8
Carshalton Boys Sports College
Department of Mathematics
Preparing students for the challenge of the AS course
Induction Booklet Summer 2016
1
Core Mathematics Transition
Core Mathematics Transition
Your AS level in mathematics will consist of two Core Mathematics modules and one
Applied Mathematics module.
AS level mathematics uses many of the skills you developed at GCSE. The big difference
is that you will be expected to recognise where you use these skills and apply them
quickly and efficiently.
In order to get off to a good start you need to be prepared. This booklet will help you
get ready for AS Core Mathematics. This work is compulsory for all students.
A set of “Mathswatch” references is provided to help you with this
work: https://www.mathswatchvle.com
Centre ID: carshaltonboys
Please use your personal login details for username and password.
Websites that can help you with AS Level topics
http://www.examsolutions.net/
http://www.mymaths.co.uk/
User Name: cbsc
Password: height
https://www.khanacademy.org/
http://www.physicsandmathstutor.com/
www.markit.education
2
CONTENTS
Section 1
Linear Equations
Section 2
Simultaneous Equations
Section 3
Factorising Quadratics
Section 4
Rearranging Formulae
Section 5
Solving Quadratic Equations
Section 6
Indices
Section 7
Surds
Section 8
Straight Line Graphs
Solutions to the exercises
Make sure you show all the relevant working out as you are working through the
topics. There are answers at the end of the Booklet to check and correct your
answers. You will be assessed on this A-A* work at the start of the term and you
need to gain at least 60% to pass.
3
Section 1: Linear Equations Mathswatch clips: 105a & 105b (old)
Clip: 135a & b (New 2015 Specs)
Exercise A: Solve these equations
1)
2x + 5 = 19
2) 5x – 2 = 13
3) 11 – 4x = 5
4)
5 – 7x = -9
5) 11 + 3x = 8 – 2x
6) 7x + 2 = 4x – 5
Exercise B: Solve the following equations.
1)
5(2x – 4) = 4
2)
4(2 – x) = 3(x – 9)
3)
8 – (x + 3) = 4
4)
14 – 3(2x + 3) = 2
Equations containing Fractions
Mathswatch clip: 163b (old)
Clip: 210 b (New 2015 Specs)
Exercise C: Solve these equations
1)
1
( x  3)  5
2
2)
2x
x
1   4
3
3
3)
y
y
3 5
4
3
4)
7x 1
 13  x
2
5)
y 1 y 1 2y  5


2
3
6
Section 2: Simultaneous Equations Mathswatch clips: 142a & b (old)
Clip: 162 (New 2015 Specs)
Solve the pairs of simultaneous equations in the following questions:
1)
x + 2y = 7
3x + 2y = 9
2)
x + 3y = 0
3x + 2y = -7
3)
3x – 2y = 4
2x + 3y = -6
4)
9x – 2y = 25
4x – 5y = 7
4
Section 3: Factorising quadratics Mathswatch clips: 140a & b, clip
141 (old)
Clip: 157 (New 2015 Specs)
Factorise these Quadratics
1)
x2  x  6
2)
x2  6 x  16
3)
2 x2  5x  2
4)
2 x 2  3x
5)
3x 2  5 x  2
6)
4 x 2  25
7)
(Factorise by taking out a common factor)
x² - 3xy – xy +3y²
Section 4: Rearranging formulae Mathswatch clip: 107 & 164
Clip: 137 (New 2015 Specs)
Exercise A: Make x the subject of each of these formulae:
1)
y = 7x – 1
3)
4y 
x
2
3
2)
y
x5
4
4)
y
4(3x  5)
9
5
Rearranging equations involving squares and square roots
Exercise B: Make t the subject of each of the following
wt
32r
1)
P
3)
1
V   t2h
3
2)
P
wt 2
32r
4)
P
2t
g
More difficult examples
Exercise C: Make x the subject of these formulae:
1)
ax  3  bx  c
2)
3( x  a)  k ( x  2)
3)
y
2x  3
5x  2
4)
x
x
 1
a
b
Section 5: Solving Quadratic Equations Mathswatch clips: 140 a & b,
161a
Clip: 157 & 191 (New 2015 Specs)
For ax 2  bx  c  0 ,
formula
x
 b  b 2  4ac
2a
this is known as the quadratic
1) Use factorisation to solve the following equations:
a)
x2 + 3x + 2 = 0
b)
x2 – 3x – 4 = 0
2) Find the roots of the following equations:
a)
x2 + 3x = 0
c)
b)
c) x2 = 15 – 2x
x2 – 4x = 0
4 – x2 = 0
3) Solve the following equations either by factorising or by using the formula:
a)
6x2 - 5x – 4 = 0
b)
8x2 – 24x + 10 = 0
6
4) Use the formula to solve the following equations to 3 significant figures. Some of the equations
can’t be solved.
a)
x2 +7x +9 = 0
b)
6 + 3x = 8x2
c)
4x2 – x – 7 = 0
d)
x2 – 3x + 18 = 0
Section 6: Indices Mathswatch clip 156
Clip: 154 (New 2015 Specs)
Basic rules of indices
a m  a n  a m n
am
 a m  a n  a mn
n
a
a0  1
a 1 
1
a
a 
1
2
m n
a  a
a
m
n
 n am 
 a mn
 a
n
m
Exercise A: Simplify the following:
1)
b  5b5 =
2)
3c 2  2c5 =
3)
b2 c  bc3 =
4)
2n6  (6n2 ) =
5)
8n8  2n3 =
6)
d 11  d 9 =
7)
a 
8)
 d 
3 2
(Remember that b  b1 )
=
4 3
=
7
More complex powers
Exercise B: Find the value of:
1)
41/ 2
2)
271/ 3
3)
 19 
4)
52
5)
180
6)
7 1
7)
272 / 3
8)
2
 
3
9)
82 / 3
10)
 0.04 
1/ 2
2
1/ 2
 8 
11)  
 27 
2/3
 1
12)  
 16 
3 / 2
Simplify each of the following:
13)
2a1/ 2  3a5 / 2
14)
x3  x 2
15)
x y 
2
4 1/ 2
8
Section 7: Surds Mathswatch Clips 157 and 158
Clip: 207 a, b & c (New 2015 Specs)
Basic Rules
a
a  b  ab
b

a
b
Simplify the surds
1) 12
2) 125
3) 48
4) 72
5) 27
Expand and simplify
1)
2)

6
2 3 5

2 8

3) 4( 5  3)
4) (2  3 )(1  3)
5) (3  5 )(3  2 5)
6) (2  5) (2  3 )
7) (1  2 )(1  3 )
8) (8  2 )(8  2 )
9) ( 3  5 )( 3  5 )
Rewrite the following expressions with rational denominators
1)
2)
3)
3
5
4
8
9
48
2 1
2
3 1
5)
5
4
6) 
3 2
4)
7)
8)
9)
10)
1
3 1
4
6 2
7
7 2
3
5 1
11)
3 1
5
12)
5 1
5 3
9
Section 8: Straight line graphs Mathswatch clips 113, 114, 128 & 143)
Clip: 96, 97, 100 (New 2015 Specs)
Linear functions can be written in the form y = mx + c, where m and c are constants.
A linear function is represented graphically by a straight line, m is the gradients and c is the yintercept of the graph.
Example: Draw the graph of y = 2x + 1
Solution:
Step 1: Make a table of values
Step 2: Use your table to draw the straight line graph
Here are some examples of linear functions not all of them in the form y = mx + c. You need to be
confident into rearranging the functions making y the subject in order to identify the gradient and yintercept.
1) y = 2x + 3
2) 3x - 2y + 1 = 0
3
1
so y  x 
2
2
3
gradient=
2
1
y-intercept=
2
gradient= 2
y-intercept= 3
1)
2)
3) 4y - x = 3
1
3
so y  x 
4
4
1
gradient=
4
3
y-intercept=
4
3)
To find the y-axis crossing, substitute x = 0 into the linear equation and solve for y.
To find the x-axis crossing, substitute y = 0 into the linear equation and solve for x.
10
Example: Rewrite the equation 3y - 2x = 5 into the form y = mx + c, find the gradient and the yintercept
Solution:
Step 1: Add 2x to both sides (so that the x term is positive):
3y = 5 + 2x
Step 2: Divide by 3 both sides:
y
Step 3: Identify the gradient and y-intercept
gradient=
2
3
2
5
x
3
3
y-intercept=
5
3
Example: Find the gradient of the line which passes through the points A (1, 4) and B (-3, 2)
Solution:
Step 1: Use the x and y values of A ( x1 , y1 ) and B ( x2 , y 2 )
y  y1
Step 2: find the gradient m  2
x 2  x1
m
24 2 1


 3 1  4 2
Finally, you need to be able to find the equation of a line from a graph.
Example: Find the equation of the straight line which passes through the point (3, 2) and has a
gradient 5
11
Exercise A: Plot the graph of each function taking the given values:
a) y= x - 3 ( x = -2 to 4)
b) y=- x + 4 ( x = -2 to 5)
c) y = 2x – 3 ( x = -1 to 5)
d) y= -3x + 5 ( x = -2 to 3)
Exercise B: Rewrite the equations below into the form y = mx + c, find the gradient and the yintercept:
a) 3x – 2y – 2 = 0
b) x + 2y – 8 =0
c) 5 = 4x – 2y
Exercise C: Work out the gradient between the sets of coordinates:
a)
b)
c)
d)
e)
f)
A ( 0, 2) and B( 3, 6)
A ( 1, 0) and B( 3, -2)
A ( 1, -3) and B( 2, -4)
A ( -4, 2) and B( 3, 5)
A ( 1, 0.5) and B( 5, -2)
A ( -7, -3) and B( -2, -6)
Exercise D: Find the equation of the line:
Good Luck!
- ----------------------------
12
Solutions
Section 1 Linear Equations
Ex A
1) 7 2) 3 3) 1½ 4) 2 5) -3/5
Ex B
1) 2.4 2) 5 3) 1 4) ½
Ex C
1) 7 2) 15
3) 24/7 4) 3 5) 2
Section 2 Simultaneous Equations
1) x = 1, y = 3
2) x = -3, y = 1
Section 3
6) -7/3
3) x = 0, y = -2
4) x = 3, y = 1
Factorising Quadratics
1) (x – 3) (x + 2)
2) (x + 8) (x – 2)
5) (3x -1) (x + 2)
6) (2x + 5) (2x – 5)
3) (2x + 1) (x + 2)
4) x (2x – 3)
7) (x – 3) (x – y)
Section 4 Rearranging Formulae
Ex A
y 1
7
1) x 
2) x  4 y  5
3) x  3(4 y  2)
4) x 
9 y  20
12
Ex B
1) t 
32rP
w
Ex C
1) x 
2) t  
c3
a b
32rP
w
3a  2k
k 3
2) x 
3V
h
4) t 
P2 g
2
2y  3
5y  2
4) x 
ab
ba
3) t  
3) x 
Section 5 Solving Quadratic Equations
1) a) -1, -2
b) -1, 4
c) -5, 3
2) a) 0, -3
b) 0, 4
c) 2, -2
3) a) -1/2, 4/3
d) no solutions
b) 0.5, 2.5
4) a) -5.30, -1.70
e) no solutions
f) no solutions
b) 1.07, -0.699
c) -1.20, 1.45
Section 6 Indices
Ex A
1) 5b6
2) 6c7
10) 4/9
4) -12n8
5) 4n5
6) d2
7) a6
8) -d12
9) ¼
10) 1/5
11) 64
Ex B
1) 2
2) 3
10) 0.2
3) b3c4
3) 1/3
11) 4/9
12) 64
4) 1/25
5) 1
13) 6a3
6) 1/7
7) 9
14) x
15) xy2
8) 9/4
9) ¼
13
Section 7 Surds
Ex A
1) 2 3
2) 5 5
3) 4 3
4) 6 2
5) 3 3
Ex B
1) 3 2  10
2)
12  48  2 3  4 3
3) 4 5  12
4) 2  2 3  3  3  5  3 3
5) 9  6 5  3 5  10  19  9 5
6) 4  2 3  2 5  15
7) 1  3  2  6
8) 64  8 2  8 2  2  62
9) 3  15  15  5  8  2 15
Ex C
14
Section 8
Ex A
Straight line graphs
Graphs to be drawn
Ex B : a) y 
Ex C :
3
1
5
x  1 b) y   x  4 c ) y  2 x 
2
2
2
4
3
2
b) gradient 
 1
2
1
c) gradient 
 1
1
3
d ) gradient 
7
 2 .5  5
e) gradient 

4
8
3
f ) gradient 
5
a ) gradient 
Ex D:
---------------------------------------------
15