Download AS-and-IB-Keeping-Yo..

The Hazeley Academy
Page 2
Welcome to your 7 week guide to keeping your maths brain.
In this booklet you will find 7 sections, one to complete per week
between now and September. Each section covers a topic from your
Higher GCSE course that will be encountered again in your Year 12
course. By completing the exercises gradually through your break
you should find that these techniques are still fresh when you return
to start in Year 12.
Each section starts with a review of the key techniques required for
the topic. It includes worked examples, , and hint bubbles, , to
help you when you are working through the exercises. Remember to
check your answers as you go and highlight any areas that you think
you need more help with when you come back in September.
The Hazeley Mathematics Department
The Hazeley Academy
Page 4:
Linear Equations
Page 5:
Exercise 1
Page 6:
Quadratic Equations
Page 7:
Exercise 2
Page 8:
Simultaneous Equations
Page 9:
Exercise 3
Page 10:
Inequalities
Page 11:
Exercise 4
Page 12:
Algebraic Fractions
Page 13:
Exercise 5
Page 14:
Indices
Page 15:
Exercise 6
Page 16:
Trigonometry
Page 17:
Exercise 7
Page 18:
Answers
Page 3
The Hazeley Academy
Page 4
An equation that only includes variables with a power of 1 is a linear equation. All linear
equations will have graphs that form a straight line.
Solving Linear Equations:
Always ensure that you show your steps when solving any equations.
 Example 1:
Solve 3x + 7 = 43
-2
-2
3x = 36
÷3
÷3
x = 12
Minimum methods acceptable: As e.g. with no arrows
 Example 2:
Solve 5x – 2 = 9
7
×7
×7
5x – 2 = 63
+2
+2
5x
= 65
+2
+2
x
= 13
To help decide the order:
imagine secret brackets around
the numerator of the fraction.
(5x – 2)
7
Minimum methods acceptable: As e.g. with no arrows
 Example 3:
Solve 2x + 4 = 7x – 11
-2x
-2x
Collect x terms first.
Move the SMALLEST amount
of x’s to the biggest
4 = 5x – 11
+11
+11
15 = 5x
÷3
÷3
3 = x
Minimum methods acceptable: As e.g. with no arrows
The Hazeley Academy
Page 5
Exercise 1:
1) Solve the following equations. Write your solution as a fraction when required.
a) 11q – 17 = 60
b) 6p + 10 = 52
c) 8y – 1 = 0
d) 4 + 5n = 64
e) 13 – b = 7
f) 20 – 3c = 8
2) Solve these equations involving divisions. Write your solution as a fraction when
required.
x
5 4
2
x
10
b) 3
4
a)
x 3
12
4
5x 1
d)
6 4
c)
3) Solve these equations involving unknowns on both sides. Write your solution as a
fraction when required.
a) 5m + 6 = 3m + 12
b) 2p + 4 = p – 3
c) 5q – 4 = 3 – q
d) 7 – 3x = 5 – 2x
4) Solve these equations involving brackets. Write your solution as a fraction when
required.
a) e + 3(e + 1) = 2e
b) 5(f + 6) = 35f
c) 3(2g + 1) + 2(g – 1) = 23
d) 5h – 3(h – 1) = 39
5) Solve these more complex equations. Write your solution as a fraction when
required.
a)
21
x
7
6
y
2x 1 x
c)
3
2
b) 30
12
4
2x 3
6
3 7
e)
x
5
15
f)
x 5 x 7
d)
Now check your work, answers on page 18.
The Hazeley Academy
Page 6
An equation that only includes variables with a power of 2 is a quadratic equation. All
quadratic equations will have graphs that form or shaped parabola.
Factorising Quadratic Expressions:
 Example 1:
a) Factorise x2 + 5x – 24
b) Factorise x2 – 49
x2 + 5x – 24
Product = -24 Sum = +5
-4 × 6
= 2 
-6 × 4
= -2 
-3 × 8
= +5 
c) Factorise 6x2 – x – 2
x2 + 0x – 49
6 × -2
6x2 – 1x – 2
Product = -49 Sum = +0
Product = -12 Sum = -1


-7 × 7 = 0  
-4 × 3
= -1 
2
Separate x: 6x – 4x + 3x – 2
Factorise pairs: 2x(3x – 2) +1(3x – 2)
Solution: (x – 3)(x + 8)
Solution: (x – 7)(x + 7)
Solution: (2x + 1)(3x – 2)
 Example 2:
Hence solve a) x2 + 5x – 24 = 0
(x – 3)(x + 8) = 0
So x – 3 = 0 or x + 8 = 0
x = 3 or
x = -8
b) x2 – 49 = 0
(x – 7)(x + 7) = 0
So x – 7 = 0 or x + 7 = 0
x = 7 or
x = -7
c) 6x2 – x – 2 = 0
(2x + 1)(3x – 2) = 0
So 2x + 1 = 0 or 3x – 2 = 0
x = -½ or
x=⅔
Completing the Square:
 Example 3:
Write in the form (x + a)2 + b = 0 and hence solve x2 + 6x – 16 = 0
Compare (x + a)2 = x2 + 2ax + a2 with the 1st two terms of the equation x2 + 6x
rd
2
Include 3 term of original
2a = 6
Remove extra a term
equation
a = 3 Substitute into required form: (x + 3)2 – 32 – 16 = 0
Simplify:
(x + 3)2 – 25 = 0 (Completed Square Form)
Rearrange to solve:
(x + 3)2 = 25
(x + 3) = ±5
x = -3 ± 5
x = 2 or x = -8
The Hazeley Academy
Page 7
Using the Quadratic Formula:
REMEMBER
A quadratic equation
MUST equal ZERO
before you solve it.
 Example 4:
Use the quadratic formula to solve 3x2 – 7x + 2 = 0
Substitute values from ax2 + bx + c = 0 into x =
x=
7
72 4 3 2 7
=
2 3
b
b2 4ac
2a
49 24 7 25 7 5 1
=
=
= or 2
6
3
6
6
Exercise 2:
1) Factorise the following quadratic expressions:
a) x2 – 5x + 4
b) x2 + 7x + 10
c) x2 – 2x – 15
d) 5y + 6 + y2
e) 10 – 11x + x2
f) y2 – y – 12
g) x2 – 121
h) 4x2 – 81
2) Factorise completely:
a) 4x2 – 5x – 6
b) 2x2 – 5x + 3
c) 15x2 + 31x + 10
d) 6x2 – x – 1
3) Solve the following equations by first factorising them:
a) y2 – 3y = 0
b) y2 – 3y – 4 = 0
c) 6x2 – 11x – 7 = 0
d) x2 + 2x = 35
e) 3y2 + 5y = 2
4) Solve the following equations by first completing the square, give answers to 1 dp:
a) x2 + 10x + 3 = 0
b) x2 – 8x – 2 = 0
c) x2 + 3x – 4 = 0
5) Solve the following equations using the quadratic formula, give answers to 1 dp:
a) x2 – 4x + 1 = 0
b) x2 – 5x + 1 = 0
c) 4x2 + 9x + 1 = 0
d) 4x2 – 2x = 3
e) 1 = x2 – 8x + 2
Now check your work, answers on page 18.
The Hazeley Academy
Page 8
Problems that involve more than one equation and more than one unknown that are to
be solved at the same time with the same values are known as “Simultaneous Equations”.
Linear Simultaneous Equations, Elimination Method:
 Example 1:
Solve the following pair of equations simultaneously:
 3x + 2y = 29 → × 4
 4x – 3y = -1 → × 3
Unknowns have the
SAME SIGN → SUBTRACT
DIFFERENT SIGNS → ADD
 12x + 8y = 116 –
 12x – 9y = -3
The aim is to get the same
amount of one unknown by
multiplying one or both
equations by a constant
17y = 119
y=7
Substitute into  to find x: 3x + 14 = 29
x=5
Check results in  4 × 5 – 3 × 7 = 20 – 21 = -1 
Linear Simultaneous Equations, Substitution Method:
 Example 2:
Solve the following pair of equations simultaneously:
 3x + y = 22
 5x + 2y = 40
 y = 22 – 3x →  5x + 2(22 – 3x) = 40
5x + 44 – 6x = 40
x=4
The aim is to get one of the
unknowns on its own and
then substitute it into the
other equation
Substitute into  to find y: y = 22 – 3 × 4 = 10
The Hazeley Academy
Page 9
Quadratic Simultaneous Equations:
 Example 3:
Solve the following equations simultaneously:
 y = 2x + 2
 y = x2 – 1
See Factorising
and Solving.
Page 4
2x + 2 = x2 – 1
0 = x2 – 2x – 3
0 = (x – 3)(x + 1)
x = 3 or x = -1
The aim is to substitute the
linear equation into the
quadratic equation.
Substitute into  to find y:
x = 3, y = 8
x = -1, y = 0
Exercise 3:
1) Solve the following simultaneous equations using the elimination method:
a) 2x + 5y = 24
4x + 3y = 20
b) 2a + 3b = 9
4a + b = 13
c) x – 2y = -4
3x + y = 9
d) 5x – 7y = 27
3x – 4y = 16
2) Solve the following simultaneous equations using the substitution method:
a) x + 3y = 5
2x + y = 5
b) x – y = 2
3x + y = 10
c) a + 4b = 6
8b – a = -3
d) 2x = 4 + z
6x – 5z = 18
3) Solve the following linear and quadratic simultaneous equations:
a) y = x2 – 2x
y=x+4
b) y = 7x – 8
y = x2 – x + 7
c) y = x2 – 3x + 7
5x – y = 8
d) y = 9x – 4
y = 2x2
Now check your work, answers on page 18.
The Hazeley Academy
Page 10
Equations involving greater than >, less than <, greater than or equal to or less than or
equal to are called “Inequalities”. Inequalities have a set of solutions.
Linear Inequalities:
REMEMBER
Change the direction of
the inequality when you
× or ÷ by a negative
number.
 Example 1:
Solve the following inequalities:
a) 3x + 7 < 31
3x < 24
x <8
See Solving Linear
Equations. Page 2
b) 8 – 5x 68
-5x 60
x -12
Quadratic Inequalities:
 Example 2:
x2 + 8x + 15 < 0
(x + 3)(x + 5) < 0 so x2 + 8x + 15 = 0 when x = -3 or x = -5
Now consider the graph of y = x2 + 8x + 15
Solve the following inequality:
See Solving
Quadratic equations.
Page 4
So the graph has y < 0
when x is between -5 and -3
-5
-3
Solution: -5 < x < -3
 Example 3:
Solve the following inequality:
-1
5
x2 – 4x – 5 > 0
(x – 5)(x + 1) > 0
so x2 – 4x – 5 = 0 when x = -1 or x = 5
Now consider the graph of y = x2 – 4x – 5
So the graph has y > 0
when x is less than -1 or greater than 5
Solution: x < -1 or x > 5
(Ensure you write the solutions separately in this case)
The Hazeley Academy
Page 11
Exercise 4:
1) Solve the following inequalities:
a)
b)
c)
d)
e)
f)
g)
h)
x – 3 > 10
x+1<0
2x + 1 6
5x < x + 1
3x + 1 < 2x + 5
2(x + 1) x – 7
3(x – 1) < 2(1 – x)
4 – 2x 2
2) Solve the following quadratic inequalities:
a)
b)
c)
d)
e)
f)
x2 + 7x + 12 < 0
x2 – 8x – 9 > 0
x2 – 144 0
12x2 – 16x + 5 < 0
4x2 – 3x – 10 > 0
x2 – 14x + 49 0
Now check your work, answers on page 19.
The Hazeley Academy
Page 12
Any fraction that involves an unknown is an “Algebraic Fraction”. You may be asked to
simplify expressions or solve equations involving algebraic fractions.
Simplifying Algebraic Fractions, Adding & Subtracting:
REMEMBER
Before you can add or
subtract fractions you
must have the same
denominators.
 Example 1:
Simplify the following algebraic fractions:
a)
x x
6 8
4 x 3x
24 24
b)
7x
24
REMEMBER
Only add the
numerators.
3
5
x 2 x 3
3x 3
x 2 x 3
3x 9
x 2 x 3
8x 1
x 2 x 3
c)
5x 2
x 2 x 3
5x 10
x 2 x 3
12
2
x 5 x 1
12 x 1
x 5 x 1
12 x 12
x 5 x 1
12 x 12 2 x
x 5 x 1
10 x 2
x 5 x 1
2x
x 1
2x
x 1
10
5
x 5
10
x 5
Simplifying Algebraic Fractions, Multiplying & Dividing:
 Example 2:
Simplify the following algebraic fractions:
REMEMBER
Always factorise
before you multiply
or divide.
x
x 2 3x
a) 2
x 5x 6 x 1
x
x x 3
x 2 x 3
x 1
x2
x 2 x 1
xx 2
x 2 5x
b) 2
x 8x 9
x 1
xx 2
x 1
x 9 x 1 xx 5
x 2
x 9 x 5
REMEMBER
Cancel whole brackets
from any numerator
& denominator.
REMEMBER
Never Divide: Flip the
second fraction over
& multiply.
The Hazeley Academy
Page 13
Solving Equations involving Algebraic Fractions:
 Example 3:
Solve the following equations:
x
x2 6x 8
a) x 2 5x 4
x 2
x
x 2 x 4
x 1 x 4
x 2
x
x 1
x
When you have a
x
single fraction on one
2x
side multiply up to
remove all fractions.
x
3
3
b)
7
x 1 x 2
5x 1
7
x 1 x 2
8 x 11
7
x 1 x 2
8 x 11 7 x 1 x 2
3x 2
x 1 x 2
3
3
3x 1
3x 3
3
3
1 5
2
5
8 x 11 7 x 2 21 x 14
7 x 2 13 x 3 0 (using quadratic
x = -0.27 or x = -1.59
formula)
Exercise 5:
1) Simplify the following algebraic fractions:
7a2b
35ab2
5ab
b)
15a 10 a2
18 a 3ab
c)
6a2
4ab 8a2
2ab
x 2 2x
e) 2
x 3x
x 2 3x
f) 2
x 2x 3
a)
x2
x2
x2
h) 2
x
x2
d)
g)
i)
6x
x
4x
5x
7x
x2
5
2
21
14
10
4
2) Write the following expressions as a single fraction:
x 1
3
x 3
b)
3
x 2
4
x 2
5
a)
3
4x
3
d)
4x
c)
2
5x
2
3x
e)
f)
3
x 2
2
x 3
4
x
5
x 1
3) Write the following expressions as a single fraction:
a)
x 2 3x 40
x 2 2x
x 2 5x 6
x 2 25
b)
x2
x
2
x
2x
x 2
4) Solve the following equations giving answers to two decimal places where necessary:
a)
2
x
2
x 1
3
b)
3
3
x 1
x 1
4
c)
2
4
x 2
x 1
3
Now check your work, answers on page 19.
The Hazeley Academy
Page 14
An “Index” is also known as a power. The plural of index is “Indices”.
Index
2
3
Base
Simplifying Indices:
You can only simplify indices when the bases are the same.
 The Rules:
32 × 35 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37 = 32+5
32
ya × yb = y a+b
35
56 ÷ 54 = 5 × 5 × 5 × 5 × 5 × 5 = 52 = 52 = 56 – 4
5×5×5×51
1
ya ÷ yb = y a – b
(32)5 = 32 × 32 × 32 × 32 × 32 = 32+2+2+2+2 = 310 = 32×5
(ya)b = y a×b
40 = 1
y0 = 1
9½ = 9 = 3, 8⅔ =
5– 3 =
3
8
2
= 22 = 4
y
n
= (m y)n
/m
y– n =
1
1
=
3
5
125
1
yn
 Example 1:
Simplify the following expressions: a) c4 × c7
= c4+7 = c11
b) p4 ÷ p– 6
= p4 – (-6) = p10
c) (r4)6
= r4×6 = r24
 Example 2:
Evaluate the following: a) 4– ½
1 1
=
=
4 2
b) (6½)3 × 6½
3
2
=6 ×6
½
c) 49
=
3
d) 2.25 – ½
2
49
3
3
= 6 2 +½
= 62 = 36
= 73 = 343
9
4
4
9
1
2
2
3
4
9
1
2
The Hazeley Academy
Page 15
Exercise 6:
1) Simplify the following expressions:
a) x3 × x4
e) (k½)6
b) m3 ÷ m2
f) (x– 3)– 2
c) y½ × y½
g) 2x2 × 3x2
d) w– 7 × w2
h) (2x)2 × (3x)3
2) Evaluate the following quantities:
a) 100
3
2
f)
g)
1
11
9
h)
1
8
i)
9
25
b) (5– 4) ½
¼
¼
c) 81 ÷ 16
d) 0·01½
1
3
3
8
e) 0·04½
3
1
2
2
1
2
Now check your work, answers on page 20.
The Hazeley Academy
Page 16
Trigonometry is the use of Sine, Cosine and Tangent to calculate sides and angles in
triangles.
Right-angled Triangles:
Sin(a) = opposite
hypotenuse
hypotenuse
Cos(a) = adjacent
hypotenuse
opposite the right angle
opposite
opposite the angle a
Tan(a) = opposite
adjacent
a
adjacent
next to the angle a
 Example 1:
Find the sides and angles indicated: a)
b)
5
o
x
58
6
o
c)
x
6
35
x
o
9
a)
Opp x
sin(35)
Hyp
5
35o
Adj
b)
x
5
5 sin(35) x
x 2 87
cos(58)
Hyp
Adj 6 58o x
Opp
6
x
6
cos(58)
x 11 32
x
c)
6
9
tan( x)
Opp
6
Hyp
xo
x
9 Adj
tan
1
6
9
x 33 7o
Non right-angled Triangles:
Label angles as capital letters, sides as lower case. The sides are labelled opposite their
corresponding angles.
Sine Rule:
a
sin A
b
sinB
Good Habits:
Write a list of the
angles and sides you
know.
c
sinC
Cosine Rule: a2 b2 c 2 2bc cos A
b2
c 2 a2 2ac cos B
c2
a2 b2 2ab cos C
Which Rule?
If you have a pair use
the Sine Rule.
c
A
b
B
C
a
The Hazeley Academy
Page 17
 Example 2:
Find the sides and angles indicated:
a) A
b)
A
5c
o
85
o
b
7
B
o
46
x a
C
o
b
B
A = 85 a = x
B=
b=
C = 46 c = 5
x
5
sin85 sin46
5 sin85
x
sin46
x 6 92
x
7
2
9
x2
82 8
x
82 8
2
x
o
B
x
A=
B=x
C=
o
a=4
b = 10
c=9
No pair so cosine rule:
12
sin x
sin x
12
12 sin76
sin x
15
1
x sin 0 78
x 50 9
10 2
42 92 2 4 9 cos x
100 16 81 72 cos x
100 16 81
cos x
72
1
cos 1
x 92 4 o
24
Exercise 7:
1) Find the missing sides and angles:
a)
b)
c)
7
d)
9
o
x
4
o
42
5
o
o
27
39
63
x
x
e)
f)
x
4
g)
17
o
o
51
x
h)
x
12
8
x
x
o
8
7
o
7
2) Find the missing sides and angles:
a)
b)
o
51
x
x
o
63
c)
7
x
o
95
8
o
44
o
d)
e)
o
115
4
x
6
52
5
9
o
x
8
o
f)
x
o
8
7
11
B
a
4
C
15
sin76
sin76
15
9 10
c
10
a
Pair so sine rule:
2 7 9 cos 68
9
b
A = 76 a = 15
B = x b = 12
C=
c=
No pair so cosine rule:
2
c
15
C
A = 68 a = x
B=
b=7
C=
c= 9
Pair so sine rule:
76
12
x a
C
A
d)
A
9c
68
b
c)
o
37
9
Now check your work, answers on page 20.
The Hazeley Academy
Page 18
Exercise 1:
1)
2)
a)
b)
c)
d)
e)
f)
q=7
p=7
y=⅛
n = 12
b=6
c=4
3)
a) x = 18
b) x = 28
c) x = 45
d) x =
4)
a)
b)
c)
d)
3
10
m=3
p = -7
q = 1⅙
x=2
5)
a)
b)
c)
d)
e = -1½
f=1
g = 2¾
h = 18
a)
b)
c)
d)
e)
f)
x=3
y=⅕
x=2
x=3
x=⅗
x = -4
Exercise 2:
1)
3)
a)
b)
c)
d)
e)
f)
g)
h)
(x – 1)(x – 4)
(x + 5)(x + 2)
(x – 5)(x + 3)
(x + 2)(x + 3)
(x – 1)(x – 10)
(x + 3)(x – 4)
(x – 11)(x + 11)
(2x – 9)(2x + 9)
a)
b)
c)
d)
(4x + 3)(x – 2)
(2x – 1)(x – 1)
(5x + 2)(3x + 5)
(3x + 1)(2x – 1)
4)
a) y(y – 3) = 0
y = 0 or y = 3
b) (y – 4) (y + 1) = 0
y = 4 or y = -1
c) (2x + 1)(3x – 7) = 0
x = -½ or x = 2⅓
d) (x + 7)(x – 5) = 0
x = -7 or x = 5
e) (3y – 1) (y + 2) = 0
y = ⅓ or y = -2
2)
a) (x + 5)2 – 22 = 0
x = -0∙3 or x = -9∙7
b) (x – 4)2 – 18 = 0
x = -0∙2 or x = 8∙2
c) (x +
3 2 25
) – =0
2
4
x = 1 or x = -4
5)
a)
b)
c)
d)
e)
x = 3∙7 or x = 0∙3
x = 4∙8 or x = 0∙2
x = -0∙1 or x = -2∙1
x = 1∙2 or x = -0∙7
x = 7∙9 or x = 0∙1
Exercise 3:
1)
2)
a)
b)
c)
d)
x = 2, y = 4
a = 3, b = 1
x = 2, y = 3
x = 4, y = -1
3)
a)
b)
c)
d)
x = 2, y = 1
x = 3, y = 1
a = 5, b = ¼
x = ½, z = -3
a) x = 4, y = 8
x = -1, y = 3
b) x = 3, y = 13
x = 5, y = 27
c) x = 3, y = 7
x = 5, y = 17
d) x = ½, y = ½
x = 4, y = 32
The Hazeley Academy
Page 19
Exercise 4:
1)
2)
a)
b)
c)
d)
e)
f)
g)
h)
x > 13
x < -1
x ≤ 2∙5
x<¼
x<4
x ≥ -9
x<1
x≥1
a)
b)
c)
d)
e)
f)
-4 < x < -3
x < -1, x > 9
-12 ≤ x ≤ 12
½<x<⅚
x < -1¼, x > 2
x=7
Exercise 5:
1)
2)
a)
b)
c)
d)
e)
f)
g)
h)
i)
a
5b
b
3 2a
6 b
2a
2 b 2a
b
x 2
x 3
x
x 1
x 5
x 2
x 3
x 2
x 5
x 2
a)
b)
c)
d)
e)
f)
7x 2
12
2x 9
15
23
20 x
1
12 x
7x 8
xx 2
3x 17
x 3 x 1
3)
a)
x 8 x 3
xx 5
b) 1
4)
a) x = -⅔ or x = 1
b) x = -½ or x = 2
c) x = 0 or x = 3
The Hazeley Academy
Page 20
Exercise 6:
1)
2)
a)
b)
c)
d)
e)
f)
g)
h)
7
x
m
y
w-5
k3
x6
6x4
108x5
a) 1000
b)
1
25
c) 1∙5
1
10
1
e)
5
d)
f) 1∙5
g)
3
10
h) 64
i) 1
2
3
Exercise 7:
1)
2)
a)
b)
c)
d)
e)
f)
g)
h)
x = 3∙18
x = 4∙09
x = 3∙60
x = 6∙17
x = 60∙3o
x = 28∙1o
x = 10∙29
x = 54∙3o
a)
b)
c)
d)
e)
f)
x = 10∙32
x = 4∙86
x = 11∙57
x = 37∙2o
x = 94∙1o
x =42∙6o