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Externally assessed
4 credits
Apply algebraic procedures in solving problems
Electronic technology, such as calculators
or computers, are not permitted in the
assessment of this standard.
Simplifying algebraic
expressions
In algebra, a variable is a letter such as x, used
to stand for a number. Algebraic terms and
expressions are formed by applying operations
(such as +, –, ×, ÷) to variables.
d.
The square of 2 less than x
e.
The square root of 1 more than y
f.
p is a multiple of 6. What is the next
multiple of 6 after p?
g.
A dog costs $d per month to feed. A cat
costs $27 less per month to feed. What is
the cost of feeding a dog and a cat for a
month? Simplify your answer.
h.
Mum is 32 years older than her son, Mark.
Mark is 4 years younger than his sister,
Daisy. Dad is two years older than Mum. If
Mum is m years old what is the sum of all
four ages? Simplify your answer.
i.
Hana buys 3 kg of apples at $a per kilogram
and 2 kg of pears at $p per kilogram.
Example
1. The number which is double y is 2 × y which is
written 2y. The number multiplying y, 2, is called
the coefficient of y, and 2y is called the product
of 2 and y.
2. The number which is 5 more than half x is
written 1x + 5 or x + 5.
2
2
An algebraic expression is made up of terms
added (or subtracted) together. Like terms (terms
with the same variables) can be simplified by
addition or subtraction.
AS 91027
MATHEMATICS AND STATISTICS 1.2
Example
1. 7y + 3y = 10y
2. 3x – 8x = –5x
3. 2a2 – 3b + 4a2 + b = 6a2 – 2b
i.
How much does this cost altogether?
Exercise A: Algebraic expressions
1.
Ans. p. 43
Write down as algebraic expressions:
a.
The product of 7 and x
b.
The sum of y and 4
c.
The number which is 3 more than five
times the number w
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ii.
How much change does Hana get if
she pays $d for her fruit?
iii. What restrictions are there on the
value of d ?
2 Achievement Standard 91027 (Mathematics and Statistics 1.2)
AS 91027
2.
Simplify where possible the following.
a.
5a + 7a
b.
10b – 12b + b
Multiplying and dividing terms
Any algebraic terms can be multiplied together.
Indices are used to simplify repeated multiplication,
e.g. y × y is written y2
Example
c.
3s – t – 2s + t
d.
2p + 8q + 9p – 14q
e.
7f 2
1. 3a × 4b = 12ab
[3 × 4 = 12, a × b = ab]
2. –2y.6y = –12yy
= –12y2
[dot means ×]
[using indices]
Division of algebraic terms is best written in fraction
x
form. Simplify using = 1.
x
–f–
10f 2
Note that if all factors in the numerator cancel, then
a factor of 1 will remain in the numerator.
– 2f
For example, 3a = 1 .
f.
g.
h.
i.
9ab
6ab + 4ba + 2a
x–
3b
Example
x2
1.
5 = 1
10x 2x
[dividing top and bottom by 5]
2.
10xy 2y
=
3
15x
[simplifying and cancelling ]
x
x
4w – 5 – w
If powers of variables occur, these can be written
as repeated multiplications before cancelling as
before.
3 – x2 – 5 + 4x
Example
j.
Q. Simplify
abc – bac – cba + acb
x2y3
2x 4y
A. Using repeated multiplications:
3.
x×x×y×y×y
x2y3
= 2×x×x×x×x×y
2x 4y
Find the missing term in each of the following
simplifications.
a.
2a – 5b – 4a +
x×x×y×y×y
x
= 2 × x × x × x × x × y [cancelling (twice)
x
y
and y ]
= a – 5b
y2
= 2x 2
Ans. p. 43
b.
5pq – 2qp –
c.
m + n – 2m + mn –
d.
– x2 – x3 +
e.
7d –
+ 2r = 3pq – r
–
+
= n – mn –
= 2x2 – 3x3
= d2 – d
Exercise B: Multiplying and
dividing terms
1.
Write each of the following in simplest form.
a.
7a × 2b
b.
–3x.2y
c.
5 × –3w
d.
–6a × –2b
e.
2x.3x
f.
3a × –b × 5c
g.
–y × 3y × –2y
h.
–15x.3
x
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Apply algebraic procedures in solving problems 3
2.
Simplify each of the following divisions.
10x
5
b.
6y
9y
c.
10
20x
d.
–2pq
4q
e.
8ab
2a
f.
2xy
4y
Order of operations
The correct order of operations must be followed
when simplifying expressions. The mnemonics
BEMA or BEDMAS (brackets, exponents,
multiplication and division, then addition and
subtraction) are a useful reminder.
Example
1. 3a + a × 5 = 3a + 5a = 8a
2.
g.
3.
9pqr
12rst
h.
a
a 2b
Simplify the following (write the powers as
repeated multiplications).
a.
b.
c.
d.
ab
ab4
4xy5
8x3y3
b.
c.
d.
[divide top and bottom by 5]
a.
x–x×4
b.
3x – (x + x)
c.
2×y+y×4
d.
(1 + 5) × (4x – x)
e.
2x – 8x
–x
2
f.
12x – 19x
3x + 4x
g.
–x × –2
h.
x + x × x – x2
6x
12m3np 4
4m2p2
Insert the missing numerators or
denominators.
a.
[division line means brackets]
Simplify using the correct order of operations.
9w3z
6w 4z3
2.
4.
2y + 3y (2y + 3y)
=
10
10
5y
=
10
y
=2
[× before +]
Exercise C: Order of operations
1.
2 3
3x2
5a2b
6x
p2q4
= 2x
= 1
ab
=
AS 91027
a.
2x
3
p
=
q
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Insert brackets to make correct statements.
a.
3 x × x + x = 6x 2
b.
8a ÷ 4 × a + 1 = 3
c.
x – x – 2x – x = –x
d.
y + y × y – y2 = y2
e.
2 p + 4p ÷ 3 p – p = 3
f.
12w 2 ÷ w – 2 w = –12w
g.
a – 6a – 4 a + 2a = a
h.
p + q × q – q = 0
Ans. p. 43
AS 91027
4 Achievement Standard 91027 (Mathematics and Statistics 1.2)
Exponents
A whole number exponent (or index or power) is
used to show repeated multiplication of the same
base number or variable.
xn = x × x × x × . . . × x
2.
e.
–30
f.
24 × 23
g.
–5 × 52
h.
–3 × –32
Simplify the following products.
a.
x3 × x 5
b.
3x2 × 7x4
For example, 2 × 2 × 2 × 2 × 2 is written 25 which
equals 32.
c.
–x × 2x2
d.
–2x.– 4x
y 6 means y × y × y × y × y × y [six factors of y
multiplied]
e.
5a × 2a3 × 3a2
f.
3xy × –2x3 × xy2
g.
4a2b2.3abc.a3
h.
9a4 × 2b2 × 3ab
n factors
–6 × –6 is written (–6)2 which is 36
Note that –62 = –1 × 62 = –36
x 0 = 1, for any non-zero number x
Example
1. 2x × 3x × 4x = 24xxx which is written 24x3
2. 5y2 × 3y5 = 5yy × 3yyyyy = 15y7
Note: x1 = x, for any number x
For indices with the same base:
•
multiply by adding the powers:
xa
•
×
xb
=
3.
xa + b
Simplify the following divisions.
a.
p6
p3
b.
12x8
6x3
c.
9a4
a
d.
2x3
4x5
e.
4ab9
2b7
f.
18x4y 6
24xy 8
g.
–7p
14p2q
h.
–9x
–18x3
divide by subtracting the powers:
xa = x a – b
xb
a
If b > a, use xb = b1– a
x
x
3
1
e.g. x5 = 2
x
x
Numbers are multiplied or divided in the usual way.
Example
1. x5 × x7 = x12
2. –2a3.5a4 = –10a7
3. 5abc2 × 2a2b3c = 10a3b4c3
p2
1
5. p7 = p5
Ans. p. 43
p9
4. p5 = p4
6.
4y 6
= 2y5
2y
Exercise D: Multiplying and
dividing with indices
1.
Evaluate.
a.
(–3)3
b.
–32
4.
Find the missing terms.
a.
c.
(–2)4
d.
12a2b2 = 2a
3b
40
b.
xy ×
= 2x3y3
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Apply algebraic procedures in solving problems 5
2.
To raise an exponential term to a new power,
multiply the powers:
(xa)b = xab
Express in simplest form.
a.
y16
b.
x20
c.
64x64
d.
100x2y 4
e.
a6
4
f.
p6
t10
g.
9x2
25
h.
9x2 + 16x2
All factors inside the bracket have their powers
AS 91027
Powers and roots of indices
a b d
xadybd
multiplied by the new power: x yc = cd
w
w
Example
1. (x10)3 = x30
2. (2x3y4)5 = 21 × 5x3 ×5y4 × 5 [2 means 21]
= 32x15y20
3.
2 4
x
3
2×4
=x 4
3
8
= x
81
3.
To find the square root of an index expression,
Use all your index rules to simplify the following.
a.
(2x2)3
4x
b.
(3p5)2
6p12
c.
4a5b4
(2ab)4
halve the power:
n
xn = x 2
Take the square roots of numbers in the usual way.
Example
16
1.
a16
= a2
36
2.
36x36 = 36 × x 2
= a8
= 6x18
Exercise E: Indices with powers and
roots
1.
Express in simplest form.
a.
(a5)3
b.
(2x5)2
c.
(–3x2)4
d.
(5xy2)0
e.
(–2w4x)3
f.
a2
b3
g.
4x
3y2
h.
5x4
10xy2
Ans. p. 43
6
d.
36x
12x
e.
2a2 × (3a)2
4a3
f.
(3x)2 × (5x + x)2
2
3
g.
50x5
2x9
h.
(3a2b3)4
6a2 × 9b4
3
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AS 91027
6 Achievement Standard 91027 (Mathematics and Statistics 1.2)
Basic algebraic fractions
c.
2ab
10bc
d.
12x2y
6xy
e.
42ab2
49a2b
f.
2x4
4x5
g.
9a3b2
12ab4
h.
56a
64abc
These are fractions with variables, such as
4
2x
or 5x
3
2y
Algebraic fractions obey the same rules as
numerical fractions.
•
Simplify fractions by ‘cancelling’ common
factors (using x = 1) and the laws of indices.
x
5xy7
For example: 3 simplifies to 5y 4
xy
•
2.
Add (or subtract) algebraic fractions with the
same denominator, by adding (or subtracting)
numerators and putting the result over the
denominator. For example:
Write as a single fraction and simplify if
possible.
a.
3x + x
5
5
c.
2a a
+
3
3
d.
x – x
7 10
e.
2y 3y
–
4
3
f.
3x
– x
12
8
a
is
b
g.
x2 2x2
+
5
4
Exercise F: Basic algebraic fractions
h.
3 – 1
4x x2
x 2x 3x
+
=
7
7
7
If the denominators are different, use
equivalent fractions to express each fraction
with the same denominator first. For example:
5a
a
+ a = 3a + 2a =
6
2
6
3
6
2
– 3 = 4 – 15 = –11
5x 2x 10x 10x 10x
•
Multiply algebraic fractions, by multiplying
numerators together to get the numerator of
the answer, and multiplying denominators
together to get the denominator of the
answer. For example:
b.
12y 2y
–
7
7
x × x = x2
3 4 12
Simplify answers where possible. For example:
3a × 5b2 = 15ab2 = 15b
8a2
8a3b
2b 4a3
•
To divide by a fraction, multiply by the
reciprocal (the reciprocal of the fraction
the fraction b ).
a
4x2 x 4x2 6 24x2
÷ =
× =
= 8x
3
3
3x
6
x
Ans. p. 44
1.
Simplify each of the following fractions.
a.
5x
10y
b.
6x
2
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Apply algebraic procedures in solving problems 7
3.
a.
2
4x
×
3x
5
b.
2a 3a
×
b
2b
c.
5x
× 2x
4y 15y
d.
3a 4b
×
a
2b
e.
3x2 × 2x3
9y
y
f.
7ab × c2
14a2
c
5.
g.
4.
3
12x × 10y
3x
5y 4
2
h.
5p × q
4q2 15p
e.
10a2b ÷ ab
d
cd
f.
12x3y 4
6y3
÷
5xy 2
10x
g.
a3 5ab
÷
8
b2
h.
9x5
3x4
÷
4y3 12y 4
Use the correct order of operations to simplify
the following expressions.
a.
x 2x 1
+
×
4
3
2
c.
3p p
+
×4
4 2
p
d.
7p p p
– +
2 4
8
Divide the fractions and simplify your answer
where possible.
a.
3a a
÷
b
4
b.
2x2 ÷ 4x
9
3
c.
ab
÷ d2
2c
c
d.
6x2 ÷ 3x
14
7
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b.
3y2 y 2
– ÷
3 y
2
AS 91027
Multiply the algebraic fractions and simplify
your answer where possible.
AS 91027
8 Achievement Standard 91027 (Mathematics and Statistics 1.2)
2.
Expanding brackets
Expand and simplify the following expressions.
a.
7 + 5(x – 3)
b.
8 – 2(x + 4)
c.
x – 2(3 – x)
d.
3x(4x + 1) – 6
e.
5x – 6(x + 2)
Brackets are expanded using the distributive law:
a(b + c) = ab + ac
Example
1. 5(x + 7) = 5 × x + 5 × 7
= 5x + 35
2. x(3x – 2) = 3x2 – 2x
3. –3ab2(2a + 4b) = –3ab2 × 2a – 3ab2 × 4b
= –6a2b2 – 12ab3
There may be more than two terms in the bracket.
Example
–6a(2a + b – 5) = –6a × 2a – 6a × b – 6a × –5
= –12a2 – 6ab + 30a
Some expressions involve expanding more than
one set of brackets. Take care with negative signs.
Example
1. 5(x + 2) – 6(2x – 3) = 5x + 10 – 12x + 18
= –7x + 28
2. x(x – 7) –2x(x + 3) = x2 – 7x – 2x2 – 6x
= –x2 – 13x
Ans. p. 44
3.
Expand and simplify.
a.
3(2x + 6) + 2(3x – 4)
b.
5(3x + 2) – (2x + 7)
c.
x(2x + 5) – x(3x + 4)
d.
3x(4x – 3) – 5x(2x – 3)
e.
7(2x + 5) – (8x – 1)
f.
–x(3 – x) + 2(x – 4)
Exercise G: Expanding brackets
1.
Expand the brackets.
a.
c.
e.
g.
2(x + 8)
5(2y – 3)
4x(2 + x)
b.
d.
f.
–3(x + 2)
–4(3y – 2)
6a(2a2 + a – 4)
–2xy(x2 – xy + y)
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ANSWERS
Exercise A: Algebraic
expressions
1.
2.
3.
(page 1)
2.
4.
g.
h. x
a.
3x × (x + x) = 6x2
b.
8a ÷ (4 × a) + 1 = 3
c.
x – x – (2x – x) = –x
d.
(x – 2)2
e.
g.
2d – 27
h. 4m – 58
i.
i.
3a + 2p
d.
(y + y) × y – y2 = y2
ii.
d – 3a – 2p
e.
(2p + 4p) ÷ (3p – p) = 3
iii. d ≥ 3a + 2p
f.
12w2 ÷ (w – 2w) = –12w
g.
a – (6a – 4a) + 2a = a
h.
(p + q) × (q – q) = 0
y+1
a.
12a
b. –b
d.
11p – 6q
e.
g.
x–
x2
j.
0
a.
3a
d.
3x2;
f. p + 6
–3f 2 – 3f f. 10ab + 2a
x2–
h. 3w – 5
i. 4x –
b. 3r
c. 2mn; m
e.
8d;
b. –6xy
d.
12ab
e.
6x2
g.
6y3
h.
–5x2
a.
2x
b. 2
d.
–p
2
3pq
4st
a
b
d.
3mnp2
a.
6x3
4b
h.
1
ab
y2
2x2
b.
–3x
b. x
–1
a.
–27
b. –9
c. 16
d.
1
e. –1
f. 128
g.
–125
h. 27
a.
x8
b. 21x6
c. –2x3
c. –15w
d.
8x2
e. 30a6
f. –6x5y3
f. –15abc
g.
12a6b3c
h. 54a5b3
a.
p3
b. 2x5
c. 9a3
d.
1
2x2
e. 2ab2
3
f. 3x2
g.
–1
2pq
h. 1 2
2x
a.
18ab5
b. 2x2y2
2.
c.
1
2x
f.
x
2
c.
3
2wz2
c. 4x2
4y
Exercise E: Indices with powers and
roots (page 5)
1.
pq5
Exercise C: Order of
operations
a.
1.
4.
b. 5a3b2
f.
Exercise D: Multiplying and dividing
with indices (page 4)
3.
3
e.
2
d2
14ab
a.
2.
c. s
a.
d.
1.
–4x
b. y + 4
g.
3.
e.
1
3
7x
Exercise B: Multiplying and dividing
terms (page 2)
1.
18x
a.
2x3
c. 5w + 3
d.
2.
(page 3)
c. 6y
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a.
a15
b. 4x10
c. 81x8
d.
1
e. –8w12x3
4
f. a6
g.
64x 3
27y 6
h.
a.
y8
b. x10
c. 8x32
d.
10xy2
3
e. a
2
f.
g.
3x
5
h. 5x
b
x9
8y 6
p3
t5
44 Answers
3.
3
2p2
a.
2x5
b.
d.
x2
2
e. 9a
g.
5
x2
6 8
h. 3a b
c. a
Exercise H: Expanding pairs of
brackets (page 9)
f. 324x4
1.
4
2
2
Exercise F: Basic algebraic fractions
ANSWERS
(page 6)
1.
2.
4.
5.
a
5c
f.
1
2x
b. 3x
d.
2x
6b
e. 7a
g.
3a2
4b2
h.
a.
4x
5
b. 10y
c. a
d.
3x
70
e. –y
f. 7x
13x2
20
8
15
h. 3x –2 4
g.
3.
c.
x
2y
a.
a.
3.
7
8bc
7
24
12
2
b. 3a2
c.
f. bc
2a
b
x2
6y2
d.
6
g.
8x
y
h.
a.
12
b
b. 3x
c. 2abc
d.
4x
e. 10a
3
f. 4x
g.
8a2
5b3
y
h.
x
d.
2x
3
5p
8
2.
1
12q
2
d
c
b.
4y
3
y
2.
3.
b. x2 + 2x – 80
c.
x2 + 3x – 18
d. x2 – 12x + 35
e.
2x2 + 11x + 15
f. 15x2 + 2x – 8
g.
15x2 – 29x + 12
a.
x2 + 8x + 16
b. y2 – 6y + 9
c.
x2 – 36
d. 4y2 + 20y + 25
e.
9y2 – 4
f. 9x2 – 6x + 1
a.
3x2 + 12x – 96
b. –2p2 + 16p – 32
c.
2x3 – 9x2 + 10x
d. –8 + 36a – 36a2
e.
8x
a.
7(x + 3)
b. a(3 + b)
c.
y(x – y)
d. 5x(x + 2)
e.
–3p(p – 4)
f. ab(c – b)
g.
a(a – b + c)
h. 2(4 – 2x2 + y)
a.
2x3(x2 – 2)
b. 3y2(2 + y2)
c.
x2y4(x + y)
d. 4ab2(2ab – 1)
e.
p2(p2 – p + 2)
f. 3x2(4x – 1)
g.
2x8(x2 – 2)
h. x8(1 + x8)
Exercise J: Factorising quadratics
(page 11)
2
c. 5
Exercise G: Expanding
brackets (page 8)
1.
x2 + 9x + 20
Exercise I: Factorising using the
distributive law (page 10)
1.
4x
5
e. 2x2
3y
a.
2.
a.
a.
2x + 16
b.
–3x – 6
c.
10y – 15
d.
–12y + 8
e.
8x + 4x2
f.
12a3 + 6a2 – 24a
g.
–2x3y + 2x2y2 – 2xy2
a.
5x – 8
b.
–2x
c.
3x – 6
d.
12x2 + 3x – 6
e.
–x – 12
a.
12x + 10
b.
13x + 3
c.
–x2 + x
d.
2x2 + 6x
e.
6x + 36
f.
x2 – x – 8
1.
2.
3.
4.
a.
(x + 5)(x + 2)
b. (x + 6)(x + 7)
c.
(x – 8)(x + 6)
d. (x + 8)(x – 5)
e.
(x + 4)(x – 4)
f. (5y + 1)(5y – 1)
g.
(x – 8)(x – 3)
h. (a – 7)(a + 3)
a.
3(x + 5)(x – 4)
b. 5(x + 2)(x – 2)
c.
2(x + 6)(x – 3)
d. 5(x – 4)(x + 3)
e.
2(x – 7)(x – 4)
f. 10(x + 2)(x + 1)
g.
8(x + 3)(x – 3)
h. –(x – 8)(x – 4)
a.
(2x + 1)(x + 3)
b. (3x + 2)(x + 2)
c.
(5x + 1)(x – 2)
d. (7x – 1)(x + 1)
e.
(4x + 3)(x + 2)
f. (3x – 2)(2x + 1)
g.
(11x – 3)(x – 2)
h. (2x + 5)(2 – x)
a.
(x + 4)(x – 4)
b. 4x(x + 2)
c.
(x – 4)(x + 3)
d. 3x(x – 5)
e.
cannot be factorised
© ESA Publications (NZ) Ltd, Freephone 0800-372 266, ISBN 978-1-927297-67-4
INDEX
add (fractions) 6
algebraic expression 1
algebraic fractions 12
multiply (fractions) 6
BEDMAS 3
BEMA 3
perfect squares 9
power (index) 2, 4
product 1
changing subject of formula 21
coefficient 1
difference of two squares 9
distributive law 8
divide (fractions) 6
elimination method 23
exponent 4
exponential equations 34
factorised 10
formula 20
operations 1
quadratic equations 28
quadratic expressions 11
reciprocal 6
repeated multiplication 2
simplified fraction 6
simultaneous equations 23
subject of the formula 21
substituting 20
substitution method 26
terms (algebraic) 1
index 4
indices 2
variable 1
like terms 1
linear equations 14
linear inequations 17
© ESA Publications (NZ) Ltd, Freephone 0800-372 266, ISBN 978-1-927297-67-4