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KS3 Mathematics
A2 Equations
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Contents
A2 Equations
A2.1 Solving simple equations
A2.2 Equations with the unknown on both sides
A2.3 Solving more difficult equations
A2.4 Equations and proportion
A2.5 Non-linear equations
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Equations
An equation links an algebraic expression and a number, or
two algebraic expressions with an equals sign.
For example,
x + 7 = 13 is an equation.
In an equation the unknown usually has a particular value.
Finding the value of the unknown is called solving the
equation.
x + 7 = 13
x=6
When we solve an equation we always line up the equals signs.
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Using inverse operations
In algebra, letter symbols represent numbers.
Rules that apply to numbers in arithmetic apply to letter
symbols in algebra.
For example, in arithmetic, if 3 + 7 = 10, we can use
inverse operations to write:
10 – 7 = 3
and
10 – 3 = 7
In algebra, if a + b = 10, we can use inverse operations
to write:
or
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10 – b = a
and
10 – a = b
a = 10 – b
and
b = 10 – a
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Using inverse operations
In arithmetic, if 3 × 4 = 12, we can use inverse operations
to write:
12 ÷ 4 = 3
and
12 ÷ 3 = 4
In algebra, if ab = 12, we can use inverse operations to
write:
12 = a
b
or
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a = 12
b
and
12 = b
a
and
b = 12
a
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Using inverse operations to solve equations
We can use inverse operations to solve simple equations.
For example,
x + 5 = 13
x = 13 – 5
x=8
Always check the solution to an equation by substituting the
solution back into the original equation.
If we substitute x = 8 back into x + 5 = 13 we have
8 + 5 = 13
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Using inverse operations to solve equations
Solve the following equations using inverse operations.
5x = 45
17 – x = 6
x = 45 ÷ 5
x=9
Check:
5 × 9 = 45
We always write
the letter before
the equals sign.
17 = 6 + x
17 – 6 = x
11 = x
x = 11
Check:
17 – 11 = 6
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Using inverse operations to solve equations
Solve the following equations using inverse operations.
x
7
=3
x=3×7
x = 21
Check:
21
7
=3
3x – 4 = 14
3x = 14 + 4
3x = 18
x = 18 ÷ 3
x=6
Check:
3 × 6 – 4 = 14
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What number am I thinking of…?
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What number am I thinking of …?
I’m thinking of a number.
When I multiply the number by 2 and add 5 the answer is 11.
What number am I thinking of …?
We can write this as an equation.
Instead of using ? for the number I am thinking of, let’s use
the letter x.
Start with x
Multiply by 2
and add 5
x
2x
2x + 5
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to give you 11.
2x + 5 = 11
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What number am I thinking of …?
Start with x
Multiply by 2
and add 5
x
2x
2x + 5
to give you 11.
2x + 5 = 11
We can solve this equation using inverse operations in the
opposite order.
Start with the equation:
2x + 5 = 11
Subtract 5:
2x = 11 – 5
2x = 6
Divide by 2:
x=6÷2
x=3
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Solving equations by transforming both sides
Solve this equation by transforming both sides in the same
way:
m
–1=2
4
+1 +1
Add 1 to both sides.
m
=3
4
×4 ×4
Multiply both sides by 4.
m = 12
We can check the solution by substituting it back into the
original equation:
12 ÷ 4 – 1 = 2
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Find the missing number
The number in each brick is found by adding the two numbers
above it. What are the missing values?
23
?
n
23?+ n
62
n +?62
127
We can start by calling the unknown number in the top brick n.
The unknown numbers in the next two bricks can be written in
terms of n.
We can now write an equation and solve it to find the value of n.
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Find the missing number
The number in each brick is found by adding the two numbers
above it. What are the missing values?
23
n
21
2344
+n
62
n+
8362
127
23 + n + n + 62 = 127
Simplify:
85 + 2n = 127
Subtract 85:
2n = 42
Divide by 2:
n = 21
Check this solution.
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Equation dominoes
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Contents
A2 Equations
A2.1 Solving simple equations
A2.2 Equations with the unknown on both sides
A2.3 Solving more difficult equations
A2.4 Equations and proportion
A2.5 Non-linear equations
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Balancing equations
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Constructing an equation
Ben and Lucy have the same number of sweets.
Ben started with 3 packets of sweets and ate 11 sweets.
Lucy started with 2 packets of sweets and ate 3 sweets.
How many sweets are there in a packet?
Let’s call the number of sweets in a packet, n.
We can solve this problem by writing the equation:
3n – 11 = 2n – 3
The number of
Ben’s sweets
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is the
same as
the number of
Lucy’s sweets.
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Solving the equation
Let’s solve this equation by transforming both sides of the
equation in the same way.
3n – 11 = 2n – 3
Start by writing the equation down.
-2n
Subtract 2n from both sides.
-2n
n – 11 = –3
+11
+11
n = 8
Always line up the equals signs.
Add 11 to both sides.
This is the solution.
We can check the solution by substituting it back into the
original equation:
3  8 – 11 = 2  8 – 3
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Constructing an equation
I’m thinking of a number.
When I multiply the number by 4, I get the same answer
as adding 9 to the number.
What number am I thinking of?
Let’s call the unknown number n.
We can solve this problem by writing the equation:
4n
The number
multiplied by 4
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=
n+9
is the
same as
the number
plus 9.
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Solving the equation
Let’s solve this equation by transforming both sides of the
equation in the same way.
4n = n + 9
-n
-n
Start by writing the equation down.
Subtract n from both sides.
3n = 9
Always line up the equals signs.
÷3
Divide both sides by 3.
÷3
n = 3
This is the solution.
We can check the solution by substituting it back into the
original equation:
43= 3+9
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Constructing an equation
Find the value of x.
(65 – 2x)o
(2x + 5)o
Remember, vertically opposite angles are equal.
We can solve this problem by writing the equation:
2x + 5 = 65 – 2x
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Solving the equation
Let’s solve this equation by transforming both sides of
the equation in the same way.
2x + 5 = 65 – 2x
+2x
+2x
Add 2x to both sides.
4x + 5 = 65
-5
4x
-5
=
÷4
x
60
÷4
=
Subtract 5 from both sides.
15
Divide both sides by 4.
This is the solution.
Check:
2  15 + 5 = 65 – 2  15
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Rectangle problem
The area of this rectangle is 27 cm2.
Calculate the value of x and
use it to find the height of the
rectangle.
8x – 14
2x + 1
Opposite sides of a
rectangle are equal.
We can use this fact to write
an equation in terms of x.
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Rectangle problem
The area of this rectangle is 27 cm2.
8x – 14 = 2x + 1
–2x
–2x
6x – 14 = 1
8x – 14
2x + 1
+14 +14
6x = 15
÷6
÷6
x = 2.5
If x = 2.5 we can find the height of the rectangle using
substitution:
8 × 2.5 – 14 = 20 – 14 = 6 cm
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Rectangle problem
The area of this rectangle is 27 cm2. What is its width?
y
Let’s call the width of the
rectangle y.
8x – 14
2x + 1
The dimensions of the rectangle
are 6 cm by 4.5 cm.
If the height of the
rectangle is 6 cm and the
area is 27 cm2 then we
can find the width by
writing the equation:
6y = 27
÷6
÷6
y = 4.5
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Select the correct equation
Veronica has 58p and
buys 4 chocolate bars.
Thomas has £1 and
buys 7 chocolate bars.
They both receive the same amount of change.
If c is the cost of one chocolate bar, which equation could
we use to solve this problem?
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A: 4c + 58 = 7c + 100
B: 58 – 4c = 1 – 7c
C: 58 – 4c = 100 – 7c
D: 4c – 58 = 7c – 1
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Equation match
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Contents
A2 Equations
A2.1 Solving simple equations
A2.2 Equations with the unknown on both sides
A2.3 Solving more difficult equations
A2.4 Equations and proportion
A2.5 Non-linear equations
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Equations with brackets
Equations can contain brackets. For example:
2(3x – 5) = 4x
To solve this we can
Multiply out the brackets: 6x –10 = 4x
+ 10
Add 10 to both sides:
6x = 4x + 10
- 4x
Subtract 4x from both sides:
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- 4x
2x = 10
÷2
Divide both sides by 2:
+ 10
÷2
x=5
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Equations with brackets
Sometimes we can solve equations such as:
2(3x – 5) = 4x
by first dividing both sides by the number in front of the bracket:
Divide both sides by 2:
3x – 5 = 2x
+5
Add 5 to both sides:
3x = 2x + 5
- 2x
Subtract 2x from both sides:
+5
- 2x
x=5
In this example, dividing first means that there are fewer steps.
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Balancing equations with brackets
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Solving equations involving division
Linear equations with unknowns on both sides can also involve
division.
3x + 2 = 11 – x
For example,
4
In this case we must start by multiplying both sides of the
equation by 4.
3x + 2 = 4(11 – x)
Multiply out the brackets:
3x + 2 = 44 – 4x
Add 4x to both sides:
7x + 2 = 44
Subtract 2 from both sides:
Divide both sides by 7:
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7x = 42
x=6
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Solving equations involving division
Sometimes the expressions on both sides of the equation are
divided.
4
5
For example,
=
(x + 3)
(3x – 5)
In this example, we can multiply both sides by (x + 3) and
(3x – 5) in one step to give:
4(3x – 5) = 5(x + 3)
Multiply out the brackets:
Subtract 5x from both sides:
Add 20 to both sides:
Divide both sides by 7:
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12x – 20 = 5x + 15
7x – 20 = 15
7x = 35
x=5
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Equivalent equations
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Contents
A2 Equations
A2.1 Solving simple equations
A2.2 Equations with the unknown on both sides
A2.3 Solving more difficult equations
A2.4 Equations and proportion
A2.5 Non-linear equations
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Sale!
Footballs!
Were £4,
Now only £3!
Original Price
Sale Price
We can write the ratio of
3
4
=
6
8
=
9
12
£4
£3
£8
£6
sale price
original price
=
12
16
£12 £16 £20
£9 £12 £15
for each pair of values.
=
15
20
=
0.75
What do you
The notice
ratios are
about
equal.
these ratios?
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Drawing a graph
Original Price, x £4
Sale Price, y
£3
£8
£6
£12 £16 £20
£9 £12 £15
y
Sale price of footballs
15
The points lie on a
straight line through
the origin.
Sale price, £
12
The equation of the
line is:
9
3
6
y= 4 x
or
3
0
0
4
8
12
16
20
x
y = 0.75x
Original price, £
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Constant speed
Susan walks to school at a constant speed. Altogether, it
takes her 10 minutes to walk 800 metres.
How far would she have walked in two minutes?
2
4
6
8
Time, minutes
10
Distance, metres 160 320 480 640 800
We can write the ratio of
160
2
=
320
4
=
distance
480
6
time
=
for each pair of values.
640
8
=
800
10
=
80
What do you
The notice
ratios are
about
equal.
these ratios?
The time and the distance are directly proportional.
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Distance/time graph
We can plot the points from the table onto a graph.
Distance, metres
Time, minutes
2
4
6
8
10
Distance, metres 160 320 480 640 800
y
800
Susan’s walking speed
The points lie on a straight
line through the origin.
640
The equation of the line is:
y = 80x
480
320
160
0
0
2
4
6
8
10
x
How far would Susan walk in:
a) 3 minutes? 240 metres
a) 15 minutes? 1200 metres
Time, minutes
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Mixing colours
Orange paint is made by mixing 4 litres of yellow paint with
6 litres of red paint.
1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint
To make the same shade of orange we must keep the
amount of yellow paint and red paint in direct proportion.
How many litres of each colour do you need to make:
a) 5 litres of orange paint? 2 l of yellow and 3 l of red
b) 1 litre of orange paint? 0.4 l of yellow and 0.6 l of red
c) 7 litres of orange paint? 2.8 l of yellow and 4.2 l of red
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Mixing colours
1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint
The ratio
red
yellow
is
6
4
= 1.5
We can write an equation linking the amount of red paint r to
the amount of yellow paint y as
r = 1.5y
How many litres of red paint would be needed to mix with 14
litres of yellow paint to make the same shade of orange?
r = 1.5 × 14
= 21 l
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Mixing colours
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Equations and direct proportion
When two quantities x and y are directly proportional to
each other we can link them with the symbol .
We write
yx
y is directly
proportional to x.
We can also link these variables with the equation
y = kx
where k is a constant value equal to
y
x
.
The graph of y = kx will always be a straight line through
the origin.
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Contents
A2 Equations
A2.1 Solving simple equations
A2.2 Equations with the unknown on both sides
A2.3 Solving more difficult equations
A2.4 Equations and proportion
A2.5 Non-linear equations
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Non-linear equations
A number added to its square equals 42.
What could the number be?
If we call the unknown number x we can write the following
equation:
x + x2 = 42
This is a non-linear equation. It contains powers greater than 1.
One solution to this equation is x = 6.
There is another solution to this equation.
x = –7
6 + 62 = 6 + 36 = 42
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and
–7 + (–7)2 = –7 + 49 = 42
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Non-linear equations
In a linear equation, unknowns in the equation cannot be
raised to any power other than 1.
For example, 4x + 7 = 23 is a linear equation.
In a non-linear equation, unknowns in the equation can have
indices other than 1.
For example, x4 + 2x = 20 is a non-linear equation.
In a quadratic equation, the highest index of any of the
unknowns is 2.
For example, x2 – 3x = 10 is a quadratic equation.
Quadratic equations usually have two solutions.
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Solving non-linear equations
Solve 3x2 – 5 = 22
We can solve this non-linear equation using inverse operations.
3x2 – 5 = 22
Adding 5:
Dividing by 3:
Square rooting:
3x2 = 27
x2 = 9
x = ±9
x = 3 or –3
When we find the square root in an equation we must always
give both the positive and the negative solution.
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Solving non-linear equations
Solve m – 6 =
25
m–6
We must start by multiplying by m – 6.
(m – 6)(m – 6) = 25
We can write this as:
(m – 6)2 = 25
m – 6 = 5 or –5
Square rooting:
To find both solutions we must add 6 to both 5 and –5.
m=5+6
m = 11
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or
m = –5 + 6
m=1
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Using a calculator to solve non-linear equations
25
Solve 4n3 + 5 = 33
m–6
We can solve this equation by using inverse operations.
4n3 + 5 = 33
Subtracting 5:
4n3 = 28
Dividing by 4:
n3 = 7
Cube rooting:
n = 7
3
We know that 7 is not a cube number so we can use the 3
key on a calculator.
n = 1.91 (to 2 d.p.)
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Using trial and improvement
Solve h2 + 3h = 45
We can factorize this equation to give h(h + 3) = 45
We can then solve the equation using trial and improvement.
h
5
6
5.4
5.3
5.37
5.38
h+3
8
9
8.4
8.3
8.37
8.36
h(h + 3)
40
54
45.36
43.99
44.9469
40.0844
too small
too big
too big
too small
too small
too big
44.9469 is closer to 45 than 40.0844 so h = 5.37 (to 2 d.p.)
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Using trial and improvement
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Using a spreadsheet
Solve a3 – 4a = 2
We can use a spreadsheet to solve this equation by trial and
improvement.
After heading the columns, put your first guess into cell A2.
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Using a spreadsheet
In cell A3 we can type =A2+0.1 without any spaces.
Clicking on the bottom right hand corner of this cell and
dragging down to cell A12 will enter the numbers 2.1, 2.2, 2.3
… increasing in steps of 0.1 up to 3.
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Using a spreadsheet
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Using a spreadsheet
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Using a spreadsheet
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Using a spreadsheet
We can see that the solution lies between 2.2 and 2.3.
We can now change the value in cell A2 to 2.2.
In cell A3 we can then enter =A2+0.01 without any spaces.
Clicking on the bottom right hand corner of this cell and
dragging down to cell A12 will enter the numbers 2.21, 2.22,
2.23 … increasing in steps of 0.01 up to 2.3.
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Using a spreadsheet
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Using a spreadsheet
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Solving non-linear equations
Using the results from the spreadsheet we can see that for
a3 – 4a = 2
a = 2.214 (to 3 d.p.)
Could this equation have any more solutions?
Use the spreadsheet to look at values of a3 – 4a between
–2 and 0. Increase the values of a by 0.1.
Find every solution to the equation a3 – 4a = 2.
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Using a spreadsheet
There are two
more solutions to
a3 – 4a = 2
between –1.7
and –1.6 …
… and between
–0.6 and –0.5.
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