## Learning Outcomes

If you work through this section you should be able to:

• Understand the use of equations to represent relationships of unknown values and known values.
• Form equations to represent relationships of unknown values and known values.
• Solve equations by working out the value of an unknown value by its relationship to a known value.
ACCORDION: Converted tabs - this tab opens by default
ACCORDION OPTIONS: enable collapse / start fully collapsed

This section requires some understanding of basic definitions and processes in algebra (see the Introduction to algebra section).

You may also find the Algebraic expressions section useful before tackling equations.

An equation is a statement that shows that one thing is equal to something else. You can form equations (using algebraic expressions) to show the relationship of unknown values to known values. An equation, then, combines algebraic expressions with information about the known values.

It is possible to work out the value of an unknown quantity in relation to a known quantity by solving an equation that reflects their relationship.

Having formed an algebraic expression (or expressions), then, you require a further piece of information that enables you to form an equation. The information may simply be a number.

#### Example 1

A grapefruit costs twice as much as an orange. The price of an orange is y pence.

So the grapefruit costs 2y pence.

A grapefruit and 3 oranges cost 60 pence in total.

Expressed as an equation this is:

2y + 3y = 60

#### Example 2

An apple costs 2 pence less than an orange. I buy 5 apples and 3 oranges for 94 pence. How much does an orange and an apple cost?

Let c = cost of an orange in pence.

Then, c − 2 = cost of an apple.

An expression for the cost of my purchase is, therefore:

5(c − 2) + 3c

As I am told that the cost of my purchase is 94p I can form an equation:

5 (c − 2) + 3c = 94

#### Example 3

A farmer wishes to enclose a rectangular part of his field with fencing, and it is to be 4 times as long as it is wide. How much fencing does he need if the area to be enclosed is to be 100m2?

Starting with the width, let the width be W metres.

So, the length = 4W

Area = length × width = 4W × W

As we are told that the area has to be 100m2, we can form the equation:

4W × W = 100

#### Example 4

A father is 25 years older than his daughter. Today he says he will be twice her age in 13 years. How old is his daughter at present?

Let her age be X years

Father's age = X + 25

In 13 years her age = X + 13

In 13 years his age = (X + 25) + 13

In 13 years he is twice her age so:

(X + 25) + 13 = 2 (X + 13)

We will now look at some simple equations to examine the techniques used to solve an equation. Here we will look at the case of a single equation with only one unknown quantity (i.e. using only one letter) and mainly with the letter raised to no power higher than 1. These are known as linear equations (because the expressions involved would lead to straight line graphs if we used a graphical method of solution).

The aim is to get all the terms in the unknown letter (traditionally "x" as the unknown) on their own on one side of the equals sign and the basic method is to:

"Do the same thing to both sides of the equation".

The examples should make this clear. The steps are spelt out in detail, but you do not need to show so much working once you understand what you are doing.

#### Examples

In these simple examples, the solution may be obviously correct but it is very good practice to check your solution back with the original question. For example, in the last question, check by substituting the x = 1/2 solution in the original equation:

8(1/2) + 11 = 4 + 11 = 15 − which is correct so x = 1/2 is correct.

A more complicated example, with less working shown

With equations like this, with x-terms on both sides, you can check your answer like this, substituting x = 3 on each side in turn:

Left hand side (LHS) = 4x − 1 = 4(3) − 1 = 12 − 1 = 11

Right hand side (RHS) = 3x + 2 = 3(3) + 2 = 9 + 2 = 11

So the LHS = RHS, the equation is balanced, so x = 3 is correct solution.

Have a look at this second example, and then check the answer below:

Now we will check this answer:

LHS = 3(2x + 3) = 3(2 × 5 + 3) = 3(10 + 3) = 3 × 13 = 39

RHS = 2(4x + 3) – 7 = 2(4 × 5 + 3) −7 = 2 × 23 − 7 = 46 − 7 = 39

So the equation is balanced, and x = 5 is the correct solution.

Now that we have some techniques for solving equations we shall re-visit the problems we looked at previously in 'Forming an equation'.

#### Example 1

A grapefruit costs twice as much as an orange. The price of an orange is y pence.

So the grapefruit costs 2y pence.

A grapefruit and 3 oranges cost 60 pence in total.

Expressed as an equation:

2y + 3y = 60

Solve the equation:

Equation Explanation

2y + 3y = 60

Collect the like terms in y together

5y = 60

Divide both sides by 5

y = 12

Relate the algebraic solution back to the problem, putting in the units (pence in this case):

An orange costs 12 pence.

Check the answer by substituting in the original problem.

3 oranges cost 3 × 12p = 36p and a grapefruit costs twice 12p = 24p.
36p + 24p = 60p, which is correct.

#### Example 2

An apple costs 2 pence less than an orange. I buy 5 apples and 3 oranges for 94 pence. How much does an orange and an apple cost?

c = cost of an orange in pence; c − 2 = cost of an apple.

An expression for the cost of my purchase is, therefore: 5(c−2) + 3c

Expressed as an equation:

5 (c − 2) + 3c = 94

Solve the equation:

Equation Explanation

5(c − 2) + 3c = 94

Expand the bracket

5c − 10 + 3c = 94

Collect like terms

8c − 10 = 94

8c = 104

Divide both sides by 8

c = 13

Relate the algebraic solution back to the problem:

The cost of an orange is 13p; an apple is 2p less so an apple is 11p.

Check:

5 apples and 3 oranges cost 5 × 11 + 3 × 13 = 55 + 39 = 94p, which is correct.

An apple and an orange cost 11 + 13 = 24p.

#### Example 3

A farmer wishes to enclose a rectangular part of his field with fencing, and it is to be 4 times as long as it is wide. How much fencing does he need if the area to be enclosed is to be 100m2?

Starting with the width, let the width be W metres.

So, the length = 4W

Area = length × width = 4W × W

Expressed as an equation:

4W × W = 100

Solve:

Equation Explanation

4W × W = 100

Carry out the multiplication

4W2 = 100

Divide both sides by 4

W2 = 25

Take the square root of both sides

W = ± 5

Relate back to problem:

W is a measurement in metres so in this case one of the solutions (−5) is obviously not meaningful so we conclude the width is 5 metres.

This means the length (4 times as great) is 20 metres.

Check:

Area = length × width = 20 × 5 = 100m2, which is correct.

How much fencing does he need?

This will be 2(length + width) = 2(20 + 5) = 2 × 25 = 50 metres.

Note: This is the only non-linear equation in this section - it had a squared term, W2. Such terms very often lead to two solutions. It may be, as in this case, that one is clearly impossible, or both solutions are possible but only one is relevant to your problem or both solutions are valid and there are two possible answers to your problem. The important thing is to check your numerical answer back against the real-world problem from which you formed the equation in the first place.

#### Example 4

A father is 25 years older than his daughter. Today he says he will be twice her age in 13 years. How old is his daughter at present?

Let her age be x years.

Father's age = x + 25

In 13 years her age = x + 13

In 13 years his age =(x + 25) + 13

In 13 years he is twice her age so:

Expressed as an equation: (x + 25) + 13 = 2 (x + 13)

Solve:

Equation Explanation

(x + 25) + 13 = 2(x + 13)

Expand the brackets

x + 25 + 13 = 2x + 26

Collect like terms

x + 38 = 2x + 26

Take 26 from both sides

x + 12 = 2x

Take x from both sides

12 = x

Relate back to problem:

So our answer is that at present the daughter is 12 years old.

Check:

In 13 years time she will be 25 years old.

Her father is 25 years older so he is 37 and in 13 years time will be 50.

So correct - in 13 years time he will be twice as old as his daughter.