Download Chapter 2

Algebra revision pack
Summer 2013
Mathematics Department
Holiday Booklet for students intending to study A level
Mathematics or A level Mathematics and Further Mathematics.
Students who have chosen Mathematics and AS Further
Mathematics will have a different scheme to work through.
Having chosen to study Mathematics you will need to be well prepared for the challenges ahead.
The A level course is much more rigorous than IGCSE and is algebra heavy. It is possible to get an
A* in the latter and still not be ready for the start of September.
This booklet forms part of your preparation for successful progress through your AS course. It is
vitally important that you spend some time working through the questions in this booklet over the
summer break before you start your course in September. All the topics in this booklet have been
covered before at IGCSE. Each chapter contains examples to refresh your memory, as well as
exercises. You should complete this booklet by the first mathematics class you have in September.
(You will be expected to show evidence of completing the work). You will also have a test in the
second week to assess your basic mathematical skill.
There is a practice test for you to complete at the end of the booklet. If you do not pass the test in
the second week you will be provided with a programme of additional work in order to bring your
basic algebra skills up to the required standard. You will then be re-tested a few weeks later.
In the first Core Maths unit calculators are NOT allowed and throughout this booklet this
rule applies.
Mr Collins
CONTENTS
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
PDC
Removing brackets
page
Linear equations and algebraic fractions
Simultaneous equations
Factorisation
Change the subject of the formula
Solving quadratic equations
Indices
Practice Paper
2
3
7
10
12
16
17
19
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Algebra revision pack
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Chapter 1: REMOVING BRACKETS
Examples
1)
4(x + 2y)
= 4x + 8y
2)
-2(2x - 5)
= (-2)(2x) + (-2)(-5)
= -4x + 10
3)
(x + 2)(x + 3) = x(x + 3) + 2(x + 3)
or
(x +2)(x + 3) = x2 + 6 + 2x + 3x
= x2 + 5x +6
or
x
x2
3x
x
3
4)
2
2x
6
(x +2)(x + 3)
= x2 + 2x +3x + 6
= x2 + 3x +6
(x - 2)(2x + 3) = x(2x + 3) - 2(2x +3)
= 2x2 + 3x – 4x - 6
= 2x2 – x – 6
or
(x - 2)(2x + 3) = 2x2 – 6 + 3x – 4x = 2x2 – x – 6
or
2x
3
EXERCISE A
x
2x2
3x
-2
-4x
-6
(2x +3)(x - 2) = 2x2 + 3x - 4x - 6
= 2x2 - x - 6
Multiply out the following brackets and simplify.
1. 6(4x + 5)
7. (x + 1)(x + 3)
2. -3(5x - 6)
8. (t + 5)(t - 1)
3. 5a – 2(3a - 4)
9. (x + 3y)(x – 2y)
4. 5y + y(2 + y)
10. 4(x - 3)(x + 1)
5. -3x – (2x + 3)
11. (y - 1)(2y + 5)
6. 5(2x - 1) – (x - 1)
12. (3 + 2x)(8 – x)
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Two Special Cases
Perfect Square:
(x + a)2 = (x + a)(x + a) = x2 + 2ax + a2
(2x - 3)2 = (2x – 3)(2x – 3) = 4x2 – 12x + 9
EXERCISE B
Difference of two squares:
(x - a)(x + a) = x2 – a2
(x - 4)(x + 4) = x2 – 42
= x2 – 16
Multiply out
1.
(x - 1)2
4.
(x + 2)(x - 2)
2.
(2x + 5)2
5.
(3x + 1)(3x - 1)
3.
(6x - 1)2
6.
(4y - 2)(4y + 2)
Chapter 2: LINEAR EQUATIONS
Example 1: Solve the equation
64 – 2x = 25
Solution: There are various ways to solve this equation. One approach is as follows:
Step 1: Add 2x to both sides (so that the x term is positive):
64 = 2x + 25
Step 2: Subtract 25 from both sides:
39 = 2x
Step 3: Divide both sides by 2:
19.5 = x
Example 2: Solve the equation 6x + 7 = 5 – 2x.
Solution:
Step 1: Begin by adding 2x to both sides
8x + 6 = 5
Step 2: Subtract 6 from each side:
8x = -1
Step 3: Divide each side by 8:
x = -1/8
Exercise A: Solve the following equations, showing each step in your working:
1)
2x + 5 = 20
2) 5x – 1 = 13
3) 11 – 4x = 1
4)
5 – 2x = -9
5) 12 + 2x = 8 – 2x
6) 7x + 2 = 4x – 5
Example 3: Solve the equation
2(3x – 2) = 20 – 3(x + 1)
Step 1: Multiply out the brackets:
6x – 4 = 20 – 3x – 3
Step 2: Simplify the right hand side:
6x – 4 = 17 – 3x
Step 3: Add 3x to each side:
9x – 4 = 17
Step 4: Add 4:
9x = 21
Step 5: Divide by 9:
x=2
PDC
1
3
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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
y
 5  11
3
Example 4: Solve the equation
Solution:
Step 1: Multiply through by 3 (the denominator in the fraction):
y  15  33
Step 2: Subtract 7:
y = 18
Example 5: Solve the equation
1
(7 x  1)  5
3
Solution:
Step 1: Multiply by 3 (to remove the fraction)
Step 2: Subtract 1 from each side
7x = 14
Step 3: Divide by 2
Example 6: Solve the equation
7 x  1  15
x=2
x 1 x  2

2
4
5
Solution:
Step 1: Find the lowest common denominator:
Step 2: Multiply both sides by the lowest common denominator
The smallest number that both 4
and 5 divide into is 20.
20( x  1) 20( x  2)

 40
4
5
5
Step 3: Simplify the left hand side:
4
20 ( x  1) 20 ( x  2)

 40
4
5
5(x + 1) + 4(x + 2) = 40
Step 4: Multiply out the brackets:
5x + 5 + 4x + 8 = 40
Step 5: Simplify the equation:
9x + 13 = 40
Step 6: Subtract 13
9x = 27
Step 7: Divide by 9:
x=3
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x2
3  5x
 1
4
6
Solution: The lowest number that 4 and 6 go into is 12. So we multiply every term by 12:
Example 7: Solve the equation x 
12 x 
12( x  2)
12(3  5 x)
 12 
4
6
Simplify
12 x  3( x  2)  12  2(3  5 x)
Expand brackets
12 x  3x  6  12  6  10 x
15 x  6  6  10 x
Simplify
Subtract 10x
Add 6
Divide by 5
5x  6  6
5x = 12
x = 2.4
Exercise C: Solve these equations
1)
3)
5)
7)
PDC
1
( x  8)  5
2
y
y
3 5
4
3
7x 1
 13  x
3
x  1 5x  3
2x 

2
3
2)
4)
6)
8)
2x
x
1   4
3
3
x2
3 x
 2
7
14
y 1 y 1 2y  5


2
3
6
5 10
2   1
x x
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Algebraic Fractions
Rules for simplifying:





Remember that the rules and methods for algebraic fractions are exactly the same as they are for simple
numeric fractions.
When adding fractions look for a common denominator and proceed as usual (sometimes you will need
to factorise to find a common denominator).
Do not expand brackets until you have to! If you leave things factorised as much as possible, you will
find it simpler to cancel things.
When you do expand always be very careful of negative signs, e.g. 3 – 2(4 – x) expands to 3 – 8 + 2x =
2x – 5.
Always simplify your answer as much as possible.
Worked Examples
1. Simplify
x  1 2x  2

4
3
We need to ‘join’ two fractions so we need to choose a common denominator. In this case it’s
easy to see that we need it to be 12
Multiply the first by
3
4
and the second by :
3
4
x  1 2 x  2 3( x  1) 4(2 x  2)



4
3
3 4
43
Expand the fractions out:
3( x  1) 4(2 x  2) 3 x  3 8 x  8



3 4
43
12
12
We can now add them together:
3 x  3 8 x  8 3 x  3  8 x  8 11x  5
.



12
12
12
12
PDC
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Questions
Exercise D:
Simplify the following:
1.
4x  3 x  2

6
4
3.
3( x  4) 2( x  3)

8
3
4.
2.
2x  3 3  2x

4
8
2
3

x  1 2x  1
Chapter 3: SIMULTANEOUS EQUATIONS
Example: Solve
2x + 5y = 16 
3x – 4y = 1

Solution: We begin by getting the same number of x or y appearing in both equation. We can get
20y in both equations if we multiply the top equation by 4 and the bottom equation by 5:
8x + 20y = 64 
15x – 20y = 5 
As the SIGNS in front of 20y are DIFFERENT, we can eliminate the y terms from the equations by
ADDING:
23x = 69
+
i.e.
x=3
Substituting this into equation  gives:
6 + 5y = 16
5y = 10
y=2
The solution is x = 3, y = 2.
Exercise A:
Solve the pairs of simultaneous equations in the following questions:
1)
x+y=7
3x + y = 15
2)
x + 3y = 0
3x + 2y = -7
3)
3x – 2y = 4
2x + 3y = -6
4)
9x – 2y = 25
4x – 5y = 7
PDC
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Algebra revision pack
5)
Summer 2013
4a + 3b = 11
5a – 4b = 6
6)
3p + 3q = 15
2p + 5q = 14
SIMULTANEOUS EQUATIONS: ONE LINEAR AND ONE QUADRATIC
Solving simultaneous equations involves finding values for x and y that satisfy more than one equation at the
same time. In the case of one linear equation and one quadratic, that probably means that there will be two
solutions (as a straight line is likely to cross a curve twice). To solve the simultaneous equations:





Rearrange the linear equation so that either the x or the y is the subject. Choose the one that avoids
fractions if you can!
Substitute this new equation into the quadratic equation and expand any brackets (and please remember
that (a  b) 2  a 2  b 2 , but (a  b) 2  a 2  2ab  b 2 )
Rearrange the new quadratic to solve it (either through factorisation or through the quadratic formula) –
please note that you will get two answers.
Take each of your two answers in turn and substitute into the original rearranged linear equation to find
the other variable.
Write your answer out so that it is clear which value of x goes with which value of y.
Worked Examples
1.
Solve the simultaneous equations
y  2 x 2  5x  7
y  x 1
The linear equation does not need to be rearranged, but can be substituted straight away into the
quadratic equation.
x  1  2x 2  5x  7
Rearrange this:
Divide by two to simplify:
Factorise:
2x 2  4x  6  0
x 2  2x  3  0
(  x  1)
(2 )
( x  3)( x  1)  0
We now have two solutions for x (–3 and 1), and for each of them we need to find the corresponding y
solution.
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Algebra revision pack
x  3 
x 1 
Summer 2013
y  ( 3)  1  4
y  (1)  1  0
Now we must present our answers so that the pairing is clear.
The solutions are x = –3, y = –4 and x = 1, y = 0 .
2. Solve the simultaneous equations
x2  y2  4
2x  y  4
First we need to rearrange the linear equation (make y the subject as this will avoid any fractions).
y  2x  4
Rearrange:
Now substitute the value of y into our quadratic equation.
x 2  (2 x  4) 2  4
Substitute:
Expand (but leave brackets!):
x 2  (4 x 2  16 x  16)  4
x 2  4 x 2  16 x  16  4
Remove brackets (watch out for signs!):
 3 x 2  16 x  20  0
Collect ‘like’ terms and rearrange (=0):
Multiply by –1 (to get +ve x2):
3 x 2  16 x  20  0
(3 x  10)( x  2)  0
We now have two solutions for x (2 and
10
3
), and for each of them we need to find the
corresponding y solution.
PDC
x2 
y  2 ( 2)  4  0
x  103
y  2( 103 )  4 

8
3
(or 2 2 3 )
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Algebra revision pack
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Now we must present our answers so that the pairing is clear.
The solutions are x = 2, y = 0 and x =
10
3
,y=
8
3
.
Exercise B:
Solve the following simultaneous equations:
1.
3.
y  x2
y  x2
2.
x 2  y 2  29
x  2 y  12
y  x 1
y  x 2  3x  4
Chapter 4: FACTORISING
Common factors
We can factorise some expressions by taking out a common factor.
Example 1:
Factorise 12x – 30
Solution:
6 is a common factor to both 12 and 30. We can therefore factorise by taking 6
outside a bracket:
12x – 30 = 6(2x – 5)
Example 2:
Factorise 6x2 – 2xy
Solution:
2 is a common factor to both 6 and 2. Both terms also contain an x.
So we factorise by taking 2x outside a bracket.
6x2 – 2xy = 2x(3x – y)
Factorise 9x3y2 – 18x2y
Example 3:
Solution:
9 is a common factor to both 9 and 18.
The highest power of x that is present in both expressions is x2.
There is also a y present in both parts.
So we factorise by taking 9x2y outside a bracket:
9x3y2 – 18x2y = 9x2y(xy – 2)
Example 4:
Factorise 3x(2x – 1) – 4(2x – 1)
Solution:
There is a common bracket as a factor.
So we factorise by taking (2x – 1) out as a factor.
The expression factorises to (2x – 1)(3x – 4)
PDC
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Exercise A
Factorise each of the following
1)
2)
3)
4)
3x + xy
2
5)
2x3 – 6x2
6)
8a5b2 – 12a3b4
7)
5y(y – 1) + 3(y – 1)
4x – 2xy
2
2
pq – p q
2
3pq - 9q
Factorising quadratics
Example 1: Factorise x2 – 9x – 10.
Solution: We need to find two numbers that multiply to make -10 and add to make -9. These
numbers are -10 and 1.
Therefore x2 – 9x – 10 = (x – 10)(x + 1).
Example 2: Factorise 6x2 + x – 12.
Solution: We need to find two numbers that multiply to make 6 × -12 = -72 and add to make 1.
These two numbers are -8 and 9.
Therefore,
6x2 + x – 12 = 6x2 - 8x + 9x – 12
= 2x(3x – 4) + 3(3x – 4)
(the two brackets must be identical)
= (3x – 4)(2x + 3)
Difference of two squares: Factorising quadratics of the form x 2  a 2
Remember that x 2  a 2 = (x + a)(x – a).
Therefore:
x 2  9  x 2  32  ( x  3)( x  3)
16 x 2  25  (4 x) 2  52  (4 x  5)(4 x  5)
Also notice that:
2 x 2  8  2( x 2  4)  2( x  4)( x  4)
and
3 x 3  48 xy 2  3 x( x 2  16 y 2 )  3 x( x  4 y )( x  4 y )
Factorising by pairing
We can factorise expressions like 2 x 2  xy  2 x  y using the method of factorising by pairing:
2 x 2  xy  2 x  y = x(2x + y) – 1(2x + y)
(factorise front and back pairs, ensuring both
brackets are identical)
= (2x + y)(x – 1)
PDC
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Exercise B
Factorise
1)
x2  x  6
8)
10 x 2  5 x  30
2)
x 2  6 x  16
9)
4 x 2  25
3)
2x2  5x  2
10)
x 2  3 x  3 xy  3 y 2
4)
2 x 2  3x
11)
4 x 2  12 x  8
5)
3x 2  5 x  2
12)
16m 2  81n 2
6)
2 y 2  17 y  21
13)
4 y 3  9a 2 y
7)
7 y 2  10 y  3
14)
8( x  1) 2  2( x  1)  10
Chapter 5: CHANGING THE SUBJECT OF A FORMULA
Example 1:
Make x the subject of the formula y = 4x + 3.
Solution:
Subtract 3 from both sides:
Divide both sides by 4;
So x 
y = 4x + 3
y – 3 = 4x
y 3
x
4
y 3
is the same equation but with x the subject.
4
Example 2:
Make x the subject of y = 2 – 5x
Notice that in this formula the x term is negative.
y = 2 – 5x
Add 5x to both sides
y + 5x = 2
Subtract y from both sides
5x = 2 – y
2 y
Divide both sides by 5
x
5
Solution:
PDC
(the x term is now positive)
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5( F  32)
is used to convert between ° Fahrenheit and ° Celsius.
9
We can rearrange to make F the subject.
5( F  32)
C
9
Multiply by 9
(this removes the fraction)
9C  5( F  32)
9C  5F  160
Expand the brackets
9C  160  5F
Add 160 to both sides
9C  160
Divide both sides by 5
F
5
9C  160
Therefore the required rearrangement is F 
.
5
The formula C 
Example 3:
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(3 x  5)
9
Rearranging equations involving squares and square roots
Example 4: Make x the subject of x 2  y 2  w2
Solution:
Subtract y 2 from both sides:
x 2  y 2  w2
x 2  w2  y 2 (this isolates the term involving x)
Square root both sides:
x   w2  y 2
Remember that you can have a positive or a negative square root. We cannot simplify the answer
any more.
Example 5: Make a the subject of the formula t 
Solution:
Multiply by 4
Square both sides
PDC
t
1 5a
4 h
1 5a
4 h
5a
h
5a
16t 2 
h
4t 
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Algebra revision pack
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16t 2 h  5a
16t 2 h
a
5
Multiply by h:
Divide by 5:
Exercise B:
Make t the subject of each of the following
1)
wt
P
32r
2)
wt 2
P
32r
3)
1
V   t2h
3
4)
P
5)
Pa 
6)
r  a  bt 2
w(v  t )
g
2t
g
More difficult examples
Sometimes the variable that we wish to make the subject occurs in more than one place in the
formula. In these questions, we collect the terms involving this variable on one side of the equation,
and we put the other terms on the opposite side.
Example 6:
Make t the subject of the formula a  xt  b  yt
Solution:
a  xt  b  yt
Start by collecting all the t terms on the right hand side:
Add xt to both sides:
a  b  yt  xt
Now put the terms without a t on the left hand side:
Subtract b from both sides:
a  b  yt  xt
Factorise the RHS:
a  b  t ( y  x)
Divide by (y + x):
ab
t
yx
So the required equation is
t
ab
yx
Example 7: Make W the subject of the formula T  W 
Wa
2b
Solution: This formula is complicated by the fractional term. We begin by removing the fraction:
2bT  2bW  Wa
Multiply by 2b:
2bT  Wa  2bW
Add 2bW to both sides:
(this collects the W’s together)
Factorise the RHS:
2bT  W (a  2b)
Divide both sides by a + 2b:
PDC
W 
2bT
a  2b
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Exercise C
Make x the subject of these formulae:
1)
3)
PDC
ax  3  bx  c
2x  3
y
5x  2
2)
4)
3( x  a )  k ( x  2)
x
x
 1
a
b
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Chapter 6: SOLVING QUADRATIC EQUATIONS
Method 1: Factorising
Example 1 : Solve x2 –3x + 2 = 0
Factorise
(x –1)(x – 2) = 0
Either (x – 1) = 0 or (x – 2) = 0
So the solutions are x = 1 or x = 2
Note: The individual values x = 1 and x = 2 are called the roots of the equation.
Example 2:
Solve x2 – 2x = 0
Factorise:
x(x – 2) = 0
Either x = 0 or (x – 2) = 0
So x = 0 or x = 2
Method 2: Using the formula
Recall that the roots of the quadratic equation ax 2  bx  c  0 are given by the formula:
x
 b  b 2  4ac
2a
Example 3: Solve the equation 2 x 2  5  7  3 x
Solution: First we rearrange so that the right hand side is 0. We get 2 x 2  3 x  12  0
We can then tell that a = 2, b = 3 and c = -12.
Substituting these into the quadratic formula gives:
3  32  4  2  (12) 3  105

(this is the surd form for the solutions)
2 2
4
If we have a calculator, we can evaluate these roots to get: x = 1.81 or x = -3.31
x
EXERCISE A
1) Use factorisation to solve the following equations:
b)
x2 – 3x – 4 = 0
a)
x2 + 3x + 2 = 0
c)
x2 = 15 – 2x
2) Find the roots of the following equations:
a)
x2 + 3x = 0
b)
x2 – 4x = 0
c)
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
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
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Algebra revision pack
Summer 2013
Chapter 7: INDICES
Basic rules of indices
y 4 means y  y  y  y .
4 is called the index (plural: indices), power or
exponent of y.
There are 3 basic rules of indices:
1)
2)
a m  a n  a mn
a m  a n  a mn
e.g.
e.g.
3)
(a m ) n  a mn
e.g.
34  35  39
38  36  32
3 
2 5
 310
Further examples
y 4  5 y3  5 y 7
4a 3  6a 2  24a 5
2c 2   3c 6   6c8
24d 7  3d 2 
24d 7
 8d 5
2
3d
(multiply the numbers and multiply the a’s)
(multiply the numbers and multiply the c’s)
(divide the numbers and divide the d terms i.e. by subtracting
the powers)
Exercise A
Simplify the following:
1)
b  5b 5
2)
3c  2c
3)
b c  bc
4)
2n6  (6n 2 )
2
2
5
3
5)
8n 8  2 n 3
6)
d 11  d 9
7)
a 
8)
 d 
3 2
4 3
More complex powers
Zero index:
a0  1 .
This result is true for any non-zero number a.
Negative powers
A power of -1 corresponds to the reciprocal of a number, i.e. a 1 
Therefore
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5 1 
1
a
1
5
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Algebra revision pack
Summer 2013
Fractional powers:
Fractional powers correspond to roots:
a1/ 2  a
a1/ 3  3 a
a1/ 4  4 a
In general:
a1/ n  n a
Therefore:
81/ 3  3 8  2
251/ 2  25  5
100001/ 4  4 10000  10
Exercise B:
Find the value of:
1)
41/ 2
27
3)
 19 
4)
5)
6)
7)
8 2 / 3
8)
 0.04 
9)
 8 
 
 27 
10)
 1 
 
 16 
1/ 3
2)
1/ 2
2
2
 
3
1/ 2
5 2
180
2/3
3 / 2
Simplify each of the following:
11)
2a1/ 2  3a 5 / 2
12)
x y 
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4 1/ 2
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Algebra revision pack
Summer 2013
Practice Booklet Test (Time allowed 30 minutes)
You may NOT use a calculator
 b  b 2  4ac
If ax + bx + c = 0 then x =
2a
2
1.
Expand and simplify
(a) (2x + 3)(2x – 1)
(b) (a + 3)2
(c) 4x(3x – 2) – x(2x + 5)
2.
Factorise
(a) x2 – 7x
(c) 2x2 + 5x – 3
(d) 6t2 – 13t + 5
3.
Simplify
4x3 y
(a)
8x 2 y 3
4.
5.
(b) y2 – 64
(b)
3x  2 4 x  1
+
3
6
Solve the following equations
h  1 3h
(a)
+
=4
(b) x2 – 8x = 0
4
5
x2
(d) x 
 6 x  10
3
(c) p2 + 4p = 12
Simplify the following expression:
2x 1 x  5

3
2
6.
Write each of the following as single powers of x and / y
x5
1
(a) 4
(b) (x2y)3
(c)  2 (d) ( x 2 y 4 ) 1
x
x
7.
Work out the values of the following, giving your answers as fractions
1
(a) 4-2
8.
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(b) 100
 8 3
(c)  
 27 
Solve the simultaneous equations
2
(d) (1 ) 3
3
3x – 5y = -11
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Algebra revision pack
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5x – 2y =
7
9.
Rearrange the following equations to make x the subject
x2
1
(b) V = πx2h
(c) y =
(a) v2 = u2 + 2ax
3
x 1
10.
Solve 5x2 – x – 1 = 0 giving your solutions in surd form
11.
Solve x 2  y 2  26
x y  4
Scores:
90% - Very good
80% - Go over the questions you answered incorrectly and try again.
70% - Your algebra skills need a good polish! Go over the questions which you answered
incorrectly and try again.
Any score less than 70% is a poor score and indicates that you will probably struggle with A
level mathematics.
If you did score < 70% I suggest you work through the first nineteen pages of the C1 *Revise
for Core Mathematics 1 which should be clearly seen on the relevant school web page. You
should also consider the following book as a further resource:
Head Start to AS-Level Maths
Published by CGP Workbooks
ISBN: 978 1 84146 993 5
Cost: £4.95 or (Amazon a little cheaper)
*Permission has been granted by Pearson’s for displaying Revise for Core Mathematics
1 Endorsed by Edexcel with coordinators: Pledger, Attwood, Macpherson, Moran,
Peetran, Staley and Wilkins.
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