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Unit 2
Contents
A1 Algebraic manipulation
A
A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.4 Factorizing quadratic expressions
A A1.5 Algebraic fractions
Multiplying terms
Simplify:
x+x+x+x+x
= 5x
x to the power of 5
Simplify:
x×x×x×x×x
= x5
x5 as been written using index notation.
The number n is called the index or
power.
xn
The number x is called the base.
Multiplying terms involving indices
We can use index notation to simplify expressions.
For example,
3p × 2p =
q2 × q3 =
6p2
3×p×2×p=
q5
q×q×q×q×q=
3r × r2 =
3×r×r×r=
2t × 2t =
(2t)2
or
3r3
4t2
Multiplying terms with the same base
When we multiply two terms with the same base the indices are added.
For example,
a4 × a2 =
(a × a × a × a) × (a × a)
=a×a×a×a×a×a
= a6
= a (4 + 2)
In general,
xm × xn = x(m + n)
Dividing terms
Remember, in algebra we do not usually use the division sign, ÷.
Instead, we write the number or term we are dividing by underneath like a fraction.
For example,
(a + b) ÷ c
is written as
a+b
c
Dividing terms
Like a fraction, we can often simplify expressions by cancelling.
For example,
n3
n3 ÷ n2 =
6p2 ÷ 3p =
n2
=
=n
6p2
n×n×n
n×n
3p
=
= 2p
2
6×p×p
3×p
Dividing terms with the same base
When we divide two terms with the same base the indices are subtracted.
For example,
a5 ÷ a2 =
4p6 ÷ 2p4 =
a×a×a×a×a
= a×a×a=
a×a
2
4×p×p×p×p×p×p
2×p×p×p×p
In general,
xm ÷ xn = x(m – n)
a3 = a (5 – 2)
= 2×p×p=
2p2 = 2p(6 – 4)
Multiplying numbers in
index form
When we multiply two numbers written in index form and with the same base we can
see an interesting result.
For example:
34 × 32 =
(3 × 3 × 3 × 3) × (3 × 3)
=3×3×3×3×3×3
= 36
73 × 75 =
= 3(4 + 2)
(7 × 7 × 7) × (7 × 7 × 7 × 7 × 7)
=7×7×7×7×7×7×7×7
= 78
= 7(3 + 5)
When we multiply two numbers with the same base the indices are added. In general, xm
What
n =you
+ n)
× xdo
x(mnotice?
Dividing numbers in index
form
When we divide two numbers written in index form and with the same base we can see
another interesting result.
For example:
45 ÷ 42 =
56
÷
54
=
4×4×4×4×4
4×4
5×5×5×5×5×5
5×5×5×5
= 4×4×4=
= 5×5=
43 = 4(5 – 2)
52 = 5(6 – 4)
When we divide two numbers with the same base the indices are subtracted. In general,
What
notice?
– n)
xm ÷doxnyou
= x(m
The power of 1
Find the value of the following using your calculator:
61
471
0.91
–51
01
Any number raised to the power of 1 is equal to the number itself. In general, x1
=x
Because of this we don’t usually write the power when a number is raised to the
power of 1.
The power of 0
Look at the following division:
64 ÷ 64 =
1
Using the second index law,
64 ÷ 64 = 6(4 – 4) =
60
60 = 1
That means that:
Any non-zero number raised to the power of 0 is equal to 1.
For example,
100 = 1
3.4520 = 1
723 538 5920 = 1
Index laws
Here is a summery of the index laws you have met so far:
xm × xn = x(m + n)
xm ÷ xn = x(m – n)
(xm)n = xmn
x1 = x
x0 = 1 (for x = 0)
Negative indices
Look at the following division:
32
÷
34
3×3
=
1
=
3×3×3×3
3×3
=
1
32
Using the second index law,
32 ÷ 34 = 3(2 – 4) =
3–2 =
That means that
Similarly,
6–1 =
1
6
7–4 =
3–2
1
32
1
74
and
5–3 =
1
53
Reciprocals
A number raised to the power of –1 gives us the reciprocal of that number.
The reciprocal of a number is what we multiply the number by to get 1.
1
The reciprocal of a is
a
The reciprocal of
is
We can find reciprocals on a calculator using the
a
b
b
a
key.
x-1
Finding the reciprocals
Find the reciprocals of the following:
1) 6
2)
1
The reciprocal of 6 =
3
7
3) 0.8
The reciprocal of
0.8 =
or
4
5
or
6
=
3
7
7
3
The reciprocal of
0.8–1 = 1.25
6
–1
3
or
=
1
6-1 =
=
7
4
5
5
4
7
3
= 1.25
Fractional indices
1
Indices can also be fractional. Suppose we have
1
1
9 2× 9 =2
9
In general,
91 =
=
9
Because 3 × 3 = 9
x 12= x
In general,
But,
1
2
9 × 9 = 9
But,
Similarly,
1+
2
9 2.
1
1
8 3× 8 × 38 =
3
1
3
8
+1 + 1 = 1
3
3
3
3
8 × 8 × 8 = 8
x 13 = x3
3
81 =
8
Because
2×2×2=8
Fractional indices
What is the value of 25 ?
We can think of
as 25
25
3
2
1 ×3
2
.
Using the rule that (xa)b = xab we can write
25
×13
2
= (25)3
= (5)3
= 125
In general,
m
n
x n= (x)m
3
2
Evaluate the following
1
2
1) 49
2
3
2) 1000
3)
8-
4)
64-
5) 4
1
3
5
2
2
3
1000
8- 1
64-
4
5
2
2
3
8
=
7
= (3√1000)2 =
1
=
3
2
3
= √49 =
1
2
49
=
1
3
1
64
= (√4)5 =
2
3
1
3√8
=
102 =
=
1
(3√64)2
25 =
32
100
1
2
=
1
42
=
1
16
Index laws for fractional indices
Here is a summery of the index laws for fractional indices.
x = x
1
2
x = x
1
n
n
x = xm or (x)m
m
n
n
n
Contents
N2 Powers, roots and standard form
A N2.1 Powers and roots
A N2.2 Index laws
A N2.3 Negative indices and reciprocals
A N2.4 Fractional indices
A
N2.5 Surds
A N2.6 Standard form
Surds
The square roots of many numbers cannot be found exactly.
For example, the value of √3 cannot be written exactly as a fraction or a decimal.
The value of √3 is an irrational number.
For this reason it is often better to leave the square root sign in and write the number as
√3.
√3 is an example of a surd.
Which one of the following is not a surd?
√2, √6 , √9 or √14
9 is not a surd because it can be written exactly.
Multiplying surds
What is the value of √3 × √3?
We can think of this as squaring the square root of three.
Squaring and square rooting are inverse operations so,
√3 × √3 = 3
In general, √a × √a = a
What is the value of √3 × √3 × √3?
Using the above result,
√3 × √3 × √3 = 3 × √3
= 3√3
Like algebra, we do not use
the × sign when writing
surds.
Multiplying surds
Use a calculator to find the value of √2 × √8.
What do you notice?
√2 × √8 = 4
(= √16)
4 is the square root of 16 and 2 × 8 = 16.
Use a calculator to find the value of √3 × √12.
√3 × √12 = 6
6 is the square root of 36 and 3 × 12 = 36.
In general, √a × √b = √ab
(= √36)
Dividing surds
Use a calculator to find the value of √20 ÷ √5.
What do you notice?
√20 ÷ √5 = 2
(= √4)
2 is the square root of 4 and 20 ÷ 5 = 4.
Use a calculator to find the value of √18 ÷ √2.
√18 ÷ √2 = 3
(= √9)
3 is the square root of 9 and 18 ÷ 2 = 9.
In general, √a ÷ √b =

a
b
Simplifying surds
We are often required to simplify surds by writing them in the form a√b. For example,
Simplify √50 by writing it in the form a√b.
Start by finding the largest square number that divides into 50.
This is 25. We can use this to write:
√50 = √(25 × 2)
= √25 × √2
= 5√2
Simplifying surds
Simplify the following surds by writing them in the form a√b.
1) √45
2) √24
3) √300
√45 = √(9 × 5)
√24 = √(4 × 6)
√300 = √(100 × 3)
= √9 × √5
= √4 × √6
= √100 × √3
= 3√5
= 2√6
= 10√3
Adding and subtracting surds
Surds can be added or subtracted if the number under the square root sign is the same.
For example,
Simplify √27 + √75.
Start by writing √27 and √75 in their simplest forms.
√27 = √(9 × 3)
√75 = √(25 × 3)
= √9 × √3
= √25 × √3
= 3√3
= 5√3
√27 + √75 = 3√3 + 5√3 =
8√3
Perimeter and area problem
The following rectangle has been drawn on a square grid.
Use Pythagoras’ theorem to find the length and width of the rectangle and hence find its
perimeter and area in surd form.
1
6
2√10
3
√10
Width = √(32 + 12)
= √(9 + 1)
2
= √10 units
Length = √(62 + 22)
= √(36 + 4)
= √40
= 2√10 units
Perimeter and area problem
The following rectangle has been drawn on a square grid.
Use Pythagoras’ theorem to find the length and width of the rectangle and hence find its
perimeter and area in surd form.
1
2√10
3
√10
Perimeter = √10 + 2√10 +
6
√10 + 2√10
2
= 6√10 units
Area = √10 × 2√10
= 2 × √10 × √10
= 2 × 10
= 20 units2
Rationalizing
the
denominator
When a fraction has a surd as a denominator we usually rewrite it so that the denominator
is a rational number.
This is called rationalizing the denominator.
Simplify the fraction
5
√2
Remember, if we multiply the numerator and the denominator of a fraction by the same
number the value of the fraction remains unchanged.
In this example, we can multiply the numerator and the denominator by √2 to make the
denominator into a whole number.
Rationalizing
the
denominator
When a fraction has a surd as a denominator we usually rewrite it so that the denominator
is a rational number.
This is called rationalizing the denominator.
5
Simplify the fraction
√2
×√2
5
√2
=
×√2
5√2
2
Rationalizing the denominator
Simplify the following fractions by rationalizing their
denominators.
1)
2
2)
√3
√2
√5
×√3
2
√3
=
×√3
3)
3
4√7
×√5
2√3
√2
3
√5
=
×√5
×√7
√10
3
5
4√7
=
×√7
3√7
28
Raising a power to a power
Sometimes numbers can be raised to a power and the result raised to another power.
For example,
(43)2 =
43 × 43
= (4 × 4 × 4) × (4 × 4 × 4)
= 46
= 4(3 × 2)
When a number is raised to a power and then raised to another power, the powers are
WhatIndogeneral,
you notice?
multiplied.
(xm)n = xmn
Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A
A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.4 Factorizing quadratic expressions
A A1.5 Algebraic fractions
Expanding two brackets
With practice we can expand the product of two linear expressions in fewer steps. For
example,
(x – 5)(x + 2) =
x2 + 2x
– 5x
– 10
= x2 – 3x – 10
Notice that –3 is
the sum of –5
and 2 …
… and that –10
is the product
of –5 and 2.
Matching quadratic expressions 1
Matching quadratic expressions 2
Squaring expressions
Expand and simplify: (2 – 3a)2
We can write this as,
(2 – 3a)2 = (2 – 3a)(2 – 3a)
Expanding,
(2 – 3a)(2 – 3a) =
2(2 – 3a) – 3a(2 – 3a)
= 4 – 6a
– 6a
= 4 – 12a + 9a2
+ 9a2
Squaring expressions
In general,
(a + b)2 = a2 + 2ab + b2
The first term
squared …
… plus 2 × the
product of the
two terms …
For example,
(3m + 2n)2 = 9m2 + 12mn + 4n2
… plus the
second term
squared.
Squaring expressions
The difference between two squares
Expand and simplify (2a + 7)(2a – 7)
Expanding,
2a(2a – 7) + 7(2a – 7)
(2a + 7)(2a – 7) =
= 4a2
– 14a
+ 14a
– 49
= 4a2 – 49
When we simplify, the two middle terms cancel out.
This is the difference between two
squares.
In general,
(a + b)(a – b) = a2 – b2
Matching the difference between two squares
Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A
A1.3 Factorization
A A1.4 Factorizing quadratic expressions
A A1.5 Algebraic fractions
Factorizing
quadratic
expressions
Quadratic expressions of the form x + bx + c can be factorized if they can be written using
2
brackets as
(x + d)(x + e)
where d and e are integers.
If we expand (x + d)(x + e) we have,
(x + d)(x + e) = x2 + dx + ex + de
= x2 + (d + e)x + de
Comparing this to x2 + bx + c we can see that:
The sum of d and e must be equal to b, the coefficient of x.
The product of d and e must be equal to c, the constant term.
Factorizing quadratic expressions 1
Matching quadratic expressions 1
Factorizing
quadratic
expressions
Quadratic expressions of the form ax + bx + c can be factorized if they can be written using
2
brackets as
(dx + e)(fx + g)
where d, e, f and g are integers.
If we expand (dx + e)(fx + g)we have,
(dx + e)(fx + g)= dfx2 + dgx + efx + eg
= dfx2 + (dg + ef)x + eg
Comparing this to ax2 + bx + c we can see that we must choose d, e, f and g such that:
a = df,
b = (dg + ef)
c = eg
Factorizing quadratic expressions 2
Matching quadratic expressions 2
Factorizing the difference between two squares
A quadratic expression in the form
x2 – a2
is called the difference between two squares.
The difference between two squares can be factorized as follows:
x2 – a2 = (x + a)(x – a)
For example,
9x2 – 16 = (3x + 4)(3x – 4)
25a2 – 1 = (5a + 1)(5a – 1)
m4 – 49n2 = (m2 + 7n)(m2 – 7n)
Factorizing the difference between two squares
Matching the difference between two squares
Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.4 Factorizing quadratic expressions
A
A1.5 Algebraic fractions
Algebraic fractions
3x
4x2
and
2a
are examples of algebraic fractions.
3a + 2
The rules that apply to numerical fractions also apply to algebraic fractions.
For example, if we multiply or divide the numerator or the denominator of a fraction by
the same number or term we produce an equivalent fraction.
For example,
3x
4x2
=
3
4x
=
6
8x
=
3y
4xy
=
3(a + 2)
4x(a + 2)
Simplifying algebraic fractions
We simplify or cancel algebraic fractions in the same way as numerical fractions, by
dividing the numerator and the denominator by common factors. For example,
Simplify
6ab
3ab2
=
=
6ab
3ab2
2
6×a×b
3×a×b×b
2
b
Simplifying algebraic fractions
Sometimes we need to factorize the numerator and the denominator before we can
simplify an algebraic fraction. For example,
2a + a2
Simplify
2a + a2
8 + 4a
8 + 4a
=
=
a (2 + a)
4(2 + a)
a
4
Simplifying algebraic fractions
b2 – 36 is the
difference between
two squares.
b2 – 36
Simplify
b2 – 36
3b – 18
(b + 6)(b – 6)
=
3b – 18
3(b – 6)
b+6
=
3
If required, we can write this as
b
3
+
6
3
=
b
3
+ 2
Manipulating algebraic fractions
Remember, a fraction written in the form
a+b
a
can be written as
c
b
+
c
c
However, a fraction written in the form
c
a+b
c
cannot be written as
+
a
c
b
For example,
1+2
3
=
1
3
+
2
3
but
3
1+2
=
3
1
+
3
2
Multiplying and dividing algebraic
We can multiply and divide algebraicfractions
fractions using the same rules that we use for
numerical fractions.
In general,
a
b
and,
a
c
=
d
÷
b
×
c
=
d
a
b
For example,
3p
4
×
ac
bd
×
d
=
c
3
6p
2
=
(1 – p)
4(1 – p)
2
ad
bc
=
3p
2(1 – p)
Multiplying and dividing algebraic
fractions
2
4
This is the
What is
÷
3y – 6
?
y–2
reciprocal
of
4
y–2
2
3y – 6
÷
4
y–2
=
=
=
2
3y – 6
2
3(y – 2)
1
6
×
×
y–2
4
y–2
4
2
Adding algebraic fractions
We can add algebraic fractions using the same method that we use for numerical fractions.
For example,
1
What is
+
a
2
?
b
We need to write the fractions over a common denominator before we can add them.
1
a
+
2
b
=
b
2a
+
ab
=
ab
ab
In general,
a
b
+
c
d
=
b + 2a
ad + bc
bd
Adding algebraic
fractions
3
y
What is
+
x
2
?
We need to write the fractions over a common denominator before we can add them.
3
x
+
y
2
=
=
=
3×2
+
x×2
6
+
2x
6 + xy
2x
xy
2x
y×x
2×x
Subtracting algebraic fractions
We can also subtract algebraic fractions using the same method as we use for numerical
fractions. For example,
p
What is
–
3
q
2
?
We need to write the fractions over a common denominator before we can subtract them.
p
3
–
q
2
=
2p
3q
–
6
6
=
In general,
a
b
–
c
d
=
ad – bc
bd
2p – 3q
6
Subtracting algebraic
fractions
3
2+p
What is
2+p
4
–
3
2q
=
=
=
4
(2 + p) × 2q
2q(2 + p)
8q
2q(2 + p) – 12
8q
q(2 + p) – 6
4q
2q
–
?
3×4
–
4 × 2q
4
=
–
2q × 4
12
8q
6
Contents
A4 Inequalities
A
A4.1 Representing inequalities on number lines
A A4.2 Solving linear inequalities
A A4.3 Inequalities and regions
A A4.4 Inequalities in two variables
A A4.5 Quadratic inequalities
Inequalities
An inequality is an algebraic statement involving the symbols
>, <, ≥ or ≤
For example,
x>3
means
‘x is greater than 3’.
x < –6
means
‘x is less than –6’.
x ≥ –2
means
‘x is greater than or equal to –2’.
x ≤ 10
means
‘x is less than or equal to 10’.
Sometime two inequalities can be combined in a single statement. For example,
If x > 3 and x ≤ 14 we can write
3 < x ≤ 14
Representing inequalities on number
Suppose x > 2. There are infinitely manylines
values that x could have.
x could be equal to
3,
7.3, 54 , 18463.431 3…
11
It would be impossible to write every solution down.
We can therefore represent the solution set on a number line as follows:
–3
–2
–1
0
1
2
3
4
5
6
7
A hollow circle, , at 2 means that this number is not included and the arrow at the end of
the line means that the solution set extends in the direction shown.
Representing inequalities on number
Suppose x ≤ 3. Again, there are infinitelylines
many values that x could have.
x could be equal to
3,
–1.4, –94 , –7452.802 …8
17
We can represent the solution set on a number line as follows,
–3
–2
–1
0
1
2
3
4
5
6
A solid circle, , at 3 means that this number is included and the arrow at the end of
the line means that the solution set extends in the direction shown.
7
Representing inequalities on number
lines
Suppose –1 ≤ x < 4. Although x is between
two values, there are still infinitely many values
that x could have.
x could be equal to
2,
–0.7, –3 , 1.648953 …16
17
We can represent the solution set on a number line as follows:
–3
–2
–1
0
1
2
3
4
5
6
7
A solid circle, , is used at –1 because this value is included and a hollow circle, , is used
at 4 because this value is not included.
The line represents all the values in between.
Integer solutions
In the examples that we have looked at so far we have assumed that the value of x can be
any real number.
Sometimes we are told that x can only be an integer, that is a positive or negative whole
number.
For example,
–3 < x ≤ 5
List the integer values that satisfy this inequality.
The integer values that solve this inequality are
–2, –1, 0, 1, 2, 3, 4, 5.
Integer solutions
Write down an inequality that is obeyed by the following set of integers:
–4, –3, –2, –1, 0, 1.
There are four possible inequalities that give this solution set,
–5 < x < 2
–4 ≤ x < 2
–5 < x ≤ 1
–4 ≤ x ≤ 1
Remember when we use < and > the values at either end are not included in the solution
set.
Contents
A4 Inequalities
A A4.1 Representing inequalities on number lines
A
A4.2 Solving linear inequalities
A A4.3 Inequalities and regions
A A4.4 Inequalities in two variables
A A4.5 Quadratic inequalities
Solving
linear
inequalities
Look at the following inequality,
x+3≥7
What values of x would make this inequality true?
Any value of x greater or equal to 4 would solve this inequality.
We could have solved this inequality as follows,
x+3≥7
subtract 3 from both sides:
x+3–3≥7–3
x≥4
The solution has one letter on one side of the inequality sign and a number on the
other.
Solving linear inequalities
Like an equation, we can solve an inequality by adding or subtracting the same value to
both sides of the inequality sign.
We can also multiply or divide both sides of the inequality by a positive value. For
example,
Solve 4x – 7 > 11 – 2x
add 7 to both sides:
4x > 18 – 2x
add 2x to both sides:
6x > 18
divide both sides by 6:
x>3
How could we check this solution?
To verify that
is the solution to
Checking solutions
x>3
4x – 7 > 11 – 2x
substitute a value just above 3 into the inequality and then substitute a value just below
3.
If we substitute x = 4 into the inequality we have
4 × 4 – 7 > 11 – 2 × 4
16 – 7 > 11 – 8
9>3
This is true.
If we substitute x = 2 into the inequality we have,
4 × 2 – 7 > 11 – 2 × 2
8 – 7 > 11 – 4
1>7
This is not true.
Solving combined linear inequalities
The two inequalities 4x + 3 ≥ 5 and 4x + 3 < 15 can be written as a single combined
inequality.
5 ≤ 4x + 3 < 15
We can solve this inequality as follows:
subtract 3 from each part:
2 ≤ 4x < 12
divide each part by 4:
0.5 ≤ x < 3
We can illustrate this solution on a number line as
–1
–0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Solving combined linear inequalities
Some combined inequalities contain variables in more than one part. For example,
x – 2 ≤ 3x + 2 ≤ 2x + 7
Treat this as two separate inequalities,
and
x – 2 ≤ 3x + 2
3x + 2 ≤ 2x + 7
– 2 ≤ 2x + 2
x+2≤7
– 4 ≤ 2x
x≤5
–2≤x
We can write the complete solution as –2 ≤ x ≤ 5 and illustrate it on a number line as:
–3
–2
–1
0
1
2
3
4
5
6
7
Overlapping solutions
Solve the following inequality and illustrate the solution on a number line.
2x – 1 ≤ x + 2 < 7
Treating as two separate inequalities,
and
2x – 1 ≤ x + 2
x+2<7
x–1≤2
x<5
x≤3
If x < 5 then it is also ≤ 3. The whole solution set is therefore given by x ≤ 3. This is can
be seen on the number line:
–3
–2
–1
0
1
2
3
4
5
6
7
Solutions in two parts
Solve the following inequality and illustrate the solution on a number line:
4x + 5 < 3x + 5 ≤ 4x + 3
Treating as two separate inequalities,
and
4x + 5 < 3x + 5
3x + 5 ≤ 4x + 3
4x < 3x
5≤x+3
x<0
2≤x
x≥2
We cannot write these solutions as a single combined inequality. The solution has two
parts.
–3
–2
–1
0
1
2
3
4
5
6
7
Contents
A4 Inequalities
A A4.1 Representing inequalities on number lines
A A4.2 Solving linear inequalities
A
A4.3 Inequalities and regions
A A4.4 Inequalities in two variables
A A4.5 Quadratic inequalities
Vertical regions
Inequalities can be represented by regions on a graph.
A region is an area where all the points obey a given rule.
Suppose we want to find the region where
x>2
This means that we want to show the area of a graph where the x-coordinate of every
point is greater than 2.
Give the coordinates of three points that would satisfy
this condition.
For example (4, 1), (6, 5), and (3, –2)
Vertical
regions
We can represent all the points where the x-coordinate is equal to 2 with the line x = 2.
The region where x > 2 does not
include points where x = 2 and so we
draw this as a dotted line.
y
x<2
5
4
3
2
1
–6 –5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
x>2
x
1 2 3 4 5 6
The region to the right of the line x =
2 contains every point where x > 2.
The region to the left of the line x = 2
contains every point where x < 2.
Horizontal regions
Suppose we want to find the region where
y≤3
This means that we want to show the area of a graph where the y-coordinate of every
point is less than or equal to 3.
Give the coordinates of three points that would satisfy
this condition.
For example, (5, 1), (–3, –4), and (0, 2)
We can represent all the points where the y-coordinate is equal to 3 with the line y = 3.
Horizontal regions
The region where y ≤ 3 includes points where y = 3 and so we draw y = 3 as a solid line.
y
y≥3
The region below the line y = 3
contains every point where y ≤ 3.
5
4
3
2
1
–6 –5 –4 –3 –2 –1 0
–1
–2
y≤3
–3
–4
–5
x
1 2 3 4 5 6
The region above the line y = 3
contains every point where y ≥ 3.
Horizontal and vertical regions
When several regions are shown on
the same graph it is usual to shade out the unwanted
combined
regions.
This is so that the required area where the regions overlap can easily be identified.
For example, to show the region
where –4 < x < –1 and –1 < y ≤ 3,
y
5
4
3
2
1
–6 –5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
x
1 2 3 4 5 6
1) Shade out the regions
x < –4 and x > –1.
2) Shade out the regions
y < –1 and y ≥ 3.
The unshaded region satisfies both –4
< x < –1 and –1 < y ≤ 3.
Horizontal and vertical regions
combined
Contents
A4 Inequalities
A A4.1 Representing inequalities on number lines
A A4.2 Solving linear inequalities
A A4.3 Inequalities in two variables
A
A4.4 Inequalities in two variables
A A4.5 Quadratic inequalities
Inequalities in two variables
Linear inequalities can be given in two variables x and y.
For example,
x+y<3
The solution set to this inequality is made up of pair of values. For example,
x=1
and
y=1
x=4
and
y = –5
x = –1
and
y=0
These solutions are usually written as coordinate pairs as
(1, 1), (4, –5) and (–1, 0).
The whole solution set can be represented using a graph.
Inequalities in two variables
We can represent all the points where the x-coordinate and the y-coordinate add up to 3
with the line x + y = 3.
The region where x + y < 3 does not
include points where x + y = 3 and so we
draw this as a dotted line.
y
5
4
3
2
1
–6 –5 –4 –3 –2 –1 0
–1
–2
–3
x+y<3
–4
–5
x+y>3
x
1 2 3 4 5 6
The region below the line x + y = 3
contains every point where x + y < 3.
The region above the line x + y = 3
contains every point where x + y > 3.
Inequalities in two variables
When a line is sloping it may not always be obvious which side of the line gives the
required region.
We can check this by choosing a point (not on the line) and substituting the x- and yvalues of the point into the inequality representing the required region.
If the point satisfies the inequality then it is in the region. If it does not satisfy the
inequality it is not in the region.
The easiest point to substitute is usually the point at the origin, that is the point (0, 0).
Inequalities in two variables
For example,
Is the point (0, 0) in the region 4y – 3x > 2
Substituting x = 0 and y = 0 into 4y – 3x > 2 gives,
4×0–3×0>2
0>2
0 is not greater than 2 and so the point (0, 0) does not lie in the required region.
The region representing 4y – 3x > 2 is therefore the region that does not contain the point
at the origin.
Inequalities in two variables
Combining inequalities in two
variables 1
Combining inequalities in two
variables 2
Real-life
problems
A ferry cannot hold more than 30 tonnes. If it holds x cars weighing 1 tonne
each and y lorries weighing 3 tonnes each write down an inequality in x and y.
x + 3y ≤ 30
If 20 cars were already on board how many more lorries
could the ferry carry?
Substituting into x + 3y < 30 and solving for y,
20 + 3y ≤ 30
subtract 20 from both sides:
divide both sides by 3:
3y ≤ 10
y ≤ 3.3 (to 1 d.p.)
The ferry can hold 3 more lorries.
Contents
A7 Sequences
A A7.1 Generating sequences from rules
A
A7.2 Linear sequences
A A7.3 Quadratic sequences
A A7.4 Geometric sequences
A A7.5 Other types of sequence
The nth term of a linear sequence
Suppose we are given a linear sequence and asked to find the nth term, un, of the
sequence. For example,
Find the nth term of the sequence 4, 7, 10, 13, 16, …
This sequence continues by adding 3 each time and so the common difference d is 3.
We compare the terms in the sequence to the multiples of 3.
Position
Multiples of
3
2
1
×3
3
×3
6
+1
Term
4
3
×3
9
+1
7
4
5
×3
12
+1
10
un = 3n + 1.
n
×3
×3
15
+1
13
…
3n
+1
16
+1
…
3n + 1
The nth term of a linear sequence
Find the nth term of the sequence 5, 3, 1, –1, –3, …
This sequence continues by subtracting 2 each time and so the common difference d
is –2.
We compare the terms in the sequence to the multiples of –2.
Position
Multiples of
–2
2
1
× –2
–2
× –2
–4
+7
Term
5
3
× –2
–6
+7
3
4
5
× –2
–8
+7
1
un = 7 – 2n.
× –2
–2n
+7
–3
n
× –2
–10
+7
–1
…
+7
…
7 – 2n
Contents
A7 Sequences
A A7.1 Generating sequences from rules
A A7.2 Linear sequences
A
A7.3 Quadratic sequences
A A7.4 Geometric sequences
A A7.5 Other types of sequence
Quadratic
sequences
When the second row of differences produces a constant number the sequence is called
a quadratic sequence.
This is because the rule for the nth term of the sequence is a quadratic expression of the
form
un = an2 + bn + c
where a, b and c are constants and a ≠ 0.
The simplest quadratic sequences is the sequence of square numbers.
1,
4,
9,
+3
16,
+5
+2
25
+7
+2
+9
+2
The constant second difference is 2 and the nth term is n2.
The nth term of a quadratic
sequence
Find the nth term of the sequence, 4, 9, 18, 31, 48, …
Let’s start by looking at the first and second differences.
4,
9,
18,
+5
31,
+9
+4
48
+13
+4
+17
+4
The second differences are constant and so the nth term is in the form un = an2 + bn + c.
Let’s find a, b and c.
The second difference is 4, so we know 2a = 4
a=2
The nth term of a quadratic
sequence
Find the nth term of the sequence, 4, 9, 18, 31, 48, …
Let’s start by looking at the first and second differences.
3,
4,
9,
18,
+5
+1
+4
31,
+9
+4
48
+13
+4
+17
+4
The value of c is the same as the value for the 0th term.
We can find this by continuing the pattern in the differences backwards from the first
term.
The 0th term is 3, so:
c=3
The nth term of a quadratic
sequence
Find the nth term of the sequence, 4, 9, 18, 31, 48, …
Putting a = 2 and c = 3 into un =an2 + bn + c gives us
un = 2n2 + bn + 3. We can use this to write an expression for the first term:
un = 2n2 + bn + 3
u1 = 2 × 12 + b × 1 + 3
u1 = 2 + b + 3
u1 = b + 5
The first term in the sequence is 4, so:
4=b+5
b = –1
The nth term of a quadratic
sequence
Once we have found the values of a, b, and c we can use them in un = an2 + bn + c to
give the nth term.
We have found that for the sequence 4, 9, 18, 31, 48, …
a = 2, b = –1 and c = 3,
un = 2n2 – n + 3
We can check this rule by substituting a chosen value for n into the formula and making
sure that it corresponds to the required term in the sequence.
For example, when n = 5 we have,
u5 = 2 × 52 – 5 + 3
= 48
Check the rule for other terms in the sequence.
The nth term of a quadratic
sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
This is the sequence of triangular numbers.
1,
3,
6,
+2
10,
+3
+1
15
+4
+1
+5
+1
The second differences are constant and so the nth term is in the form un =an2 + bn + c.
Let’s find a, b and c.
The second difference is 1, so 2a = 1
a=½
The nth term of a quadratic
sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
This is the sequence of triangular numbers.
0,
1,
3,
6,
+2
+1
+1
10,
+3
+1
15
+4
+1
+5
+1
The value of c is the same as the value for the 0th term.
We can find this by continuing the pattern in the differences backwards from the first
term.
The 0th term is 0, so:
c=0
The nth term of a quadratic
sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
Putting a = ½ and c = 0 into un =an2 + bn + c gives us
un = ½n2 + bn. We can use this to write an expression for the first term:
un = ½n2 + bn
u1 = ½ × 12 + b × 1
u1 = ½ + b
The first term in the sequence is 1, so
1=½+b
b=½
The nth term of a quadratic
sequence
Once we have found the values of a, b, and c we can use them in an2 + bn + c to give
the nth term.
We have found that for the sequence 1, 3, 6, 10, 15, …
a = ½, b = ½ and c = 0;
un =
n2
n
+
2
2
or
un =
n2 + n
2
Checking, when n = 5, we have
u5 = ½(52 + 5)
= 15
Check the rule for other terms in the sequence.
Fraction sequences
The terms in the following sequence are all fractions:
2 ,
3
3 ,
6
4 ,
9
5 ,
12
6 , …
15
Find the next two terms in the sequence.
Look at the sequence formed by the numerators and the sequence formed by the
denominators separately.
+1
2 ,
3
+3
+1
3 ,
6
+3
+1
4 ,
9
+3
+1
5 ,
12
+1
6 ,
15
+3
+1
7 ,
18
+3
8 , …
21
+3
Fraction sequences
The terms in the following sequence are all fractions,
2 ,
3
3 ,
6
4 ,
9
5 ,
12
6 , …
15
Find the formula for the nth term of the sequence.
The sequence formed by the
numerators is:
The sequence formed by the
numerators is:
u1 = 2
u1 = 3
u2 = 3
u2 = 6
u3 = 4
u3 = 9
un =
n+1
un =
3n