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In order to access this I need to be confident with:

Powers and roots

BIDMAS

Negative numbers

Decimals

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GCSE Maths Algebra Algebraic Expressions Substitution

Substitution

Here is everything you need to know about substitution for GCSE maths (Edexcel, AQA and OCR). You’ll learn how to substitute positive numbers, negative numbers and decimals into various algebraic expressions.

Look out for the substitution worksheets and exam questions at the end.

What is substitution?

Substitution means replacing the variables (letters) in an algebraic expression with their numerical values. We can then work out the total value of the expression.

We can substitute values into formulae to help us work out many different things. Examples range from the formula for the area of a triangle:

\[A=\frac{1}{2} b h\]

to the formula for calculating BMI:

\[B M I=\frac{w}{h^{2}}\]

What is substitution?

This lesson is part of our series on algebraic expressions. You may find it helpful to start with the main algebraic expressions lesson for a summary of what to expect and then also work through the following:

Substitution worksheets

Get your free substitution worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Substitution worksheets

Get your free substitution worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

How to substitute a value into an algebraic expression

In order to substitute into an algebraic expression:

Rewrite the expression substituting each letter with its given numerical value.
Calculate the total value of the expression. Remember that you must apply BIDMAS.

Explain how to substitute a value into an algebraic expression in 2 steps

Substitution examples

Example 1: basic expression, one variable

Find the value of 3b + 4 when b = 10.

Here b = 10 so we substitute the b in 3b + 4 for 10.

\[3\times 10 + 4\]

Remember that 3b means 3 ✕ b.

2We can now work out the value of the expression:

\[3\times 10 + 4 = 34\]

Example 2: basic expression, one variable

Find the value of

\[20-\frac{m}{5}\]

when m = 35.

Here m = 35 so we substitute the m in the expression for 35.

\[20-\frac{m}{5}\]

\[\begin{aligned} 20-\frac{35}{5} \end{aligned}\]

Remember that

\[\frac{35}{5}\]

means 35 ÷ 5.

We can now work out the value of the expression:

\[\begin{aligned} 20-\frac{35}{5} &=20-7 \\ &=13 \end{aligned}\]

Example 3: basic expression, two variables

Find the value of 3x − 5y when x = 5 and y = 4.

$x$ is $5$ and $y$ is $4$ so

\[3\times 5 – 5\times 4\]

Work out the value of the expression:

\[\begin{aligned} 3 \times 5-5 \times 4 &=15-20 \\ &=-5 \end{aligned}\]

Notice that we have applied BIDMAS – multiplying is done before subtracting.

Example 4: brackets

Find the value of 2p(q − r) when p = 5, q = 15 and r = 6.

Substituting $p$ for $5$ , $q$ for $15$ and $r$ for $6$ .

\[2\times 5\left ( 15-6 \right )\]

Work out the value of the expression:

\[\begin{aligned} 2 \times 5 \times(15-6) &=2 \times 5 \times 9 \\ &=90 \end{aligned}\]

Example 5: powers

Find the value of

\[s^{2}(20-\sqrt{t})\]

when s = 2 and t = 9.

Substituting $s$ for $2$ and $t$ for $9$ :

\[2^{2}(20-\sqrt{9})\]

Work out the value of the expression:

\[\begin{aligned} 2^{2} \times(20-\sqrt{9}) &=2^{2} \times(20-3) \\ &=2^{2} \times 17 \\ &=4 \times 17 \\ &=68 \end{aligned}\]

Substituting decimals and negative numbers

We substitute decimals in the same way as any other number.

When substituting negative numbers we need to be particularly careful and remember the rules for operations with negatives.

In particular, remember that:

+ + = +
+ − = −

− − = +
− + = −

So 7 + − 3 = 4 and 6 − − 5 = 11.

These rules also apply for multiplying and dividing negative numbers.

Another important point to remember is that squared means multiplied by itself.

\[\begin{aligned} (-5)^{2} &=-5 \times-5 \\ &=25 \end{aligned}\]

Notice that when you square a negative number you get a positive answer.

Example 6: decimals

Find the value of 4g(10 − 2h) when g = 2.5 and h = 1.8.

Substituting $g$ for $2.5$ and $h$ for $1.8$ :

\[4\times 2.5(10-2\times 1.8)\]

Work out the value of the expression:

\[\begin{aligned} 4 \times 2.5 \times(10-2 \times 1.8) &=4 \times 2.5 \times(10-3.6) \\ &=4 \times 2.5 \times 6.4 \\ &=64 \end{aligned}\]

Example 7: negative numbers

Find the value of 5z² + 7y when z = − 10 and y = − 3.

Substituting $z$ for $- 10$ and $y$ for $- 3$ :

\[5\times (-10)^{2} +7\times -3\]

We have put the − 10 in brackets here as we need to make sure that we are squaring − 10 and not subtracting 10 squared.

Work out the value of the expression:

\[\begin{aligned} 5 \times(-10)^{2}+7 \times-3 &=5 \times 100+7 \times-3 \\ &=500+-21 \\ &=479 \end{aligned}\]

Example 8: negative numbers

Find the value of

\[\frac{4 p\left(3 q^{2}-70\right)}{6+p}\]

when p = − 4 and q = − 5.

Substituting $p$ for $- 4$ and $q$ for $- 5$ :

\[\frac{4 \times-4\left(3 \times(-5)^{2}-70\right)}{6+-4}\]

Work out the value of the expression:

\[\begin{aligned} \frac{4 \times-4\left(3 \times(-5)^{2}-70\right)}{6{+-4}} &=\frac{4 \times-4(3 \times 25-70)}{6+-4} \\ &=\frac{4 \times-4 \times 5}{6+-4} \\ &=\frac{-80}{2} \\ &=-40 \end{aligned}\]

Using formulae

We substitute into formulae in exactly the same way. We use formulae to work different things out.

Example 9: formulae

Speed is calculated using the formula

\[S=\frac{D}{T}\]

where D is distance and T is time.

Find the speed at which a car travelled if it took 2 hours to travel a distance of 100 miles.

Here D = 100 and T = 2

Substituting into the formula:

\[S=\frac{100}{2}\]

Notice that we leave the S on the left-hand side.

Work it out:

\[S=50mph\]

Example 10: formulae

The quadratic formula

\[x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\]

\[x=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\]

can be used to solve quadratic equations.

Find the two possible values of x when a = 1, b = − 2 and c = − 15.

Substituting into both formulae:

\[x=\frac{2+\sqrt{(-2)^{2}-4 \times 1 \times-15}}{2 \times 1} \quad \text { or } \quad x=\frac{2-\sqrt{(-2)^{2}-4 \times 1 \times-15}}{2 \times 1} \mid\]

Work them out:

\[\begin{array}{l} x=\frac{2+\sqrt{(-2)^{2}-4 \times 1 \times-15}}{2 \times 1} \\ x=\frac{2+\sqrt{4–60}}{2} \\ x=\frac{2+\sqrt{64}}{2} \\ x=\frac{2+8}{2} \\ x=\frac{10}{2} \\ x=5 \end{array}\]

\[\begin{array}{l} x=\frac{2-\sqrt{(-2)^{2}-4 \times 1 \times-15}}{2 \times 1} \\ x=\frac{2-\sqrt{4–60}}{2} \\ x=\frac{2-\sqrt{64}}{2} \\ x=\frac{2-8}{2} \\ x=\frac{-6}{2} \\ x=-3 \end{array}\]

Common misconceptions

These are some of the common misconceptions around substitution:

The meaning of ab

Remember in algebraic expressions, ab means a multiplied by b, not the number b written after the number a. So if a = 3 and b = 5, ab = 3 ✕ 5, not 35.

Not applying BIDMAS

BIDMAS rules need to be followed here as in normal numerical calculations.

Mistakes with negative numbers

The most common mistake is thinking that a negative number squared gives a negative answer. Remember that squared means multiplied by itself and a negative number multiplied by a negative number gives a positive answer.

Practice substitution questions

65

-3

25

-5

7x-10=7 \times5-10=35-10=25

28

32

13

7

\frac{y}{4}+8=\frac{20}{4}+8=5+8=13

-6.5

56.5

17

5

7p-2q=7(8.5)-2(1.5)=59.5-3=56.5

360

630

60

6

6g(3h-9)=6(10)\times(3(5)-9)=60\times(15-9)=60\times6=360

10

7

4.4

44

\frac{m^{2}+\sqrt{n}}{10}=\frac{8^{2}+\sqrt{36}}{10}=\frac{64+6}{10}=\frac{70}{10}=7

-40

56

24

4

3e^{2}+8=3(-4)^{2}+8=3(16)+8=56

$5$ Newtons

$9$ Newtons

$8$ Newtons

$6$ Newtons

$F=ma\\$
$F=2\times3\\$
$F=6$

$10$ Joules

$40$ Joules

$20$ Joules

$100$ Joules

$KE=\frac{1}{2} m v^{2}\\$
$KE=\frac{1}{2}\times 5\times 4^{2}\\$
$KE=\frac{1}{2}\times5 \times 16\\$
$KE=40$

Substitution GCSE questions

1. Work out the value of $3x + 2y$ when $x = 6$ and $y = 4$

(1 mark)

Show answer

3\times6 + 2 \times4 = 26

(A1)

2. Work out the value of $\frac{4 a-10 b}{5}$ when $a = 5$ and $b = - 2$

(2 marks)

Show answer

$4\times5 = 20$ and $10 \times− 2 = − 20$

(M1)

\frac{20–20}{5}=\frac{40}{5}

= 8

(A1)

3. The cost, $C$ , of hiring a canoe is given by the formula

$C = 20 + 2.5h$

where $h$ is the number of hours the canoe is to be hired for.

Calculate the cost of hiring the canoe for $6$ hours.

(1 mark)

Show answer

C = 20 + 2.5\times 6

(M1)

C = 20 + 15

C = 35

(A1)

Learning checklist

Substitute positive numbers, negative numbers and decimals into basic algebraic expressions

Substitute into expressions involving brackets and powers

Substitute into formulae

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