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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.

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:

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

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:

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

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

DOWNLOAD FREEIn 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.

Find the value of

- Here
b = 10 so we substitute theb in3b + 4 for10 .

\[3\times 10 + 4\]

Remember that

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

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

Find the value of

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

when

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

We can now work out the value of the expression:

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

Find the value of

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.

Find the value of

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:

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

Find the value of

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

when

Substituting s for 2 and t for 9:

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

Work out the value of the expression:

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

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

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.

Find the value of

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:

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

Find the value of ^{2} + 7y

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:

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

Find the value of

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

when

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:

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

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

Speed is calculated using the formula

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

where

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

Work it out:

\[S=50mph\]

The quadratic formula

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

or

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

can be used to solve quadratic equations.

Find the two possible values of

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:

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

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

These are some of the common misconceptions around substitution:

**The meaning of**ab

Remember in algebraic expressions,

**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.

1. Find the value of 7x-10 when x=5

65

-3

25

-5

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

2. Find the value of \frac{y}{4}+8 when y=20

28

32

13

7

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

3. Find the value of 7p-2q when p=8.5 and q=1.5

-6.5

56.5

17

5

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

4. Find the value of 6g(3h-9) when g=10 and h=5

360

630

60

6

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

5. Find the value of \frac{m^{2}+\sqrt{n}}{10} when m=8 and n=36

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

6. Find the value of 3e^2+8 when e= -4

-40

56

24

4

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

7. Force can be calculated using the formula F = ma where m = mass and a = acceleration. Calculate the force on an object with a mass of 2kg and an acceleration of 3m/s^{2} .

5 Newtons

9 Newtons

8 Newtons

6 Newtons

F=ma\\

F=2\times3\\

F=6

8. Kinetic energy can be calculated using the formula KE=\frac{1}{2} m v^{2} , where m = mass and v = velocity. Calculate the kinetic energy of an object with mass 5kg moving at a velocity of 4m/s .

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

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)**

- 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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