Ch 05

Expanding

5
Belinda works for an
advertising company that
produces billboard
advertising. The cost of a
billboard is based on the
area of the sign and is $50
per square metre. If we
increase the length of the
sign by 2 m and the height
of the sign by 3 m, can you
write a rule for the
increase in the cost of the
billboard?
This chapter shows you
how to manipulate
algebraic terms and
expressions to place them
in the most useful form.

144 Maths Quest 9 for Victoria

Expanding single
brackets
In the previous chapter on introductory
algebra, we learned that algebra is a type of
language.
In this chapter we can take the idea fur-
ther and look at some of the ‘advanced
features’ of this language. At first we will
rely on using common numerical examples
to illustrate the techniques, but after a
while, when your confidence has increased,
the techniques will become easier to under-
stand using algebra only.

What is expanding?
Consider the English word: won’t. It really
stands for will not. In going from won’t to
will not we have expanded the word, but
the meaning remains unchanged. It is the
same with expanding in algebra; we go
from a more compact form (won’t) to an
expanded form (will not). Consider the
following example with numbers.
3(4 + 5)
How can we find its value?
We know from our work on order of operations in chapter 1 that we do brackets first,
so that:
3(4 + 5)
= 3(9)
Now the brackets mean multiply, so:
3(4 + 5)
= 3(9)
= 27
Consider now an alternative way of expanding the original expression, temporarily
‘ignoring’ order of operations.
3(4 + 5)
= 3(4) + 3(5)
This may seem unusual, or even incorrect, but it isn’t. (It is not!)
3(4 + 5)
= 3(4) + 3(5)
= 12 + 15
= 27
You can try this with any numbers you like, but the result is the same. This expan-
sion is valid and correct.
Expansion means to multiply everything inside the brackets by what is directly
outside the brackets.

Chapter 5 Expanding 145
WORKED Example 1
Expand the following expressions.
a 5(4 + 3) b 5(x + 3) c 5(x − y) d −a(x − y)
THINK WRITE
a 1 Write the expression. a 5(4 + 3)
2 Expand the brackets. = 5(4) + 5(3)
3 Multiply out the brackets. = 20 + 15
= 35
4 Check that the result is valid by simplifying the 5(4 + 3)
brackets in the original expression ﬁrst. = 5(7)
= 35, so it is indeed valid.
b 1 Write the expression. b 5(x + 3)
2 Expand the brackets. = 5(x) + 5(3)
3 Multiply out the brackets. = 5x + 15
c 1 Write the expression. c 5(x − y)
2 Expand the brackets. = 5(x) + 5(−y)
3 Multiply out the brackets. = 5x − 5y
(Remember that a positive term multiplied
by a negative term makes a negative term.)
d 1 Write the expression. d −a(x − y)
2 Expand the brackets. = −a(x) − a(−y)
3 Multiply out the brackets. = −ax + ay
(Remember that a negative term multiplied
by a negative term makes a positive term.)

Note: It doesn’t matter what is immediately outside the brackets. It may be a number or
a pronumeral or both. The following expansions are a little more complex.

WORKED Example 2
Expand each of the following.
a 5x(6y − 7z) b −4y(2x + 3w) c x(2x + 3y)
THINK WRITE
a 1 Write the expression. a 5x(6y − 7z)
2 Expand the brackets. = 5x(6y) + 5x(−7z)
3 Multiply out the brackets. = 30xy − 35xz
(Multiply number parts and pronumeral parts
separately and write pronumerals for each term
in alphabetical order.)
b 1 Write the expression. b −4y(2x + 3w)
2 Expand the brackets. = −4y(2x) − 4y(3w)
3 Multiply out the brackets. = −8xy − 12wy
c 1 Write the expression. c x(2x + 3y)
2 Expand the brackets. = x(2x) + x(3y)
3 Multiply out the brackets. = 2x2 + 3xy
(Remember that x multiplied by itself gives x2.)


Expanding and collecting like terms
With more complicated expansions, like terms may need to be collected after the
expansion of the bracketed part. Remember that like terms contain the same
pronumeral parts. You ﬁrst expand the brackets, then collect the like terms.

WORKED Example 3
Expand and simplify by collecting like terms.
a 4(x − 4) + 5 b x(y − 2) + 5x c x(y − z) + 5x d 7x + 6(y − 2x)
THINK WRITE
a 1 Write the expression. a 4(x − 4) + 5
2 Expand the brackets. = 4(x) + 4(−4) + 5
3 Multiply out the brackets. = 4x − 16 + 5
4 Collect any like terms. = 4x − 11
b 1 Write the expression. b x(y − 2) + 5x
2 Expand the brackets. = x(y) + x(−2) + 5x
3 Multiply out the brackets. = xy − 2x + 5x
4 Collect any like terms. = xy + 3x
c 1 Write the expression. c x(y − z) + 5x
2 Expand the brackets. = x(y) + x(−z) + 5x
3 Multiply out the brackets. = xy − xz + 5x
4 Collect any like terms. There are no like terms.
d 1 Write the expression. d 7x + 6(y − 2x)
2 Expand the brackets. = 7x + 6(y) + 6(−2x)
3 Multiply out the brackets. = 7x + 6y − 12x
4 Collect any like terms. = −5x + 6y

remember
remember
1. Expansion means to multiply everything inside the brackets by what is directly
outside the brackets.
2. After expanding brackets, simplify by collecting any like terms.

reads
L Sp he 5A Expanding single brackets
et
EXCE

Expanding
single WORKED 1 Expand the following expressions.
brackets Example
1
a 3(x + 2) b 4(x + 3) c 5(m + 4) d 2(p + 5)
d
hca e 4(x + 1) f 7(x − 1) g −4(y + 6) h −5(a + 1)
Mat

Expanding
i −3(p − 2) j −(x − 1) k −(x + 3) l −(x − 2)
single m 3(2b − 4) n 8(3m − 2) o −6(5m − 4) p −3(9p − 5)
brackets
WORKED 2 Expand each of the following.
Example
ogram a x(x + 2) b y(y + 3) c a(a + 5) d c(c + 4)
2
GC pr

e x(4 + x) f y(5 + y) g m(7 − m) h q(8 − q)
Expanding
i 2x(y + 2) j 5p(q + 4) k −3y(x + 4) l −10p(q + 9)
m −b(3 − a) n −7m(5 − n) o −6a(5 − 3a) p −4x(7 − 4x)

WORKED 3 Expand and simplify by collecting like terms. 5.1
Example
a 2(p − 3) + 4 b 5(x − 5) + 8 c −7(p + 2) − 3 HEET

SkillS
3
d −4(3p − 1) − 1 e 6x(x − 3) − 2x f 2m(m + 5) − 3m
g 3x(p + 2) − 5 h 4y(y − 1) + 7 i −4p(p − 2) + 5p
j 5(x − 2y) − 3y − x k 2m(m − 5) + 2m − 4 l −3p(p − 2q) + 4pq − 1
m −7a(5 − 2b) + 5a − 4ab n 4c(2d − 3c) − cd − 5c o 6p + 3 − 4(2p + 5)
p 5 − 9m + 2(3m − 1)

Oops! Any errors?
Here are 6 expressions that someone has simplified. Have any errors been made?
a 5(x − 1) = 5x − 1
b 3x + 6x = 9 + x
c 4(3x) = 12x
d 8x − 3x = 5
e −2(x − 7) = −2x −14
f x(x + 5) = 2x + 5x
1 Which expressions have been simplified correctly?
2 Explain why someone might make the errors you have found.
3 Correct any errors you find by rewriting the expression on the right-hand side of
the equals sign.
4 Choose 2 values for x and evaluate the left- and right-hand sides to check
whether they are now equivalent.

The Bagels game
In the game of Bagels, a player is to determine a 3-digit number (no digit repeated)
by making educated guesses. After each guess, a clue is given about the guess.
Here are the clues.
bagels: no digit correct
pico: one digit is correct but in the wrong position
fermi: one digit is correct and in the correct position
1 In each of the problems below, a number of guesses have been made with the
clue for each guess shown to its right. From the given set of guesses and clues,
determine the 3-digit number.
a 123 bagels b 908 bagels
456 pico 134 pico
789 pico 387 pico fermi
075 pico fermi 256 fermi
087 pico 237 pico pico
??? ???
2 Now try this game with a partner. One person is to decide on a 3-digit number
and provide clues to the other person who is guessing what the 3-digit number is.


What is a metre?
metre?
Expand the brackets in the
expressions given to find the puzzle’s
answer code.

e – 2ae 6e – 3 e 2 e – 2ae a 2e – ae 2 5e – 10 8 – 4e –4a – 6 e – 2ae 2ae – 3e 8 – 4e 5e – 10 6a – 2 a 2e – ae 2 6a – 2

6e – 3e 2 8 – 4e 2ae – 3e a 2 – ae e – 2ae e 2 + e e 2 + e e – 2ae a 2 – 4a 2ae – 3e 6e – 3e 2 3a + ae a 2 – 4a – 4a – 6

2 2
6e – 3e 2 3a + ae 8 – 4e 5e – 10 e – 2ae a 2e – ae 2 6e – 3e 2 6a – 2 2ae – 3e 6 – 3a 8 – 4e –4a – 6 2ae – 2ae 2 a – 4a a – ae

6e – 3 e 2 3a + ae 8 – 4e 2ae – 3e a 2 – 4a 2ae – 2ae 2 6e – 3 e 2 3a + ae 3ae + 6e a 2 – 4a e2 + e 8 – 4e

6e – 3 e 2 a 2 – 4a 6e – 3 e 2 3a + ae 8 – 4e 8 – 4e 5a – 10ae 4 a 2 – 4a 6a – 2 6e – 3 e 2 a 2 – 4 a 2ae – 2ae 2

= – a (e – a )
=
= 2(3 a – 1)
= = – e (–2 a + 3)
=
= 3(2 – a )
= = – a (4 – a )
=
= 5(e – 2)
= = 3 e (a + 2)
=
= –4(e – 2)
= = 5 a (1 – 2 e )
=
= –2(2 a + 3)
= = –2 ae (e – 1)
=
= a (3 + e )
= = ae (a – e )
=
= e (1 – 2 a )
= = 3 e (2 – e )
=
= e (e + 1)
=
= –4 a (– a + 1)
=

Expanding two brackets
When expanding an expression that contains two (or more) brackets, the steps are the
same as before.
Step 1 Expand each bracket (working from left to right).
Step 2 Collect any like terms.

WORKED Example 4
Expand and simplify the following expressions.
a 5(x + 2y) + 6(x − 3y) b −5x(y − 2) + y(x + 3)
c 7y(x − 2y) + y2(x + 5) d −5xy(1 + 2y) + 6x(y + 4x)
THINK WRITE
a 1 Write the expression. a 5(x + 2y) + 6(x − 3y)
2 Expand each bracket. = 5(x) + 5(2y) + 6(x) + 6(−3y)
3 Multiply out the brackets. = 5x + 10y + 6x − 18y
4 Collect any like terms. = 11x − 8y
b 1 Write the expression. b −5x(y − 2) + y(x + 3)
2 Expand each bracket. = −5x(y) − 5x(−2) + y(x) + y(3)
3 Multiply out the brackets. = −5xy + 10x + xy + 3y
4 Collect any like terms. = −4xy + 10x + 3y
c 1 Write the expression. c 7y(x − 2y) + y2(x + 5)
2 Expand each bracket. = 7y(x) + 7y(−2y) + y2(x) + y2(5)
3 Multiply out the brackets. = 7xy − 14y2 + xy2 + 5y2
4 Collect any like terms. = 7xy − 9y2 + xy2
d 1 Write the expression. d −5xy(1 + 2y) + 6x(y + 4x)
2 Expand each bracket. = −5xy(1) − 5xy(2y) + 6x(y) + 6x(4x)
3 Multiply out the brackets. = −5xy −10xy2 + 6xy + 24x2
4 Collect any like terms. = xy − 10xy2 + 24x2

As you can see, there is really no difference between the questions in this section and
the previous section; just a little more complex ‘bookkeeping’ is required. Be careful
when collecting like terms which may only appear to be like. For example, 4x2y and
4xy2 are not like terms at all.

remember
remember
To expand an expression:
1. expand each bracket (working from left to right)
2. collect any like terms.


5B Expanding two brackets

5.2 WORKED 1 Expand and simplify the following expressions.
HEET Example
a 2(x + 2y) + 3(2x − y) b 4(2p + 3q) + 2(p − 2q)
SkillS

4a
c 7(2a + 3b) + 4(a + 2b) d 5(3c + 4d) + 2(2c + d)
e −4(m + 2n) + 3(2m − n) f −3(2x + y) + 4(3x − 2y)
d
g −2(3x + 2y) + 3(5x + 3y) h −5(4p + 2q) + 2(3p + q)
hca
i 6(a − 2b) − 5(2a − 3b) j 5(2x − y) − 2(3x − 2y)
Mat

Expanding k 4(2p − 4q) − 3(p − 2q) l 2(c − 3d) − 5(2c − 3d)
two m 7(2x − 3y) − (x − 2y) n −5(p − 2q) − (2p − q)
brackets o −3(a − 2b) − (2a + 3b) p 4(3c + d) − (4c + 3d)
ogram WORKED 2 Expand and simplify the following expressions.
Example
GC pr

a a(b + 2) + b(a − 3) b x(y + 4) + y(x − 2)
Expanding 4b, c, d c c(d − 2) + c(d + 5) d p(q − 5) + p(q + 3)
e 3c(d − 2) + c(2d − 5) f 7a(b − 3) − b(2a + 3)
g 2m(n + 3) − m(2n + 1) h 4c(d − 5) + 2c(d − 8)
i 3m(2m + 4) − 2(3m + 5) j 5c(2d − 1) − (3c + cd)
k −3a(5a + b) + 2b(b − 3a) l −4c(2c − 6d) + d(3d − 2c)
m 6m(2m − 3) − (2m + 4) n 2p(p − 4) + 3(5p − 2)
o 7x(5 − x) + 6(x − 1) p −2y(5y − 1) − 4(2y + 3)
3 multiple choice
a What is the equivalent of 3(a + 2b) + 2(2a − b)?
A 5a + 6b B 7a + 4b C 5(3a + b) D 7a + 8b E 12a − 12b
b What is the equivalent of −3(x − 2y) − (x − 5y)?
A −4x + 11y B −4x − 11y C 4x + 11y D 4x + 7y E 3x + 30y
c What is the equivalent of 2m(n + 4) + m(3n − 2)?
A 3m + 4n − 8 B 5mn + 4m C 5mn + 10m D 5mn + 6m E 6mn − 16m

QUEST
S
M AT H

GE

1 Find the next 3 terms of this algebraic sequence.
EN

x + 3y, 2x + y, 3x + 4y, 5x + 5y, 8x + 9y, 13x + 14y, ….
CH L 2 Mind-reading tricks often use algebra as a base. Try the following
AL mind-reading trick. Use algebra to explain why the trick works.
Double the number of the month in which you were born. Subtract
16 from your answer. Subtract 20 from your result, then multiply by 10.
Finally, add the day of the month in which you were born to your
answer. The number you end up with shows the month and day you
were born. For example, if you were born on June 17, your answer will
be 617.
Try this trick on another person. ‘Read’ the person’s mind by stating
the month and day they were born from the number they tell you at the
end of the calculation.

Expanding pairs of brackets
In the previous section we expanded expressions with two brackets which were separated
by a + or − sign, such as 5(x + 2y) + 6(x − 3y). In this section we begin to look at
expressions where there are two brackets being multiplied together, such as (x + 2y)
(x − 3y). These require a more careful analysis and technique.
Let us again refer to a numerical example:
(2 + 6)(7 − 3)
The ‘traditional’ approach, using order of operations from chapter 1, results in:
(2 + 6)(7 − 3)
= (8)(4)
= 32
Consider the ‘alternative’ approach. First multiply the 2 by the second bracket and
then the 6 by the second bracket.
(2 + 6)(7 − 3)
= 2(7 − 3) + 6(7 − 3)
= 2(7) + 2(−3) + 6(7) + 6(−3)
= 14 − 6 + 42 − 18
= 32
Again, we end up with identical results, no matter which method is used. It may
appear unnecessarily long for numeric examples (and, indeed, it is) but it works well
for algebraic expressions.
When multiplying expressions within brackets, multiply each term in the first
bracket by each term in the second bracket.

WORKED Example 5
Expand and simplify each of the following expressions.
a (6 − 5)(15 + 3) b (x − 5)(x + 3) c (x + 2)(x + 3) d (2x + 2)(2x + 3)
THINK WRITE
a 1 Write the expression. a (6 − 5)(15 + 3)
2 Expand by multiplying each of the = 6(15 + 3) − 5(15 + 3)
terms in the first bracket by each of
the terms in the second bracket.
3 Expand each of the remaining brackets. = 6(15) + 6(3) − 5(15) − 5(3)
4 Simplify. = 90 + 18 − 75 − 15
= 18
5 Check the results using the order of (6 − 5)(15 + 3)
operations method. = (1)(18)
= 18
b 1 Write the expression. b (x − 5)(x + 3)
2 Expand by multiplying each of the = x(x + 3) − 5(x + 3)
terms in the first bracket by each of
the terms in the second bracket.
3 Expand each of the remaining = (x)(x) + x(3) − 5(x) − 5(3)
brackets. = x2 + 3x − 5x − 15
4 Collect like terms. = x2 − 2x − 15 Continued over page


THINK WRITE
c 1 Write the expression. c (x + 2)(x + 3)
2 Expand by multiplying each of the terms in the first = x(x + 3) + 2(x + 3)
bracket by each of the terms in the second bracket.
3 Expand each of the remaining brackets. = x(x) + x(3) + 2(x) + 2(3)
= x2 + 3x + 2x + 6
4 Collect like terms. = x2 + 5x + 6
d 1 Write the expression. d (2x + 2)(2x + 3)
2 Expand by multiplying each of the terms in the first = 2x(2x + 3) + 2(2x + 3)
bracket by each of the terms in the second bracket.
3 Expand each of the remaining brackets. = 2x(2x) + 2x(3) + 2(2x) + 2(3)
= 4x2 + 6x + 4x + 6
4 Collect like terms. = 4x2 + 10x + 6

Note: The last step in each question (collection of like terms), is most important, and
often forgotten. Without completing this step, the expansion is not fully correct.

An alternative method
Part of the problem in expanding pairs of brackets is a bookkeeping one — keeping
track of which terms have been multiplied by which. An alternative approach uses a
diagram to keep track of the various multiplication operations.

WORKED Example 6
Use a diagrammatic technique to expand (2x + 3y)(4x − 5z).
THINK WRITE
1 Write the expression and add 4 curved lines 2
connecting each term, according to the pattern shown.
1
2 Number each of the curved lines.
(2x + 3y)(4x – 5z)
3
4
3 Perform each of the multiplications, in order of the 1: 2x(4x) = 8x2
numbers on the lines. 2: 2x(−5z) = −10xz
3: 3y(4x) = 12xy
4: 3y(−5z) = −15yz
4 Write the expression and its expansion by placing the (2x + 3y)(4x − 5z)
terms on a single line. = 8x2 − 10xz + 12xy − 15yz
5 Collect like terms if necessary (none in this example).

There is nothing magical or special about this method, but it forces you to keep track of
the 4 multiplications required in the expansion.
This method is often given the name FOIL, where the letters stand for: O
First — multiply the first term in each bracket. F
Outer — multiply the 2 outer terms. (a + b) (c + d)
Inner — multiply the 2 inner terms. I
L
Last — multiply the last term of each bracket.

remember
remember
1. When multiplying expressions within pairs of brackets, multiply each term in
the first bracket by each term in the second bracket, then collect the like terms.
2. You can use a ‘diagrammatic method’ (or FOIL) to help you keep track of
which terms are to be multiplied together.

5C Expanding pairs of brackets

WORKED 1 Expand and simplify each of the following expressions. Math
Example
a (a + 2)(a + 3) b (x + 4)(x + 3) c (y + 3)(y + 2)

cad
5
d (m + 4)(m + 5) e (b + 2)(b + 1) f (p + 1)(p + 4) Expanding
pairs of
g (a − 2)(a + 3) h (x − 4)(x + 5) i (m + 3)(m − 4) brackets
j (y + 5)(y − 3) k (y − 6)(y + 2) l (x − 3)(x + 1)
m (x − 3)(x − 4) n (p − 2)(p − 3) o (x − 3)(x − 1)
WORKED 2 Use a diagrammatic technique to expand the following. GC pro
Example
a (2a + 3)(a + 2) b (3m + 1)(m + 2) (6x + 4)(x + 1)

gram
6
c
d (c − 6)(4c − 7) e (7 − 2t)(5 − t) f (1 − x)(9 − 2x) Expanding
g (2 + 3t)(5 − 2t) h (7 − 5x)(2 − 3x) i (5x − 2)(5x − 2)
3 Expand and simplify each of the following.
a (x + y)(z + 1) b (p + q)(r + 3) c (2x + y)(z + 4)
d (3p + q)(r + 1) e (a + 2b)(a + b) f (2c + d)(c − 3d)
g (x + y)(2x − 3y) h (4p − 3q)(p + q) i (3y + z)(x + z)
j (a + 2b)(b + c) k (3p − 2q)(1 − 3r) l (7c − 2d)(d − 5)
m (4x − y)(3x − y) n (p − q)(2p − r) o (5 − 2j)(3k + 1)
GAME
4 multiple choice

time
a The equivalent of (x + 7)(x − 2) is: Expanding
A x2 + 5x − 14 B 2x + 5 C x2 − 5x − 14 D x2 + 5x + 14 E x2 − 5x + 14 — 001
b What is the equivalent of (4 − y)(7 + y)?
A 28 − y2 B 28 − 3y + y2 C 28 − 3y − y2 D 11 − 2y E 28 + 3y − y2 SHE 5.1
ET
Work
c The equivalent of (2p + 1)(p − 5) is:
A 2p2 − 5 B 2p2 − 11p − 5 C 2p2 − 9p − 5 D 2p2 − 6p − 5 E 2p2 + 9p − 5

QUEST
S
M AT H

GE

1 Xavier left for school in the morning. One quarter of the way to school,
he passed a post office. The clock on the outside of the post office
EN

showed 7.44 am. Halfway to school, he passed a convenience store. The
CH L time shown there was 7.53 am. If Xavier continues walking at the same
AL
speed, at what time will he get to school?
2 The human heart beats about 105 times each day. Approximately how
many times does the heart beat in an 80-year lifetime?
3 There are 12 people trying out for a tennis team. Five of them are girls. What
percentage of the possible doubles teams could be mixed double teams?


What has area got to do with
expanding?
1 Draw a square of side length x. x 3
2 What is the area of this square?
Consider the rectangle at right which has x
a length 3 units longer and a width 2 units
wider than the square you have just drawn.
Notice that it has been divided up into
4 regions.
3 Find the area of each of the 4 regions. 2
4 What is the total area of the rectangle?
5 Write an expression for the length of
the rectangle.
6 Write an expression for the width of the rectangle.
7 Using the relationship, area = length × width and your answers to parts 5 and
6, write an expression for the area of the rectangle.
8 Relate your answers to parts 4 and 7. What have you noticed?
9 Show, using a diagram and areas of 4 regions, how to obtain the expanded
expression for (x + 5)(x + 7).
10 Show, using a diagram and areas of 4 regions, how to obtain the expanded
expression for (a + b)(c + d).
11 Challenge: Show, using a diagram and areas of 4 regions, how to obtain the
expanded expression for (x − 2)(x + 3).

1
1 Expand 5(x + 3).
2 Expand z(3 − 7z).
3 Expand 2(p − 7q) + 3p − 5q and simplify by collecting like terms.
4 Expand and simplify 5(a + 2b) + 2(3a + b).
5 Expand and simplify m(n + 1) + n(m − 4).
6 multiple choice
If K(a − 3b) − (a + 10b) = 3a − 22b, then the missing number is:
A 1 B 2 C 3 D 4 E 5
7 Expand and simplify (a + 1)(a + 4).
8 Expand and simplify (p − 7)(6p − 3).
9 Expand and simplify (4 − 3j)(4k + 2).
10 True or false? (x + 4)(x + 10) = x2 + 14x + 40

Expansion patterns
Although the techniques learned in the previous section are perfectly adequate for all
expansions of pairs of brackets, there are some ‘special’ cases where the expansion is
particularly simple and can be done very quickly if you recognise the pattern. After
comparing the result with that obtained using previous methods, perhaps you will adopt
these ‘short cuts’.

Difference of two squares rule
The ﬁrst pattern we will examine is obtained as a result of expanding a pair of brackets
to produce a ‘difference of two squares’.
That is, we produce two terms which are perfect squares (can be expressed as a
number and/or a pronumeral squared) where one term is subtracted from the other.
Consider expanding (x + 3)(x − 3).
(x + 3)(x − 3) = x(x − 3) + 3(x − 3)
= x(x) + x(−3) + 3(x) + 3(−3)
= x2 − 3x + 3x − 9
= x2 − 9
Notice how the ‘middle terms’, −3x + 3x cancel each other out. This is the key to the
pattern and will always happen. (Can you prove this?) Note: The terms left over are the
squares of each of the original terms.
In other words, (x + 3)(x − 3) = x2 − 32.
Notice the pattern of terms in the pair of brackets which produce the difference of
two squares.
Here are some more examples.
(x + 5)(x − 5) (x + 4)(x − 4) (x + h)(x − h) (2x + 7)(2x − 7)
= x2 − 52 = x2 − 42 = x2 − h2 = (2x)2 − 72
= x2 − 25 = x2 − 16 = 4x2 − 49
Therefore, when we recognise this pattern we merely have to write the squares of
each of the two terms and place a minus sign between them.
(a + b)(a − b) = a2 − b2

WORKED Example 7
Use the difference of two squares rule, if possible, to expand and simplify each of the
following.
a (x + 8)(x − 8) b (6 − x)(6 + x) c (2x − 3)(2x + 3) d (5 + 3x)(5 − 3x)
THINK WRITE
a 1 Write the expression. a (x + 8)(x − 8)
2 Check that the expression can be
written as the difference of two
squares by comparing it with
(a + b)(a − b). It can.
3 Write the answer as the difference of = x2 − 82
two squares using the formula = x2 − 64
(a + b)(a − b) = a − b , where a = x
2 2
Continued over page
and b = 8.


THINK WRITE
b 1 Write the expression. b (6 − x)(6 + x)
2 Check that the difference of two squares rule can be
used. It can, because (6 − x)(6 + x) means the same
as (6 + x)(6 − x).
3 Write the answer as the difference of two squares = 62 − x2
using the formula = 36 − x2
(a + b)(a − b) = a2 − b2, where a = 6 and b = x.
c 1 Write the expression. c (2x − 3)(2x + 3)
used. It can.
3 Write the answer as the difference of two squares = (2x)2 − 32
using the formula = 4x2 − 9
(a + b)(a − b) = a2 − b2, where a = 2x and b = 3.
d 1 Write the expression. d (5 + 3x)(5 − 3x)
used. It can.
3 Write the answer as the difference of two squares = 52 − (3x)2
using the formula = 25 − 9x2
(a + b)(a − b) = a2 − b2, where a = 5 and b = 3x.

Expanding identical brackets (perfect squares)
The next pattern worth examining is the expansion of identical brackets. That is, each
bracket is the same, such as (x + 3)(x + 3), which can be written as (x + 3)2. Again, we
can express this using the two symbols a and b. So a perfect square could be written in
the form (a + b)(a + b).
Let us try this with a set of numbers ﬁrst; using the pair of brackets (3 + 4)(3 + 4)
which equals 49.
(3 + 4)(3 + 4) = 3(3 + 4) + 4(3 + 4)
= 3(3) + 3(4) + 4(3) + 4(4) (Can you see a pattern here?)
= 9 + 12 + 12 + 16
= 49
Now let’s try it with pronumerals.
(x + 3)(x + 3) = x(x + 3) + 3(x + 3)
= x(x) + x(3) + 3(x) + 3(3)
= x2 + 3x + 3x + 9
= x2 + 6x + 9
There is a pattern in these expansions, fairly similar to the one we just learned.
(a + b)(a + b) = a2 + 2ab + b2
There is an additional result that occurs when there is a minus sign instead of a plus
sign:
(a − b)(a − b) = a2 − 2ab + b2
The difference between the two patterns is quite small: the minus sign in the brackets
results in a single minus sign in the ‘middle term’, −2ab.
This pattern can also be described in words.
Square the ﬁrst term, add the square of the last term, then add (or subtract) twice
their product.

WORKED Example 8
Use the identical brackets (perfect squares) technique to expand and simplify the
following.
a (x + 1)(x + 1) b (x − 2)2 c (2x + 5)2 d (4x − 5y)2
THINK WRITE
a 1 Write the expression. a (x + 1)(x + 1)
2 Recognise the pattern of identical (x)(x) = x2
brackets: square the ﬁrst term.
3 Square the last term. (1)(1) = 1
4 Add (because of the + sign in the 2(x)(1) = 2x
bracket) twice the product.
5 Apply the formula: (x + 1)(x + 1) = x2 + 2x + 1
(a + b)(a + b) = a2 + 2ab + b2.
b 1 Write the expression and express it b (x − 2)2 = (x − 2)(x − 2)
as a pair of brackets.
2 Recognise the pattern of identical (x)(x) = x2
3 Square the last term. (−2)(−2) = 4
4 Subtract (because of the − sign in the −2(x)(2) = −4x
bracket) twice the product.
5 Apply the formula: (x − 2)2 = x2 − 4x + 4
(a − b)(a − b) = a2 − 2ab + b2.
c 1 Write the expression and express it c (2x + 5)2
as a pair of brackets. = (2x + 5)(2x + 5)
2 Recognise the pattern of identical (2x)(2x) = 4x2
3 Square the last term. (5)(5) = 25
4 Add twice the product. 2(2x)(5) = 20x
5 Apply the formula: (2x + 5)2 = 4x2 + 20x + 25
(a + b)(a + b) = a2 + 2ab + b2.
d 1 Write the expression and express it d (4x − 5y)2
as a pair of brackets. = (4x − 5y)(4x − 5y)
2 Recognise the pattern of identical (4x)(4x) = 16x2
3 Square the last term. (−5y)(−5y) = 25y2
4 Subtract twice the product. −2(4x)(5y) = −40xy
5 Apply the formula: (4x − 5y)2 = 16x2 − 40xy + 25y2
(a − b)(a − b) = a2 − 2ab + b2.

remember
remember
1. The difference of two squares rule is: (a + b)(a − b) = a2 − b2.
2. The identical brackets (perfect squares) rules are:
(a + b)(a + b) = a2 + 2ab + b2
(a − b)(a − b) = a2 − 2ab + b2.


5D Expansion patterns

reads
L Sp he
WORKED 1 Use the difference of two squares rule, if possible, to expand and simplify each of the
Example
et
EXCE

following.
7a, b
Expanding a (x + 2)(x − 2) b (y + 3)(y − 3) c (m + 5)(m − 5)
(ax + b)(ax – b) d (a + 7)(a − 7) e (x + 6)(x − 6) f (p − 12)(p + 12)
g (a + 10)(a − 10) h (m − 11)(m + 11) i (p − q)(p + q)
hca
d WORKED 2 Use the difference of two squares rule, if possible, to expand and simplify each of the
Example
Mat

following.
7c, d
Expansion a (2x + 3)(2x − 3) b (3y − 1)(3y + 1) c (5d − 2)(5d + 2)
patterns d (7c + 3)(7c − 3) e (2 + 3p)(2 − 3p) f (1 − 9x)(1 + 9x)
g (5 − 12a)(5 + 12a) h (3 + 10y)(3 − 10y) i (2b − 5c)(2b + 5c)
WORKED 3 Use the identical brackets (perfect squares) rules to expand and simplify each of the
Example
following.
8a, b
a (x + 2)(x + 2) b (a + 3)(a + 3) c (b + 7)(b + 7)
d (c + 9)(c + 9) e (m + 12)2 f (n + 10)2
g (x − 6)2 h (y − 5)2 i (9 − c)2
j (8 + e) 2
k (x + y) 2
l (u − v)2
reads
L Sp he
WORKED 4 Use the identical brackets (perfect squares) rules to expand and simplify each of the
Example
et
EXCE

following.
8c, d
Expanding a (2a + 3)2 b (3x + 1)2 c (2m − 5)2 d (4x − 3)2
(ax + b)2 e (5a − 1) 2
f (7p + 4) 2
g (9x + 2) 2
h (4c − 6)2
i (3 + 2a) 2
j (5 + 3p) 2
k (2 − 5x) 2
l (7 − 3a)2
m (9x − 4y) 2
n (8x − 3y) 2
o (9x − 2y) 2
p (7x − 4y)2

Using expanding formulas to square
large numbers
Can you evaluate 9972 without a calculator and in less than 90 seconds? We would be
able to evaluate this using long multiplication, but it would take a fair amount of time
and effort. Mathematicians are always looking for quick and simple ways of solving
problems.
What if we consider the expanding formula which produces the difference of two
squares?
(a + b)(a − b) = a2 − b2
Adding b2 to both sides gives (a + b)(a − b) + b2 = a2 − b2 + b2.
Simplifying and swapping sides gives a2 = (a + b)(a − b) + b2.
We can use this new formula and the fact that multiplying by 1000 is an easy
operation to evaluate 9972.
1 If a = 997, what should we make the value of b become so that (a + b) equals 1000?
2 Substitute these a and b values into the formula to evaluate 9972.
3 Try this method to evaluate the following.
a 9952 b 9902 c 992 d 99912 e 99 9982
4 Can you use the expanding formulas (a + b) = a + 2ab + b2 or
2 2

(a − b)2 = a2 − 2ab + b2 to evaluate 9972? Explain your method for this.
5 List three examples of your own and show how you were able to evaluate them
using the method from part 4.

More complicated expansions
Although we have covered many expansion problems and patterns, these represent only
a small proportion of the possible algebraic expressions that can be expanded. Never-
theless, the techniques we have learned so far are very useful — more complicated
expansions just require more ‘bookkeeping’. The most important of these bookkeeping
functions is the collection of like terms after expansion.
Expanding more than two brackets
There are several possible combinations, such as expanding 3 brackets, 4 brackets, and so on.

WORKED Example 9
a (x + 3)(x + 4) + 4(x − 2) b (x − 2)(x + 3) − (x − 1)(x + 2)
c (x + 2)(x − 2) − (x + 2)(x + 2) d 2(x + 3)(x − 4) + (x − 2)2
THINK WRITE
a 1 Write the expression. a (x + 3)(x + 4) + 4(x − 2)
2 Expand and simplify the first pair of (x + 3)(x + 4)
brackets. = x2 + 4x + 3x + 12
= x2 + 7x + 12
3 Expand the last bracket. 4(x − 2)
= 4x − 8
4 Add the two results. (x + 3)(x + 4) + 4(x − 2)
= x2 + 7x + 12 + 4x − 8
5 Collect like terms. = x2 + 11x + 4
b 1 Write the expression. b (x − 2)(x + 3) − (x − 1)(x + 2)
2 Expand and simplify the first pair of (x − 2)(x + 3)
brackets. = x2 + 3x − 2x − 6
= x2 + x − 6
3 Expand and simplify the second pair of (x − 1)(x + 2)
brackets. = x2 + 2x − x − 2
= x2 + x − 2
4 Subtract all of the second result from the first (x − 2)(x + 3) − (x − 1)(x + 2)
result. (So, place the second result in a bracket.) = x2 + x − 6 − (x2 + x − 2)
Remember that −(x + x − 2) = −1(x + x − 2).
2 2

5 Collect like terms. = x2 + x − 6 − x2 − x + 2
= −4
c 1 Write the expression. c (x + 2)(x − 2) − (x + 2)(x + 2)
2 Expand the first pair of brackets. It is a (x + 2)(x − 2)
difference of two squares expansion. = x2 − 4
3 Expand the second pair of brackets. It is an (x + 2)(x + 2)
identical bracket expansion. = x2 + 4x + 4
4 Subtract the two results. (x + 2)(x − 2) − (x + 2)(x + 2)
= x2 − 4 − (x2 + 4x + 4)
5 Collect like terms. = x2 − 4 − x2 − 4x − 4
= −4x − 8
Continued over page


THINK WRITE

d 1 Write the expression. d 2(x + 3)(x − 4) + (x − 2)2
2 Expand and simplify the ﬁrst pair of 2(x + 3)(x − 4)
brackets. Then multiply by the = 2(x2 − 4x + 3x − 12)
coefﬁcient of 2 outside the pair. = 2(x2 − x − 12)
= 2x2 − 2x − 24
3 Expand the second pair of brackets. (x − 2)2
It is an identical bracket expansion. = x2 − 4x + 4
4 Add the two results. 2(x + 3)(x − 4) + (x − 2)2
= 2x2 − 2x − 24 + (x2 − 4x + 4)
5 Collect like terms. = 2x2 − 2x − 24 + x2 − 4x + 4
= 3x2 − 6x − 20

remember
remember
1. Brackets or pairs of brackets that are added or subtracted must be expanded
separately.
2. Always collect any like terms following an expansion.

More complicated
5E expansions

HEET
5.3 WORKED
Example
1 (x + 3)(x + 5) + (x + 2)(x + 3) 2 (x + 4)(x + 2) + (x + 3)(x + 4)
SkillS

9
3 (x + 5)(x + 4) + (x + 3)(x + 2) 4 (x + 1)(x + 3) + (x + 2)(x + 4)
5 (p − 3)(p + 5) + (p + 1)(p − 6) 6 (a + 4)(a − 2) + (a − 3)(a − 4)
HEET
5.4
7 (p − 2)(p + 2) + (p + 4)(p − 5) 8 (x − 4)(x + 4) + (x − 1)(x + 20)
SkillS

9 (y − 1)(y + 3) + (y − 2)(y + 2) 10 (d + 7)(d + 1) + (d + 3)(d − 3)
11 (x + 2)(x + 3) + (x − 4)(x − 1) 12 (y + 6)(y − 1) + (y − 2)(y − 3)
hca
d
13 (x + 2) + (x − 5)(x − 3)
2
14 (y − 1)2 + (y + 2)(y − 4)
Mat

More 15 (p + 2)(p + 7) + (p − 3)2 16 (m − 6)(m − 1) + (m + 5)2
complicated 17 (x + 3)(x + 5) − (x + 2)(x + 5) 18 (x + 5)(x + 2) − (x + 1)(x + 2)
expansions
19 (x + 3)(x + 2) − (x + 4)(x + 3) 20 (m − 2)(m + 3) − (m + 2)(m − 4)
ogram
21 (b + 4)(b − 6) − (b − 1)(b + 2) 22 (y − 2)(y − 5) − (y + 2)(y + 6)
GC pr

Expanding 23 (p − 1)(p + 4) − (p − 2)(p − 3) 24 (x + 7)(x + 2) − (x − 3)(x − 4)
25 (c − 2)(c − 1) − (c + 6)(c + 7) 26 ( f − 7)( f + 2) − ( f + 4)( f + 5)
27 (m + 3) − (m + 4)(m − 2)
2
28 (a − 6)2 − (a − 2)(a − 3)
29 (p − 3)(p + 1) − (p + 2)2 30 (x + 5)(x − 4) − (x − 1)2

Why
Why does the giraffe have a long neck?
giraffe hav

Expand and simplify the expressions
to find the puzzle’s answer.

= (x – 4)(x + 3) – (x + 5)(x – 3)

=
= 2(x + 3) + 3(x – 1) = 2x(x – 3) + 3x(6 – x)

= =
= 5(x 2 + 3x + 2) – 4(x 2 + 3x)

= (x – 3)(x + 2) + 6 =
= 5x(2 – x) – 3(3x + 1)

= =
= (x – 1)(x + 6) + 3(x – 4)

= x (x – 4) + 2(2x + 1) =
= 3(x 2 – x + 1) – 2(x + 2)

= =
= (x + 2) 2 + 3(x – 2)

=
= (x + 5)(x – 2) + 2(x + 4) = (x – 5)(x + 5) – (x + 4)(x – 4)

= =
= –3(x + 7) – (x + 5)(x – 8)

=
= (x – 3)(x + 3) – x (x – 2) = 5(x – 7) + 2(x 2 – 3x + 18)

= =
= (x – 6)(x – 7) + 2(7x – 21)

= = (x – 2)2 – (x + 1)2
= 5(x + 1) – 2(x – 2)

= =
= (2x – 3)(3x – 2) – 6

=
= (x + 2)(x – 2) + (x + 3)(x – 4) = (2 + x)(3 – x) + (x + 1)(x + 4)

= =
= 7(2 – 3x) + 5(4x – 3)

=
= 2(x – 7)(x + 5) + 5(x + 14) = –3(x 2 + x – 2) + 5(x – 1)

= =
= 4(2x – 6) – 3(x – 8)

=
= –2( x 2 + 3x – 6) – 12 = (x + 2)(x – 2) – (x – 1)(x + 1)

= =
= 2(x 2 + 3x – 7) – 2(x 2 + 7x – 7)

=

–x – 1 x 2 + x x 2 – x 5x + 3 –8x ’ –3x 2 + 2x + 1 x 2 + 8x – 18 –5x 2 + x – 3 –x 2 + 19 –3x + 3 x 2 + 2 –6x + 3 3x + 9 –3

–x 2 + 12x –2x 2 – 6x 3x 2 – 5x – 1 2x 2 + x 5x x 2 + 3x + 10 2x – 9 – 9 2x 2 – x – 16 2x 2 – x + 1 6x 2 – 13x x 2 + 5x – 2 6x + 10 x 2 + 7x – 2


Higher order expansions and
Pascal’s triangle
Pascal’s triangle, as shown in the ﬁgure
below, is a special arrangement of numbers
in a triangular shape. Any number is the
sum of the two numbers immediately above
it, with 1s running down the sides. The
triangle was named after Blaise Pascal
(1623–62), a French mathematician who
wrote about the triangle in 1653. However,
earlier mathematicians knew about the
‘magic’ of this triangle. Chu Shih-Chieh,
a Chinese mathematician, included an
illustration of the triangle in a book in 1303.
There are many patterns to observe
with Pascal’s triangle. 1
Let’s look at one of these. 1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 a Expand (x + 1) . 2

b Which line of Pascal’s triangle links to your answer for part 1a? Describe the
pattern you have observed.
2 a Expand (x + 1)3. This means expand (x + 1)(x + 1)(x + 1). Hint: First expand
(x + 1)2 then multiply this expansion by (x + 1).
b Which line of Pascal’s triangle links to your answer for part 2a?
3 Use the pattern you can observe to show that (x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1.
reads
L Sp he 4 Use Pascal’s triangle to expand each of the following.
a (x + 1)5 b (x + 1)6 c (x + 1)7 d (x + 1)8 e (x + 1)9 f (x + 1)10
et
EXCE

Expanding (You may need to copy the diagram above and add more lines to Pascal’s triangle.)
(x + a)n
Let’s look at what happens when we have a minus sign in the brackets.
5 Expand (x − 1)2, (x − 1)3 and (x − 1)4 by multiplying terms.
6 Describe what effect the minus sign has on the expansions.
7 Use Pascal’s triangle and your observations from question 6 to expand each of
the following.
a (x − 1)5 b (x − 1)6 c (x − 1)7 d (x − 1)8 e (x − 1)9 f (x − 1)10

Extension
Can you work out how to use Pascal’s triangle to expand each of the following?
Clearly explain how you are able to do this.
a (x + 2)3, (x + 2)4, . . . (x + 2)10 b (x + y)3, (x + y)4, . . . (x + y)10
c (2x + 3) , (2x + 3) , . . . (2x + 3)
3 4 10
d (2x − 3y)3, (2x − 3y)4, . . . (2x − 3y)10

Simplifying algebraic fractions —
addition and subtraction
In chapter 3 we spent some time simplifying algebraic fractions. This section is effec-
tively a continuation of that topic with more complicated fractions, requiring use of the
expansion techniques we have learned in this chapter.

Pronumerals in the numerator only
This type of problem is very similar to those in chapter 3.

WORKED Example 10
Simplify the following.
y+3 y–4 y+2 y–3
a ----------- + -----------
- - b ----------- – -----------
- -
2 5 7 3

THINK WRITE

y+3 y–4
a 1 Write the fractions. a ----------- + -----------
-
2 5
5( y + 3) 2( y – 4)
2 Find the lowest common denominator = ------------------- + -------------------
- -
10 10
(LCD). The lowest common multiple
(LCM) of 2 and 5 is 10.
5( y + 3) + 2( y – 4)
3 Express as a single fraction. = ---------------------------------------------
-
10
5y + 15 + 2y – 8
4 Simplify the numerator by expanding = ---------------------------------------
-
10
brackets and collecting like terms.
7y + 7
= --------------
-
10
y+2 y–3
b 1 Write the fractions. b ----------- – -----------
-
7 3
3( y + 2) 7( y – 3)
2 Find the lowest common denominator = ------------------- – -------------------
- -
21 21
(LCD). The lowest common multiple
(LCM) of 7 and 3 is 21.
3( y + 2) – 7( y – 3)
3 Express as a single fraction. = ---------------------------------------------
-
21
3y + 6 – 7y + 21
21
– 4y + 27
= ----------------------
21


Pronumerals in the denominator
In the case where there are pronumerals in the denominator, the common denominator
is taken to be the product of each of the denominators. Then, proceed as in previous
cases.

WORKED Example 11
Simplify the following.
7 7 7 3
a ----------- + -----------
- - b -------------- – --------------
- -
a+2 a–1 2b – 5 4b + 1

THINK WRITE

7 7
a 1 Write the fractions. a ----------- + -----------
- -
a+2 a–1
7(a – 1) 7(a + 2)
2 Find the LCD. The LCM of (a + 2) = --------------------------------- + ---------------------------------
- -
(a + 2)(a – 1) (a + 2)(a – 1)
and (a − 1) is (a + 2)(a − 1).
7(a – 1) + 7(a + 2)
3 Express as a single fraction. = ----------------------------------------------
-
(a + 2)(a – 1)
7a – 7 + 7a + 14
-
(a + 2)(a – 1)
14a + 7
= ---------------------------------
-
(a + 2)(a – 1)
7 3
b 1 Write the fractions. b -------------- − --------------
- -
2b – 5 4b + 1

Find the LCD. The LCM of (2b − 5) 7 ( 4b + 1 ) 3 ( 2b – 5 )
2 = ---------------------------------------- − ----------------------------------------
- -
and (4b + 1) is (2b − 5)(4b + 1). ( 2b – 5 ) ( 4b + 1 ) ( 2b – 5 ) ( 4b + 1 )

7 ( 4b + 1 ) – 3 ( 2b – 5 )
3 Express as a single fraction. = ----------------------------------------------------
-
( 2b – 5 ) ( 4b + 1 )
28b + 7 – 6b + 15
4 Simplify the numerator by expanding = ------------------------------------------
-
( 2b – 5 ) ( 4b + 1 )
22b + 22
= ----------------------------------------
( 2b – 5 ) ( 4b – 1 )

It is customary to leave the denominator as brackets, without expanding them.

Pairs of brackets in the denominator
Although it is possible to have two different pairs of brackets in the denominator of
each fraction, in this section we will consider the case where one of each pair is an
identical bracket. The common denominator will consist of each bracket that appears in
the question; the repeated bracket needs to appear only once.

WORKED Example 12
Simplify.
4 1
---------------------------------- + ----------------------------------
-
( x + 2)( x + 3) ( x – 4)( x + 3)

THINK WRITE

4 1
1 Write the fractions. --------------------------------- + ---------------------------------
- -
( x + 2)( x + 3) ( x – 4)( x + 3)
4( x – 4) 1( x + 2)
2 Find the LCD. The LCM of (x + 2)(x + 3) = -------------------------------------------------- + --------------------------------------------------
- -
( x + 2)( x + 3)( x – 4) ( x + 2)( x + 3)( x – 4)
and (x − 4)(x + 3) is (x + 2)(x + 3)(x − 4).
4( x – 4) + 1( x + 2)
3 Express as a single fraction. = --------------------------------------------------
-
( x + 2)( x + 3)( x – 4)
4x – 16 + x + 2
4 Simplify the numerator by expanding = --------------------------------------------------
-
( x + 2)( x + 3)( x – 4)
5x – 14
= --------------------------------------------------
-
( x + 2)( x + 3)( x – 4)

remember
remember
1. When adding or subtracting algebraic fractions, you must ﬁrst ﬁnd a common
denominator.
2. Pronumerals can appear in either the numerators or the denominators.
3. If the pronumerals are in the denominator, the common denominator is usually
the product of the individual denominators.

Simplifying algebraic
5F fractions — addition and
subtraction
WORKED 1 Simplify each of the following. 5.5
Example HEET
x+1 x+3 m+1 m+2 x+2 x+5
SkillS

10 a ----------- + -----------
- - b ------------ + ------------
- - c ----------- + -----------
- -
2 4 6 2 4 3
x–1 x+2 y–3 y+1 a+6 a–2
d ----------- + -----------
- e ----------- + -----------
- f ----------- + -----------
- -
2 3 5 2 5 6 Math
cad

p+2 p–3 x–4 x+1 5x + 2 2x + 1
g ----------- + -----------
- - h ----------- + -----------
- i -------------- + --------------
- - Adding and
10 5 3 5 5 15 subtracting
m + 6 2m + 1 2p – 3 p + 2 2x – 1 2x – 3 algebraic
j ------------ + ----------------
- - k --------------- + -----------
- l -------------- + --------------
- - fractions
5 2 6 3 3 2
4x + 3 3x – 1 2x – 1 4x – 5 2x – 1 x – 1
m -------------- + --------------
- - n -------------- + --------------
- - o -------------- + -----------
-
4 2 2 3 2 4


2 Simplify each of the following.
m+2 m+1 x+4 x+5 p–3 p+3
a ------------ – ------------
- - b ----------- + -----------
- - c ----------- + -----------
- -
3 4 2 7 2 3
y–4 y+2 x+5 x+1 a+3 a–6
d ----------- + -----------
- e ----------- + -----------
- - f ----------- + -----------
- -
3 5 2 3 2 5
m–1 m+3 p+2 p–3 x–3 x–2
g ------------ – ------------
- - h ----------- – -----------
- - i ----------- – -----------
8 4 15 5 4 3
y–2 y–5 x–3 x–1 4p + 1 2p + 2
j ----------- – ----------- k ----------- – ----------- l --------------- – ---------------
- -
7 3 2 4 3 9
5y + 3 3y – 2 3a – 2 2a – 1 2x – 1 3x + 1
m -------------- – --------------
- - n -------------- – --------------
- - o -------------- – --------------
- -
3 12 3 10 5 15
3 multiple choice
a+2 a+3
a What is the equivalent of ----------- + ----------- ?
- -
3 4
2a + 5 7a + 17 12a + 72 7a + 17 2a + 5
A --------------- B -----------------
- C --------------------
- D -----------------
- E --------------
-
7 7 12 12 12
2x + 1 x – 3
b What is the equivalent of -------------- + ----------- ?
-
8 4
4x 2+6 4x – 5 5x – 5 4x + 7 3x – 4
A ----------------- B --------------
- C --------------
- D --------------
- E --------------
-
8 8 4 8 12
x+3 x+2
c The equivalent of ----------- – ----------- is:
- -
3 6
x+4 3x + 8 3x – 4 5 2x + 1
A -----------
- B --------------- C --------------
- D --
- E --------------
-
6 6 6 3 9
5m – 1 m – 2
d The equivalent of --------------- – ------------ is:
- -
2 3
13m – 7 4m + 1 13m + 1 13m – 1 4m – 1
A ------------------
- B ----------------
- C -------------------
- D ------------------
- E ---------------
-
6 5 6 6 6
WORKED 4 Simplify the following.
Example
11 2 1 4 3 7 2
a ------------ + ---
- - b ----------- + --
- - c -- + -----------
- -
m+3 m x+2 x a a+4
9 5 3 2 4 5
d -- + -----------
- - e -- + ----------
- - f --- + ------------
- -
b b+3 c c–1 m m–3
7 2 2 4 4 2
g ----------- + --
- - h ----------- + --
- - i ----------- + --
- -
p–8 p p–5 p a–6 a
6 4 1 2 2 1
j ----------- + --
- - k ----------- + -----------
- - l ----------- + -----------
- -
q–1 q a+1 a+2 b+3 b+2
3 1 4 2 1 3
m ----------- + -----------
- - n ------------ + ------------
- - o ----------- + -----------
- -
x+2 x+3 m+1 m+3 p–1 p+4
5 Simplify the following.
7 2 6 2 9 1
a ------ – ------------
- - b ----- – -----------
- - c ----- – -----------
- -
2m m + 1 2x x + 3 4x x + 5
4 2 7 3 5 2
d ------------ – ---
- - e ------------ – ---
- - f ----------- – --
- -
m+5 m m+3 m p–3 p

Ch 05

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Ch 05