# Forming Expressions, Equations, and Formulae

GCSELevel 1-3Level 4-5AQAEdexcelOCRWJEC

## Forming Expressions, Equations, and Formulae

The words expression, equation and formula are sometimes wrongly used interchangeably. They do in fact have distinct definitions.

An expression is a mathematical phrase, one that does not contain an ‘$=$‘ sign. It may contain numbers or variables. For example $6x+3$

An equation may equate an expression to something. It may however only contain one unknown and as such will only be true for certain values of the variable. For example $6x+3=15$ is an equation, one that is only true when $x=2$

A formula gives the relationship between variables. Usually so that one variable can be calculated given the others. For example $A = \pi r^2$ is the area of a circle.

An identity is a relationship between two expressions. Those expressions may contain variables, for example
$\text{tan}(x)=\dfrac{\text{sin}(x)}{\text{cos}(x)}$

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## Expressions

An expression is a statement made in the language of mathematics. We ascribe meaning to the symbols we use so they may be used instead of words, and so an expression must also work as a sentence.

$4x^2-32$

means “four times ‘x’ squared minus thirty-two”. If an expression doesn’t make sense as a sentence, then it probably doesn’t make sense as an expression.

Here is some nonsense

$!x^( \times +$

This in not an expression, because it doesn’t make any sense. As a sentence this would be “factorial times ‘x’ to the power of open bracket, times plus”. It does not follow the syntax of maths.

We may form expressions from the processes we know a variable goes through, for example, if we know that the variable $x$ is added to $16$, and then all of that is multiplied by $2.5$, we can form an expression for this, which is

$2.5(x+16)$

The order of operations is important, remember BIDMAS, the reason we put $x+16$ in the brackets is because we want that to be done first.

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## Equations

An equation is when two equal expressions are joined by an equals sign. They can have at most one unknown value. This means that an equation may be solvable. For example

$14x +15 = 71$

may be solved, my manipulating the equation:

$14x = 56$
$x=4$

The equation has been solved and the unknown value, $x$ has been found.

You may have to formulate equations, it is very similar to formulating expressions. If we say that $6$ plus twelve lots of $x$ is equal to $66$, then we can formulate the equations

$6 +12x = 66$

Once again, take care of the order of operations.

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Level 1-3GCSEAQAEdexcelOCRWJEC

## Formulae

A formula gives the relationship between variables. A simple example might be the circumference of a circle

$C=\pi \times d$

Where $C$ is the circumference and $d$ is the diameter of the circle.

A formula is distinct from an equation, as an equation can have at most one variable. So instead of showing a relation between variables, an equation can be used to calculate the value of one variable. Whereas a formula will need the other variables for the unknown variable to be found.

Pythagoras’ theorem is the formula $a^2+b^2=c^2$. If the length of the base and the length of the height of a right angled triangle is known, then they can be substituted in to $a$ and $b$ to find the unknown variable, the length of the hypotenuse, $c$.

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## Forming Formulae

Formulae are created to help with making the same calculations with a change in parameters

It is expected of us to be able to create formula given some information. Firstly we must find out what the subject of the formula is, and then what that value is dependent on.

For example, if a car rental company charges $£50$ per day to rent a car, plus an additional $10\text{p}$ a mile, then we can create a formula for the cost of the car hire.

The variables the cost depends on are the number of days the car is hired (was shall call this $t$ for time), and the number of miles the car travels (we shall call this $d$ for distance). We shall then call the cost of the car hire $C$,

$C=50t+0.1d$

We multiply the number of days by $50$ and the distance in miles by $0.1$ as $10\text{p}$ is $£0.10$, this gives us $C$ in pounds.

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## Example 1: Finding Fahrenheit

When the temperature is $0\degree$ in Celsius, then it is $32 \degree$ Fahrenheit.

Then for every $1 \degree \text{C}$ increase in temperature, there is an increase by $1.8 \degree \text{F}$.

Create a formula for the temperature in Fahrenheit, in terms of the temperature in Celsius, then use that to find $100\degree \text{C}$ in Fahrenheit.

[4 marks]

As $0\degree \text{C} = 32\degree \text{F}$ we know that

$\text{F}= k \text{C} + 32$

and, as for every increase of $1\degree \text{C}$, the increase is $1.8\degree \text{F}$, so $k=1.8$ which gives us

$\text{F}=1.8\text{C} + 32$

where $\text{C}$ is the temperature in Celsius and $\text{F}$ is the temperature in Fahrenheit.

So when $\text{C}= 100\degree$

$\text{F} = 1.8 \times 100 + 32 = 212 \degree$

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## Example 2: Finding Further Formulae

The cost of a house started at $£250000$ and every year it’s value increases by $£7000$, create a formula for the cost of the house, then work out how much the house will be worth after $10$ years.

[4 marks]

Let the cost of the house $=C$, then let number of years $=t$, this is the factor that $C$ is dependent on. Then the formula for $C$ is

$C= 250000+7000t$

So when $t=10$

$C=250000+7000 \times 10 = 320000$

The price of the house after $10$ years is $£320,000$

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## Forming Expressions, Equations, and Formulae Example Questions

Question 1: Waiters at a restaurant are paid $£9.30$ an hour, they are also given tips by the customers.

a) Create a formula to calculate how much the waiters are earning.

[2 marks]

b) One waitress at the restaurant works an $8$ hour shift one night and is given $£13.50$ in tips. How much did she earn that night.

[1 mark]

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a) Let the total pay $=P$, the amount of tips received $=t$ and the hours worked $=h$,

$P=9.3h+t$

b) Where $h=8$ and $t=13.5$ we can substitute this into the formula

$P=9.3 \times 8 +13.5$

$P = £87.90$

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Question 2: A bike hire shop charges a flat rate of $£12$ to hire a bike and then $£8$ per day it is hired for.

a) Where $n$ is the number of days a bike is hired for, create a formula to work out the cost $£C$ of hiring the bike.

[2 marks]

b) From a rival bike shop the cost, $£C$ to hire a bike for $n$ days is

$C=20+6n$

How many days would you have to rent a bike for the cost to be the same at each shop.

[2 marks]

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a) The rate starts at $12$ and then for each day the bike is hired, $£8$ is added to the cost. This gives us an formula of

$C=12 + 8n$

Where $£C$ is the total cost of the rental and $n$ is the number of days rented for.

b) For shop 1:

$C=12+8n$

For shop 2:

$C=20 +6n$

So they are equal when

$12 +8n = 20 + 6n$
$12+ 2n = 20$
$2n = 8$
$n = 4$

The cost to hire a bike is the same from each shop is you hire for $4$ days.

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Question 3: A salesman is given a base pay and commission on how many products he sells. He is offered two contracts.

Contract $1$ has base pay of $£25,000$ per year and $£50$ commission per product sold.

Contract $2$ has base pay of $£30,000$ per year and a commission of $£25$ per product sold.

He believes he will sell $150$ products a year. Based on this, which contract will make him the most money?

[4 marks]

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For contract $1$, the total pay, $£P$ will be

$P = 25,000 +50n$

Where $n$ is the number of products sold, if $n=150$ then

$P = 25,000+ 50\times 150$
$P = 32,500$

So he is expected to earn $£32,500$ in a year with contract $1$

With contract $2$ the base pay is $£30,000$ and the commission is $£25$ per product sold. This gives

$P=30,000+ 25 \times n$

So if $n =150$ the expected earnings are

$P=30,000 + 25 \times 150$
$P=33750$

His expected earnings for contract $2$ are $£33750$, and so contract $2$ will make him more money.

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