small numbers you can check in your head: with $r = 2$ and $h = 2$ , that’s $80 + 6 = 86$ naira.

More: writing expressions from words

Expanding brackets

A number or letter in front of a bracket multiplies everything inside it: $3(2x - 5) = 6x - 15$ , and $-2(x - 4) = -2x + 8$ (a negative times a negative is positive).

Two brackets multiplied together need every term in the first to multiply every term in the second. The grid keeps track of all four products:

Expanding bracketsStep through the grid

(x + 3)(x + 2)

×	x	2
x	?	?
3	?	?

first bracket's number = 3second bracket's number = 2

Every term in the first bracket multiplies every term in the second: four multiplications. Watch the signs: the sign in front of a number belongs to it.

Four productsEvery term of one bracket times every term of the other

Function notation

A function is a rule, written like $f(x) = x^2 + 1$ . The letter in the bracket is the input: $f(3)$ means put 3 in place of $x$ everywhere in the rule. The input can be an expression too: $f(x - 2)$ means put $(x - 2)$ , in brackets, in place of every $x$ .

Whatever is in the bracketreplaces x everywhere in the rule

Sometimes the question runs the other way. If $f(x + 3) = x^2 - 4x$ and you want $f(5)$ , choose $x$ so that the bracket is 5: $x + 3 = 5$ gives $x = 2$ , then $f(5) = 2^2 - 4(2) = -4$ .

The inverse function $f^{-1}$ undoes $f$ . Do the opposite steps in the reverse order, or write $y = f(x)$ and make $x$ the subject: $y = 2x - 5$ gives $x = \frac{y + 5}{2}$ , so $f^{-1}(x) = \frac{x + 5}{2}$ .

The inverse machineUndo each step of f, last one first

More: function notation and inverses

A past question, step by step

Worked example · WAEC 2024

WAEC 2024 · Paper 1 · Q5

Simplify: $(2p - q)^2 - (p + q)^2$ .

Expand each square
Multiply out each bracket squared, keeping both middle terms.
$\begin{aligned} (2p - q)^2 &= 4p^2 - 4pq + q^2 \\ (p + q)^2 &= p^2 + 2pq + q^2 \end{aligned}$
Think first. What do you get for $(2p - q)^2$ and for $(p + q)^2$ ?
Take away the whole of the second
Put the second expansion in a bracket so the minus sign reaches every term:
$\begin{aligned} &4p^2 - 4pq + q^2 - (p^2 + 2pq + q^2) \\ &= 4p^2 - 4pq + q^2 - p^2 - 2pq - q^2 \end{aligned}$
Think first. The minus sign in front applies to every term of $(p + q)^2$ . What are the signs after you take it away?
Collect like terms
$= 3p^2 - 6pq$
Match the options
The options are factorised, so take out the common factor $3p$ : $3p^2 - 6pq = 3p(p - 2q)$ . The answer is A.

Your turn

WAEC 2022 · Paper 1 · Q10

Expressions, formulae & change of subject

If $a = 3$ and $b = -7$ , find the value of $\dfrac{5b + (a + b)^2}{(a - b)^2}$ .

Worked solution (try it first)

Put in $a = 3$ and $b = -7$ .
The top is $5(-7) + (3 - 7)^2 = -35 + 16$ , which is $-19$ .
The bottom is $(3 - (-7))^2 = 10^2 = 100$ .
So the value is $\dfrac{-19}{100} = -0.19$ , option C.

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