Substitution

Substituting positive and negative numbers into formulas, formulas from real life, and working backwards to find a letter.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 113 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What Substitution Means

A formula is a rule written with letters. Substitute means swap each letter for its number, then work it out.

Rule

A number next to a letter means multiply: $5 p$ is $5 \times p$ .

Example

$T = 5 p - 2 d$ with $p = 7$ and $d = 4$ : $T = 5 \times 7 - 2 \times 4 = 35 - 8 = 27$ .

Key idea

Always write the line with the numbers in it. It shows your method, even if you slip later.

Substitute, Then Follow the Order

After you swap in the numbers, work it out in this order: brackets first, then powers, then multiply and divide, then add and subtract.

Worked example

1
$P = 3 a^{2} + 2 b$ — The formula. We are told a = 4 and b = 5.
2
$P = 3 \times 4^{2} + 2 \times 5$ — Swap each letter for its number. Write this line: it earns a mark for working.
3
$P = 3 \times 16 + 2 \times 5$ — Powers first, so square only the 4: $4^{2} = 16$ . The 3 is not squared.
4
$P = 48 + 10$ — Multiply next: $3 \times 16 = 48$ and $2 \times 5 = 10$ .
5
$P = 58$ — Add last: $48 + 10 = 58$ .

Common mistake

Working out $3 \times 4 = 12$ first and then squaring. The power belongs to the 4 only.

Negative Numbers: Use Brackets

Most marks are lost here. When the number is negative, put it in a bracket before you do anything.

Worked example

1
$Q = 2 u^{2} - 5 u$ — The formula. We are told u = −3.
2
$Q = 2 (- 3)^{2} - 5 (- 3)$ — Swap u for (−3). The bracket keeps the minus sign with the 3.
3
$Q = 2 \times 9 - 5 \times (- 3)$ — Powers first: $(- 3)^{2} = (- 3) \times (- 3) = 9$ . Two negatives multiply to a positive.
4
$Q = 18 - 5 \times (- 3)$ — Multiply: $2 \times 9 = 18$ .
5
$Q = 18 + 15$ — Multiply: $- 5 \times (- 3) = + 15$ . Minus times minus is plus.
6
$Q = 33$ — Add: $18 + 15 = 33$ .

Rule

Powers of a negative: an even power is positive, $(- 2)^{2} = 4$ and $(- 2)^{4} = 16$ . An odd power is negative, $(- 2)^{3} = - 8$ .

Common mistake

Writing $- 3^{2} = - 9$ . The bracket squares the whole −3: $(- 3)^{2} = 9$ .

Brackets, Fractions and Roots

A fraction line, a bracket and a square root sign each group numbers together. Work out everything inside them first.

Worked example

1
$M = \frac{5 a - 3}{b ^{2}}$ — The formula, with a = 3 and b = −2. The fraction line works like a bracket round the top and round the bottom.
2
$M = \frac{5 \times 3 - 3}{( - 2 ) ^{2}}$ — Swap in the numbers. The −2 goes in a bracket.
3
$M = \frac{15 - 3}{( - 2 ) ^{2}}$ — Top first. Multiply: $5 \times 3 = 15$ .
4
$M = \frac{12}{( - 2 ) ^{2}}$ — Finish the top: $15 - 3 = 12$ .
5
$M = \frac{12}{4}$ — Now the bottom: $(- 2)^{2} = 4$ . Squaring a negative gives a positive.
6
$M = 3$ — Divide last: $12 \div 4 = 3$ . Top and bottom had to be finished before this.

A square root covers everything under its sign. Finish the inside before you take the root.

Worked example

1
$d = h^{2} + 2 k$ — The formula, with h = 4 and k = 10.
2
$d = 4^{2} + 2 \times 10$ — Swap in the numbers.
3
$d = 16 + 20$ — Under the root, powers and multiplying first: $4^{2} = 16$ and $2 \times 10 = 20$ .
4
$d = 36$ — Add under the root: $16 + 20 = 36$ .
5
$d = 6$ — Take the square root last, because $6 \times 6 = 36$ .

Common mistake

In $s = \frac{1}{2} a t^{2}$ only t is squared. With a = 6 and t = 4: $\frac{1}{2} \times 6 \times 16 = 48$ . Squaring $\frac{1}{2} \times 6 \times 4 = 12$ gives 144, which is wrong.

Formulas from Real Life

Many questions give a rule in words: a cost, a cooking time, a journey. Write the rule, then substitute.

A van costs $40 to hire, plus $25 for each day. Find the cost for 6 days.

Worked example

1
$C = 40 + 25 d$ — Write the rule as a formula. C is the cost in dollars and d is the number of days.
2
$C = 40 + 25 \times 6$ — d is 6, because the van is hired for 6 days.
3
$C = 40 + 150$ — Multiply before you add: $25 \times 6 = 150$ .
4
$C = 190$ — Add: $40 + 150 = 190$ . The cost is $190.

Key idea

Finish with the unit the question uses: dollars, minutes, kilograms.

Working Backwards

Some questions give the answer and ask for a letter. Substitute everything you know, then solve the equation.

Worked example

1
$C = 40 + 25 d$ — The same formula. This time the cost is given, C = 265, and we want d.
2
$265 = 40 + 25 d$ — Put 265 in for C. Now d is the only letter left.
3
$225 = 25 d$ — Take 40 from both sides: $265 - 40 = 225$ . This leaves the d term alone.
4
$d = 9$ — Divide both sides by 25: $225 \div 25 = 9$ . The van was hired for 9 days.

Common mistake

Putting 265 in for d. Check which letter the number belongs to before you swap it in.

Exams often give a second letter too, and it is often negative. Put in both numbers before you solve.

Worked example

1
$R = 3 m - 2 k$ — The formula. We know R = 23 and k = −4, and we want m.
2
$23 = 3 m - 2 \times (- 4)$ — Put in both known numbers. The −4 goes in a bracket.
3
$23 = 3 m + 8$ — $- 2 \times (- 4) = + 8$ . Minus times minus is plus.
4
$15 = 3 m$ — Take 8 from both sides: $23 - 8 = 15$ .
5
$m = 5$ — Divide both sides by 3. Check: $3 \times 5 - 2 \times (- 4) = 15 + 8 = 23$ .

Solving for a Square (Extended/Higher)

Sometimes the letter you want is squared. Get the square on its own, then take the square root.

Worked example

1
$E = \frac{1}{2} m v^{2}$ — The energy of a moving object. We know E = 100 and m = 8, and we want the speed v.
2
$100 = \frac{1}{2} \times 8 \times v^{2}$ — Put in both known numbers.
3
$100 = 4 v^{2}$ — Multiply the numbers: $\frac{1}{2} \times 8 = 4$ .
4
$25 = v^{2}$ — Divide both sides by 4 so that v² is on its own: $100 \div 4 = 25$ .
5
$v = \pm 5$ — Square root both sides. Both $5^{2}$ and $(- 5)^{2}$ make 25, so v is 5 or −5.
6
$v = 5$ — Keep the positive value: a speed cannot be negative. Check: $\frac{1}{2} \times 8 \times 25 = 100$ .

Common mistake

Taking the square root too early: $100 = 10$ is wrong here. Get $v^{2}$ on its own first.

Which Method? Read the Question

Look at what you are given and what you are asked for.

You see	You do
Positive numbers	Swap them in, then powers, multiply, add
A negative number	Put it in a bracket first
A bracket, fraction line or root	Work out inside it first
A rule in words	Write the formula, then substitute
The answer is given	Substitute, then solve for the letter
The letter you want is squared	Get it alone, square root, keep the positive

Key idea

These are usually 2 marks. One mark is for the line with the numbers swapped in.

Practise substitution now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on substitution

Each with a full worked solution.

All 113 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.