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

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. Begin with a true statement
9 + 6 = 15
Add 3 to both sides =
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 = 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
ZERO!
x – = 5 + 9
x = 14
x = 5 + 9

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 :
y – = – 3
+ 5
+ 5
We can now cancel the 5s on the left side, and at the same time work out – 3 + 5
y = 2

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:
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: x3 x3
We can now cancel the 3s on the left side, and at the same time work out 12 x 3
y = 36

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

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)
– 8
+ 3 =
– 5
Still true!
Try this for – 9 – 5 = – 12
and get – = + 12
Still true!
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!
Example 5
Solve – a + 7 = 12
Change all the signs a
– 7 =
– 12
Now add 7 to both sides, as before
a – = –
a = – 5
and get
Now check to see you’re right by substituting a = – 5 into the original equation
– – = 12 TRUE!

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

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

13. Example 8:
Solve
First we multiply by 5 to get rid of the fraction
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 = –
3a = – 23
Divide both sides by 3
a = – 7.667

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

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

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

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

18. and then proceed as usual!
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:
3a – 5 – a = a + 11 – a
What happens to the right-hand side?
3a – 5 – a = a + 11 – a
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!
2a – 5 = 11
Add 5 to both sides
2a – =
2a = 16
Divide both sides by 2
a = 8

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

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

21. Equations with fractions on both sides
Example 16
Solve
Multiply both sides by the LCD, 20. This kills the fractions
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

22. Problem solving

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

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

25. x = 17 Example 19 Find the value of x in this diagram
As these are co-interior, they are supplementary and so must add to 180º
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
= 180, so we’re correct!
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

26. Example 18 Find the value of b in this isosceles triangle
As it’s isosceles, the other bottom angle must also be (2b + 1)
The three angles add to 180, so…..
(2b + 1)º
Now check your answer by substituting 37 into the 3 angles and seeing that they add to 180.
37 – 7 = 30
2 x = 75
x 75 = 180,
so we’re correct!
b – 7 + 2(2b + 1) = 180
b – 7 + 4b + 2 = 180
Expanding brackets
5b – 5 = 180
Cleaning up left side
5b = 185
Adding 5 to both sides
b = 37
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
So Jimmy has $4.15
and
Joseph has………. 2x
Mary has 2 x $4.15 = $8.30
Now we know they total to 24…………..
x + 2x + 2x = 24
Joseph has $ $3.25
= $11.55
5x = 24
Finally make sure
they add to $24
5x = 20.75
x = 4.15

28. Let the youngest (John) be x.
Example 20
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
So John is 31
5x + 4 = 159
Mary is 2 x 31 = 62
5x = 155
Peter is = 66
– 4 from both sides
x = 31
divide both sides by 5

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

30. Example 22
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.
x + x + x x – 12 = 168
Sylvia is x – 12.
4x – 4 = 168
The twins are 43!
4x =
4x = 172
x = 43
Also, Marmaduke is 51, Sylvia is 31

31. Example 23
(3a – 5)
2(3a – 5)
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)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
Width = 49cm
Length = 98cm
18a – 30 = 294
18a = 324
Area: 49 x 98 = 4802cm2
a = 18

32. Archie has 20, Muriel 30 and Ossie 45
Example 24
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
x + 1.5x + 2.5x – 5 = 95
So Muriel’s share is 1.5x
5x – 5 = 95
5x = 100
Oswald’s share is x + 1.5x – = 2.5x – 5
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
And so is this side also
2x - 6
2x - 6
x + 14
This side is 5x – 4 – (3x + 2)
So this side has to be 20 – x + 2x – 6
= 5x – 4 – 3x – 2
= 2x - 6
= x + 14
Now add up all the sides!
12x + 20 = 176
X = 13
12x = 156

34. Example 26
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?

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 x – 26 = 4(0.8x + 20) - 8
4.4x - 6 = 3.2x
Attila is 65
Peregrine is 52
Ugly is 72
Otto is 91
1.2x = 78
x = 65

37. Coins!! The solution 29 – x . Example 27
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?
The solution
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 .

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

39. – 10x = 410 – 580 – 10x = – 170 x = 17 Number of 10c = 17
Clean up 
x = 17
Divide by – 10 
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!
Number of 10c = 17
Number of 20c = 12
Finally, check that 17 x 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 ¼ x 2πr r r r r Factorising! Area = ¼ π r 2 + r 2
We set up an equation for the perimeter….
Left side + top + bottom + quadrant = 1000m
r = m
r r r = 1000m
4r = 1000m
Factorising!
Area = ¼ π r 2 + r 2
Area = sq metres