Mathematics

1
Mathematics Intermediate Tier
Algebra
GCSE Revision
2
Intermediate Tier - Algebra revision
Contents Collecting like terms Multiplying
terms together Indices Expanding single
brackets Expanding double brackets Substitution So
lving equations Finding nth term of a
sequence Simultaneous equations Inequalities
Factorising common factors Factorising
quadratics Other algebraic manipulations Solving
quadratic equations Rearranging formulae Curved
graphs Graphs of y mx c Graphing
inequalities Graphing simultaneous
equations Solving other equations graphically
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Collecting like terms
You can only add or subtract terms in an
expression which have the same combination of
letters in them
e.g. 1. 3a 4c 5a 2ac 8a 4c 2ac
e.g. 2. 3xy 5yx 2xy yx 3x2 7xy
3x2
Simplify each of these expressions
5a 2b
x2 2x 2
cannot be simplified
13x2 4x
4a2 3a 6
ab
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Multiplying terms together
Remember your negative numbers rules for
multiplying and dividing Signs same ? ve
answer Signs different ? -ve answer
e.g. 1. 2 x 4a 8a
e.g. 2. - 3b x 5c - 15bc
e.g. 3. (5p)2 5p x 5p 25p2
Simplify each of these expressions
-6a
-56af
-12d
-36b2
30ac
4st
12s2
35y
3a2
-6a2
49a2
81k2
5
(F2)4
Indices
c0
a2 x a3
t2 ? t2
2e7 x 3ef2
4xy3 ? 2xy
x7 ? x4
b1
5p5qr x 6p2q6r
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Expanding single brackets
Remember to multiply all the terms inside the
bracket by the term immediately in front of the
bracket
e.g. 4(2a 3)
If there is no term in front of the bracket,
multiply by 1 or -1
Expand these brackets and simplify wherever
possible
3a - 12
8r2 12r
12c 30
-4a - 2
-2d - 2g
-2t - 2
cd 4c
16a 32
-10a 15
a2 - 6a
6p2 - 6p 5
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Expanding double brackets
Split the double brackets into 2 single brackets
and then expand each bracket and simplify
(3a 4)(2a 5)
3a lots of 2a 5 and 4 lots of 2a 5
3a(2a 5) 4(2a 5)
6a2 15a 8a 20
If a single bracket is squared (a 5)2 change
it into double brackets (a 5)(a 5)
6a2 7a 20
Expand these brackets and simplify
c2 8c 12
c2 14c 49
6a2 5a 4
16g2 8g 1
15a2 a 28
7p2 11p 6
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Substitution
If a 5 , b 6 and c 2 find the value
of
c2
3a
ac
4b2
144
15
4
10
(3a)2
ab 2c
c(b a)
2
225
26
a2 3b
4bc a
(5b3 ac)2
9.6
1 144 900
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Now find the value of each of these expressions
if a - 8 , b 3.7 and c 2/3
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Solving equations
Solve the following equation to find the value of
x
4x 17 7x 1
? Take 4x from both sides
17 7x 4x 1
17 3x 1
? Add 1 to both sides
17 1 3x
18 3x
? Divide both sides by 3
18 x 3
Now solve these 1. 2x 5 17 2. 5 x
2 3. 3x 7 x 15 4. 4(x 3) 20
6 x
x 6
Find x to 1 d.p. if x2 3x 200
Some equations cannot be solved in this way and
Trial and Improvement methods are required
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Solving equations from angle problems
Rule involved Angles in a quad 3600
4y 2y y 150 360 7y 150
360 7y 360 150 7y
210 y 210/7 y 300
Angles are 300,600,1200,1500
4v 5 2v 39 4v - 2v 5 39
2v 5 39 2v 39 - 5 2v
34 v 34/2 v 170
Rule involved Z angles are equal
Check (4 x 17) 5 73 , (2 x 17) 39 73
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Finding nth term of a sequence
This sequence is the 2 times table shifted a
little
2 4 6 8 10
12
5 , 7 , 9 , 11 , 13 , 15 ,.
Each term is found by the position number times
2 then add another 3. So the rule for the
sequence is nth term 2n 3
100th term 2 x 100 3 203
Find the rules of these sequences
And these sequences
2n 1 2n 4 5n 2 6n 20 7n
n2 n2 2 -2n 22 -3n 43 20n - 14

• 1, 3, 5, 7, 9,
• 6, 8, 10, 12,.
• 3, 8, 13, 18,
• 20,26,32,38,
• 7, 14, 21,28,

• 1, 4, 9, 16, 25,
• 3, 6,11,18,27.
• 20, 18, 16, 14,
• 40,37,34,31,
• 6, 26,46,66,

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Simultaneous equations
4a 3b 17 6a ? 2b 6
8a 6b 34 18a ? 6b 18

26a 52
a 52 26
a 2
8 3b 17
3b 17 - 8
So the solutions are a 2 and b 3
3b 9
b 3
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Inequalities can be solved in exactly the same
way as equations
Inequalities
14 ? 2x 8

• Add 8 to
• both sides

The difference is that inequalities can be given
as a range of results Or on a scale
14 8 ? 2x
22 ? 2x

• Divide both
• sides by 2

22 ? x 2
Here x can be equal to 11, 12, 13, 14, 15,
11 ? x

• Remember to
• turn the sign
• round as well

x ? 11
Find the range of solutions for these
inequalities
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Factorising common factors
Factorising is basically the reverse of expanding
brackets. Instead of removing brackets you are
putting them in and placing all the common
factors in front.
5x2 10xy 5x(x 2y)
Factorise the following (and check by expanding)
3(5 x) 2(a 5) a(b 5) a(a 6) 4x(2x 1)

• 15 3x
• 2a 10
• ab 5a
• a2 6a
• 8x2 4x

• 10pq 2p
• 20xy 16x
• 24ab 16a2
• ?r2 2 ?r
• 3a2 9a3

2p(5q 1) 4x(5y - 4) 8a(3b 2a) ?r(r 2) 3a2(1
3a)
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Factorising quadratics
Here the factorising is the reverse of expanding
double brackets
Factorise x2 9x - 22
x2 4x 21 (x 7)(x 3)
To help use a 2 x 2 box
x

• Factor
• pairs
• of - 22
• 1, 22
• 22, 1
• 2, 11
• 11, 2

x2
x
- 22
Factorise the following
(x 3)(x 1) (x 2)(x 1) (x 10)(x 3) (x
2)(x 6) (x 2)(x 5)

• x2 4x 3
• x2 - 3x 2
• x2 7x - 30
• x2 - 4x - 12
• x2 7x 10

Find the pair which give - 9
-11
-11x
2x
2
Answer (x 2)(x 11)
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Other algebraic manipulations
Cancelling common factors in expressions and
equations
Factorising a difference of two squares
These two expressions are equal 6v2 9x 27z
? 2v2 3x 9z 3
Fully factorise this expression 4x2 25
Look for 2 square numbers separated by a minus.
Simply Use the square root of each and a and
a to get (2x 5)(2x 5)
As are these 2(x 3)2 ? 2(x 3) (x 3)
The first equation can be reduced to the
second 4x2 8x 16 0 ? x2 2x 4 0

• Fully factorise these
• 81x2 1
• ¼ t2
• 16y2 64

• Simplify these
• 5x2 15x 10 0
• 4(x 1)(x 2)
• x 1

• Answers
• (9x 1)(9x 1)
• (½ t)(½ t)
• 16(y2 4)

Answers (a) x2 3x 2 0 (b) 4(x 2)
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Solving quadratic equations (using factorisation)
Solve this equation
x2 5x 14 0
? Factorise first
(x 7)(x 2) 0

• Now make each bracket
• equal to zero separately

x 7 0 or x 2 0
? 2 solutions
Solve these
(x 2)(x 2) (x 5)(x 2) (x 7)(x 5) (x
1)(x 6) (x 3)(x 2)
? x -2 or x -2 ? x 5 or x 2 ? x -7
or x -5 ? x -1 or x 6 ? x -3 or x 2

• x2 4x 4
• x2 - 7x 10
• x2 12x 35
• x2 - 5x - 6
• x2 x - 6

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Rearrange the following formula so that a is the
subject
Rearranging formulae
Now rearrange these
V u at
a
V
a
V
Answers
1. a P 5 4
4. h (B e)2
2. e Ar b
5. u d(E 4v)
6. p Q st 4c
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Curved graphs
There are three specific types of curved graphs
that you may be asked to recognise.
Any curve with a number /x is this shape
Any curve starting with x2 is U shaped
If you are asked to draw an accurate curved graph
(e.g. y x2 3x 1) simply substitute x values
to find y values and thus the co-ordinates
Any curve starting with x3 is this shape
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Graphs of y mx c
In the equation y mx c m the
gradient (how far up for every one along) c
the intercept (where the line crosses the y axis)
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Graphs of y mx c
Write down the equations of these lines
Answers y x y x 2 y - x 1 y - 2x
2 y 3x 1 y 4 y - 3
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Graphing inequalities
Find the region that is not covered by these
3 regions x ? - 2 y ? x y gt 3
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Graphing simultaneous equations
Solve these simultaneous equations using a
graphical method 2y 6x 12 y 2x 1
2y 6x 12
y 2x 1
Finding co-ordinates for y 2x 1 x 0 ? y
(2x0) 1 ? y 1 ? (0, 1) x 1 ? y (2x1) 1
? y 3 ? (1, 3) x 2 ? y (2x2) 1 ? y 5 ?
(2, 5)
The co-ordinate of the point where the two
graphs cross is (1, 3). Therefore, the solutions
to the simultaneous equations are
x 1 and y 3
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Solving other equations graphically
If an equation equals 0 then its solutions lie
at the points where the graph of the equation
crosses the x-axis.
e.g. Solve this equation graphically x2 x 6
0
All you do is plot the equation y x2 x 6
and find where it crosses the x-axis (the line
y0)
There are two solutions to x2 x 6 0 x -
3 and x 2
Similarly to solve a cubic equation (e.g. x3
2x2 4 0) find where the curve y x3 2x2
4 crosses the x-axis. These points are the
solutions.