Chapter 11

Solving simple equations

11.1 Introduction

3 x−4isanexampleofanalgebraic expression.
3 x− 4 =2isanexampleofanalgebraic equation(i.e.
it contains an ‘=’ sign).
An equation is simply a statement that two expressions
are equal.

Hence, A=πr^2 (where A is the area of a circle
of radiusr)
F=

9

5

C+32 (which relates Fahrenheit and
Celsius temperatures)
and y= 3 x+2(whichistheequationofa
straight line graph)
are all examples of equations.

11.2 Solving equations

To ‘solve an equation’ means ‘to find the value of the
unknown’. For example, solving 3x− 4 =2 means that
the value ofxis required.
In this example,x=2. How did we arrive atx=2?
This is the purpose of this chapter – to show how to
solve such equations.
Many equations occur in engineering and it is essential
that we can solve them when needed.
Here are some examples to demonstrate how simple
equations are solved.

Problem 1. Solve the equation 4x= 20

Dividing each side of the equation by 4 gives

4 x

4

=

20
4
i.e.x= 5 by cancelling, which is the solution to the
equation 4x=20.

Thesameoperationmustbe applied to both sides of an

equation so that the equality is maintained.

We can do anything we like to an equation,as long

as we do the same to both sides.This is, in fact, the

only rule to remember when solving simple equations

(and also when transposing formulae, which we do in

Chapter 12).

Problem 2. Solve the equation

2 x

5

= 6

Multiplying both sides by 5 gives^5

(

2 x

5

)

= 5 ( 6 )

Cancelling and removing brackets gives 2 x= 30

Dividing both sides of the equation by 2 gives

2 x

2

=

30

2

Cancelling gives x= 15

which is the solution of the equation

2 x

5

=6.

Problem 3. Solve the equationa− 5 = 8

Adding 5 to both sides of the equation gives

a− 5 + 5 = 8 + 5

i.e. a= 8 + 5

i.e. a= 13

which is the solution of the equationa− 5 =8.

Note that adding 5 to both sides of the above equation

results in the−5 moving from the LHS to the RHS, but

the sign is changed to+.

Problem 4. Solve the equationx+ 3 = 7

Subtracting 3 from both sides gives x+ 3 − 3 = 7 − 3

i.e. x= 7 − 3

i.e. x= 4

which is the solution of the equationx+ 3 =7.

DOI: 10.1016/B978-1-85617-697-2.00011-9