1. Equations
An equation is any mathematical statement
that contains an = sign.
6 + 4 = 10
5 – 9 = – 4 8 + 4 = 15 – 3
6 x 4 = 24
27 3 = 9 y + 4 = 10
2a + 4 = 12
are all examples of
equations!

2. If we begin with a true equation
9 + 6 = 15
We can do anything we like (add, multiply,
subtract or divide) to the numbers on
either side of the = sign as long as we do
the same thing to both sides!

3. 9 + 6 = 15
Begin with a true statement
Add 3 to both sides
9 + 6 + 3 = 15 + 3 18 = 18 still true!
Subtract 8 from both sides
9 + 6 – 8 = 15 – 8 7 = 7 still true!
Multiply both sides by 5
(9 + 6) x 5 = 15 x 5 75 = 75
still true!
Divide both sides by 3
(9 + 6) 3 = 15 3 5 = 5
still true!

4. This is a very useful process when there is
an unknown (like x) on one side, and we wish
to isolate it to solve the equation.
Solve x – 9 = 5
We want x on its own on the left of the
= sign, so we aim to get rid of the – 9.
The opposite of – 9 is + 9, so we ADD 9
to both sides
x – 9 + 9 = 5 + 9
x = 14x = 5 + 9
ZERO!

5. Example 1: Solve y – 5 = – 3
We need to remove the 5 by “undoing” the
minus. The opposite of minus is add, so we
ADD 5 to both sides :
We can now cancel the 5s on the left
side, and at the same time work out – 3 + 5
y = 2
y – 5 = – 3+ 5 + 5

6. Example 2: Solve 5y = 30
As this is really 5 times y = 30, to isolate the
y we need to remove the 5 by “undoing” the
times. The opposite of times is divide, so we
DIVIDE both sides by 5:
5
30
5
5 y
We can now cancel the 5s on the left side,
and at the same time work out 30 5
y = 6

7. Example 3: Solve y/3 = 12
As this is really y divided by 3 = 12, we need
to remove the 3 by “undoing” the divide. The
opposite of divide is multiply, so we
MULTIPLY both sides by 3:
12
3
y
We can now cancel the 3s on the left
side, and at the same time work out 12 x 3
y = 36
x3x3

8. EXAMPLE 4: Solve 3y + 5 = 26
To get x alone, first we need to remove the 5,
then the 3. Begin by taking 5 from both sides
3y + 5 = 26
3y = 21
Now we divide both sides by 3
y = 7
ZERO
– 5– 5
3
21
3
3 y
Cancel the 3s on the left side

9. And now for a really useful trick!
Suppose we begin with 8 – 3 = 5
You’re allowed to change all the signs (the sign
in front of every term)
Still true!
Try this for 2 – 9 – 5 = – 12
and get – 2 + 9 + 5 = + 12
Still true!
– 8 + 3 = – 5
But remember you must change ALL the signs

10. This trick is really useful in equations where
there is a negative in front of the letter!
Solve – a + 7 = 12
Change all the signs
Now add 7 to both sides, as before
and get
a – 7 = – 12
a – 7 + 7 = – 12 + 7
a = – 5
Now check to see you’re right by substituting
a = – 5 into the original equation
– – 5 + 7 = 12 TRUE!

11. Example 6
Solve 5 – 2a = – 9
Change all the signs as the “a” has a minus in
front
Now add 5 to both sides, as before
– 5 + 2a = 9
– 5 + 2a + 5 = 9 + 5
2a = 14
Now check to see you’re right by substituting
a = 7 into the original equation
5 – 2 x 7 = – 9 TRUE!
a = 7

12. Solve 7
4
a
Since this is the same as a 4 = 7,
we do the opposite of divide, i.e.
multiply by 4
a = 28
x 4 x 4
4
a
= 7
Cancel the 4s on the left

13. Example 8:
Solve
6
5
37 a
First we multiply by 5 to get rid of the fraction
6
5
37 a
x 5 x 5
Cancel the 5s on the left
7 – 3a = 30
Seeing there’s a minus in front of the a, we can
change all signs
– 7 + 3a = – 30
Add 7 to both sides
– 7 + 3a + 7 = – 30 + 7
3a = – 23
Divide both sides by 3
a = – 7.667

14. Example 9:
Solve
6
5
3
4
a
First we subtract 4 from both sides
Cancel the 4s on the left
Multiply both sides by 5
3a = 10
Divide both sides by 3
a = 10/3
2
5
3a
464
5
3
4
a
– 4 – 4

15. Example 10:
Solve 4
7
2
9
a
Add 9 to both sides
Multiply both sides by 7
Divide both sides by 2
a = 45.5
13
7
2a
Sign change
4
7
2
9
a
2a = 91

16. Solve 2(x – 5) = 15
Expand the brackets
2x – 10 = 15
Add 10 to both sides
2x – 10 + 10 = 15 + 10
2x = 25
Divide both sides by 2
x = 25/2 or 12.5
Example 11……….

17. Solve 2(x – 5) + 3(2x + 1) = 28
Expand the brackets
2x – 10 + 6x + 3 = 28
Add 7 to both sides
8x – 7 + 7 = 28 + 7
8x = 35
Divide both sides by 8
x = 35/8 or 4.375
Example 12……….
Clean up left side
8x – 7 = 28

18. Equations with an unknown on both sides
Example 13 Solve 3a – 5 = a + 11
The aim is to get the a on one side only, so try
taking a from both sides:
What happens to the right-hand side?
3a – 5 – a = a + 11 – a
3a – 5 – a = a + 11 – a
11
Add 5 to both sides
2a – 5 + 5 = 11 + 5
2a = 16
Divide both sides by 2
a = 8
2a – 5 =
Back at the start, we
could have taken 3a from
both sides instead of just
a. This would have given
– 5 = 11 – 2a
and then proceed as usual!

19. Equations with an unknown on both sides
Example
14
Solve 9 – a = 12 + 3a
The aim is to get only one term with a.
So try adding a to both sides:
What happens to the left-hand side?
9 - a + a = 12 + 3a + a
9 – a + a = 12 + 3a + a
12 + 4a
Take 12 from both sides
9 – 12 = 12 + 4a – 12
– 3 = 4a
Divide both sides by 4
a = – ¾
9 =

20. Equations with fractions on both sides
Example
15 Solve
4
2
3
13 xx
Multiply both sides by the LCD, 12. This kills the fractions.
Put brackets around the numerators.
1
12
4
)2( x
1
12
3
)13( x
Cancel
4(3x + 1) = 3(2 – x)
Expand
12x + 4 = 6 – 3x
15x = 2
x = 2/15
4 3

21. Equations with fractions on both sides
Example
16 Solve
4
5
5
)24(3 xx
Multiply both sides by the LCD, 20.
This kills the fractions
1
20
4
5 x
1
20
5
)24(3 x
Cancel and make sure brackets are around numerators
4 3(4 – 2x) =5(5 - x)
Expand
48 – 24x = 25 – 5x
23 = 19x
x = 23/19

23. Example 17
A rectangular field is 5m
longer than it is wide
Its perimeter is 200m.
Find its dimensions (width and length).
Key Strategy ….. Always let x equal the smallest part
So, Let x equal the width.
So the length is… x + 5 and
The four sides total to 200…………..
x + x + x + 5 + x + 5 = 200
4x + 10 = 200
4x = 190
x = 47.5
So the width is
47.5m
Length is 47.5 + 5 =
52.5m
Finally make sure
they add to 200
x
x + 5
Instead of writing x + 5
twice, you could have
written 2(x + 5). This
becomes 2x + 10 when you
get rid of the brackets!

24. Example 18
Another rectangular field is
12m longer than it is wide
Its perimeter is 1km.
Find its dimensions (width and length).
Let x equal the width.
So the length is… x + 12 and
The four sides total to 1000…………..
4x + 24 = 1000
4x = 976
x = 244
So the width is
244m
Length is 244 + 12 =
256m
Finally make sure
they add to 1000
x
x + 12

25. Example 19
Find the value of x in this diagram
(4x – 3)º
(7x – 4)º As these are co-
interior, they are
supplementary and so must
add to 180º
4x – 3 + 7x – 4 = 180
Clean up left side
11x – 7 = 180
Add 7 to both sides
11x = 187
Divide both sides by 11
x = 17
It is wise to check
your answer by
substituting 17 into
both angles and seeing
that they add to 180.
7 x 17 – 4 = 115
4 x 17 – 3 = 65
115 + 65 = 180, so
we’re correct!

26. Example 18
Find the value of b in this isosceles triangle
(2b + 1)º
(b – 7 )º As it’s isosceles, the other bottom
angle must also be (2b + 1)
The three angles add to 180, so…..
b – 7 + 2(2b + 1) = 180
b – 7 + 4b + 2 = 180
5b – 5 = 180
5b = 185
b = 37
Now check your answer
by substituting 37 into
the 3 angles and seeing
that they add to 180.
37 – 7 = 30
2 x 37 + 1 = 75
30 + 2 x 75 = 180,
so we’re correct!
(2b + 1)º
Expanding brackets
Cleaning up left side
Adding 5 to both sides
Dividing both sides by 5

27. Example 19
Jimmy, Mary and Joseph have $24 between them.
Mary has twice the amount Jimmy has.
Joseph has $3.25 more than Mary.
How much do they each have?
Key Strategy ….. Always let x equal the smallest share
So, Let x equal Jimmy’s amount as he has the least.
So Mary has……… 2x and
Joseph has………. 2x + 3.25
Now we know they total to 24…………..
x + 2x + 2x + 3.25 = 24
5x + 3.25 = 24
5x = 20.75
x = 4.15
So Jimmy has
$4.15
Mary has 2 x $4.15 =
$8.30
Joseph has
$8.30 + $3.25
= $11.55Finally make sure
they add to $24

28. Mary is twice as old as John, and 4 years younger
than Peter. The sum of their ages is 159. How old
are they?
Let the youngest (John) be x.
So Mary’s age is 2x
& Peter’s age is 2x + 4
Now we add them up, knowing it will equal 159.
x + 2x + 2x + 4 = 159
5x + 4 = 159
5x = 155
x = 31
So John is 31
Mary is 2 x 31 = 62
Peter is 62 + 4 = 66– 4 from both sides
divide both sides by 5
Example 20

29. The isosceles triangle and the square have the
same perimeter. Find x as a mixed numeral
2x – 3
x + 5
Triangle’s perimeter
= x + 5 + 4x + 2
= x + 5 + 2(2x + 1)
= 5x + 7
Square’s perimeter
= 4(2x - 3)
= 8x - 12
5x + 7 = 8x – 12
5x + 7 – 5x = 8x – 12 – 5x
7 = 3x – 12
7 + 12 = 3x – 12 + 12
19 = 3x
3
1
6x
Example 21

30. Twins Bessie and Albert have a brother, Marmaduke, 8
years older than they are, and they have a sister, Sylvia,
who is 12 years younger than they are. Together their
ages add to 168. Use algebra to find the twins’ ages.
Let the twins’ ages be x.
Marmaduke is x + 8.
Sylvia is x – 12.
x + x + x + 8 + x – 12 = 168
4x – 4 = 168
4x = 168 + 4
4x = 172
x = 43
The twins are 43!
Also, Marmaduke is 51, Sylvia is 31

31.  (3a – 5)cm
Robbie the rectangle is twice as long as he
is wide. His perimeter is 294 cm.
Calculate his dimensions and his area.
6(3a – 5) = 294
18a – 30 = 294
18a = 324
a = 18
Width = 49cm
Length = 98cm
Area: 49 x 98 = 4802cm2
If his width is (3a – 5)
then his length is twice
that, so must be 2(3a – 5).
This means all sides must
total 6(3a – 5)
(3a – 5)
2(3a – 5)
2(3a – 5)

32. Archibald, Muriel and Oswald come across a
bag of 95 marbles. They divide them up in
such a way that Muriel has 50% more than
Archibald, and Oswald has five fewer than
Archibald and Muriel combined. How many
does each have?
Let x be Archibald’s share
So Muriel’s share is 1.5x
Oswald’s share is x + 1.5x – 5
= 2.5x – 5
x + 1.5x + 2.5x – 5 = 95
5x – 5 = 95
5x = 100
x = 20
Archie has 20, Muriel 30 and Ossie 45

33. Example 25
5x - 4
3x + 2
20 – x
The perimeter of this
shape is 176 cm. Find
the value of x
This side is 5x – 4 – (3x + 2)
2x - 6
And so is this
side also
2x - 6
= 5x – 4 – 3x – 2
= 2x - 6
So this side has to
be 20 – x + 2x – 6
= x + 14
x + 14
Now add up all the sides!
12x + 20 = 176
12x = 156 X = 13

34. Attila, Otto, Peregrine and Ugly are cousins.
Peregrine is two decades younger than Ugly, and
Peregrine’s age is 80% of Attila’s age. Otto, the
eldest, is 26 years younger than the total of
Attila’s and Peregrine’s ages.
Their ages total eight less than four times Ugly’s
age. How old are they?
Example 26

35. Try letting Attila’s age = x, only because it says
“Peregrine’s age is 80% of Attila’s age” making it
easy to write Peregrine’s age as 0.8x.
So…..
Let Attila’s age = x
Peregrine’s age = 0.8x
Ugly’s age = Peregrine’s age + 20
= 0.8x + 20
Otto’s age = Attila + Peregrine - 26
= x + 0.8x – 26
= 1.8x – 26

36. Their ages total eight less than four times Ugly’s age.
Attila = x
Peregrine = 0.8x
Ugly = 0.8x + 20
Otto = 1.8x – 26
x + 0.8x + 0.8x + 20 + 1.8x – 26 = 4(0.8x + 20) - 8
4.4x - 6 = 3.2x + 80 - 8
1.2x = 78
x = 65
Attila is 65
Peregrine is 52
Ugly is 72
Otto is 91

37. Little Jimmy has a number of 10c and 20c coins in his piggybank.
His 29 coins total to $4.10,
How many of each kind of coin does he have?
Let x be the number of 10 cent coins.
Then, since there are 29 coins altogether, we can let the number of 20c coins
be
29 – x .
Example 27

38. so we now have that there are….
• x coins each valued at 10c, and…
• (29 – x) coins each valued at 20c
The x coins each valued at 10 cents must be worth a total of 10x
and the (29 – x) coins each valued at 20 cents must be worth a total of
20(29 – x)
We know these values total to 410, so
10x + 20(29 – x) = 410
10x + 580 – 20x = 410
580 – 10x = 410
Expand
Clean up

39. – 10x = 410 – 580
– 10x = – 170
x = 17
Remember, x was the number of 10 cent coins, so
there are 17 ten-cent coins. There were 29 coins
altogether, so there must be (29 – x) i.e.29 – 17 = 12
twenty-cent coins!
Clean up
Divide by – 10
Number of 10c = 17
Number of 20c = 12
Finally, check that 17 x 10 + 12 x 20 = 410

40. Example 28 – This uses FACTORISING
The diagram represents a path enclosing a park.
The curved section is a quadrant of a circle, radius r m.
The longest side is twice the width of the park. The
perimeter is 1km. Calculate the area of the park in m2
Circumference of a circle
C = 2π r
Area of a circle
A = π r 2

41. r
rr
r
r
¼ x 2πr
We set up an equation for the perimeter….
Left side + top + bottom + quadrant = 1000m
r + r + 2r + = 1000mr
2
4r + = 1000mr
2
1000
2
4 r
2
4
1000
r
r = 179.50755m
Area = ¼ π r 2 + r 2
Area = 57531 sq metres
Factorising!