MQ10 VIC ch 01 Page 17 Tuesday, November 20, 2001 10:49 AM
17
QUEST
GE
S
EN
M AT H
Chapter 1 Rational and irrational numbers
CH
AL
L
1 Find three numbers, w, x and y, none of which are perfect squares or
zero and that make the following relationship true.
w+ x = y
2 a If 132 = 169, 1332 = 17 689 and 13332 = 1 776 889, write down the
answer to 13 3332 without using a calculator or computer.
b If 192 = 361, 1992 = 39 601 and 19992 = 3 996 001, write down the
answer to 19 9992 without using a calculator or computer.
c Can you find another number between 13 and 19 where a similar
pattern can be used?
Simplifying surds
Some surds, like some fractions, can be reduced to simplest form.
Only square roots will be considered in this section.
Consider:
Now,
Taking
9 and
36 = 6
36 = 9 × 4, so we could say:
9×4 =6
4 separately:
9× 4 =3×2=6
If both 9 × 4 = 6 and 9 × 4 = 6, then 9 × 4 = 9 × 4 .
This property can be stated as: ab = a × b and can be used to simplify surds.
8 =
=
4×2
4× 2
= 2 × 2 which can be written as 2 2 .
A surd can be simplified by dividing it into two square roots, one of which is the
highest perfect square that will divide evenly into the original number.
WORKED Example 8
Simplify each of the following.
a 40
b 72
THINK
a 1 Write the surd and divide it into two
parts, one being the highest perfect
square that will divide into the surd.
2 Write in simplest form by taking the
square root of the perfect square.
WRITE
a 40 =
b
b
1
2
Write the surd and divide it into two
parts, one being the highest perfect
square that will divide into the surd.
Write in simplest form by taking the
square root of the perfect square.
4 × 10
= 2 10
72 =
36 × 2
=6 2
MQ10 VIC ch 01 Page 18 Tuesday, November 20, 2001 10:49 AM
18
Maths Quest 10 for Victoria
22 can not be simplified because no perfect square divides exactly into 22.
If a smaller perfect square is chosen the first time, the surd can be simplified in more
than one step.
72 =
4 × 18
= 2 18
= 2× 9× 2
= 2×3 2
=6 2
This is the same answer as found in worked example 8(b) but an extra step is included.
When dividing surds into two parts, it is critical that one is a perfect square.
For example, 72 = 24 × 3 is of no use because an exact square root can not be
found for either part of the answer.
WORKED Example 9
Simplify 6 20 .
THINK
WRITE
6 20 = 6 × 4 × 5
1
Write the expression and then divide
the surd into two parts, where one
square root is a perfect square.
2
Evaluate the part which is a perfect
square.
= 6×2 5
3
Multiply the whole numbers and write
the answer in simplest form.
= 12 5
Sometimes it is necessary to change a simplified surd to a whole surd. The reverse process is applied here where the rational part is squared before being placed back under
the square root sign.
WORKED Example 10
Write 5 3 in the form
a.
THINK
WRITE
1
Square the whole number part, then
express the whole number as a square
root.
52 = 25 so 5 =
2
Write the simplified surd and express
it as the product of 2 square roots, one
of which is the square root in step 1 .
5 3 =
3
Multiply the square roots to give a
single surd.
25
25 × 3
=
25 × 3
=
75
MQ10 VIC ch 01 Page 19 Tuesday, November 20, 2001 10:49 AM
Chapter 1 Rational and irrational numbers
19
WORKED Example 11
Ms Jennings plans to have a climbing frame that is in the shape of a large cube with sides
2 metres long built in the school playground.
H
a Find the length of material required to join the opposite vertices
G
of the face which is on the ground.
F
E
b Find the exact length of material required to strengthen the frame
by joining a vertex on the ground to the vertex which is in the air
D
C
and which is furthest away.
c Find an approximate answer rounded to the nearest cm.
A
B
THINK
WRITE
a
a D
1
Draw a diagram of the face, mark in
the diagonal, the appropriate
measurements and label the vertices.
C
2m
A
2
B
2m
2
2
AC = AB + BC
Use Pythagoras’ theorem to find the
length of the diagonal.
2
= 22 + 22
=8
AC =
8
=2 2
b
3
Answer the question in a sentence.
1
Draw a diagram of the triangle
required, label the vertices and mark
in the appropriate measurements.
2 2 metres of material is required.
b
G
2m
A
2
Use Pythagoras’ theorem to find the
length of the diagonal.
2 2m
2
2
AG = CG + AC
Simplify the surd.
4
Write your answer in a sentence.
c Round the answer to 2 decimal places.
2
= 22 + ( 2 2 ) 2
= 12
AG =
3
C
12
=2 3
The length of material required is 2 3
metres.
c The approximate length to the nearest cm is
3.50 metres.
MQ10 VIC ch 01 Page 20 Tuesday, November 20, 2001 10:49 AM
20
Maths Quest 10 for Victoria
remember
1. To simplify a surd, divide it into two square roots, one of which is a perfect
square.
2. Not all surds can be simplified.
3. ab = a × b
4. Some perfect squares to learn are: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 . . .
1D
1.11
SkillS
HEET
WORKED
Example
8
Mat
d
hca
Simplifying
surds
WORKED
Example
EXCE
et
reads
L Sp he
Simplifying
surds
9
WORKED
Example
10
GC pr
ogram
Surds
Simplifying surds
1 Simplify each of the following.
a
20
b
8
c
18
d
49
e
30
f
50
g
28
h
108
i
288
j
48
k
500
l
162
m
52
n
55
o
84
p
98
q
363
r
343
s
78
t
160
6 64
d 7 50
e
10 24
i
j
12 242
d 4 5
e
8 6
i
j
13 2
2 Simplify each of the following.
a 2 8
b 5 27
c
f
g 4 42
h 12 72
5 12
3 Write each of the following in the form
a.
a 2 3
b 5 7
c
f
g 4 2
h 12 5
3 10
9 45
6 3
10 6
4 multiple choice
a
1000 is equal to:
A 31.6228
B 50 2
C 50 10
D 10 10
b
80 in simplest form is equal to:
A 4 5
B 2 20
C 8 10
D 5 16
c Which of the following surds is in simplest form?
A 60
B
C 105
D 117
147
d Which of the following surds is not in simplest form?
A 102
B
C 116
D 118
110
e 6 5 is equal to:
A 900
B
C 150
D 180
30
f Which one of the following is not equal to the rest?
A 128
B 2 32
C 8 2
D 4 8
g Which one of the following is not equal to the rest?
A 4 4
B 2 16
C 8
D 16
h 5 48 is equal to:
A 80 3
B 20 3
C 9 3
D 21 3
E 100 10
E 10
E
432
E
122
E 13.42
E 64 2
E
64
E 15 16
MQ10 VIC ch 01 Page 21 Tuesday, November 20, 2001 10:49 AM
21
Chapter 1 Rational and irrational numbers
5 Challenge: Reduce each of the following to simplest form.
a
675
b
1805
c
1792
d
578
e
a2c
f
bd 4
g
h 2 jk 2
h
f3
6 A large die with sides measuring 3 metres is to be placed in
front of the casino at Crib Point. The die is placed on one of
11
its vertices with the opposite vertex directly above it.
a Find the length of the diagonal of one of the faces.
b Find the exact height of the die.
c Find the difference between the height of the die and the
height of a 12-metre wall directly behind it. Approximate
the answer to 3 decimal places.
WORKED
Example
7 A tent in the shape of a tepee is being used as a cubby house.
The diameter of the base is 220 cm and the slant height is
250 cm.
a How high is the tepee? Write the answer in simplest surd
form.
b Find an approximation for the height of the tepee in centimetres, rounding the answer to the nearest centimetre.
250 cm
220 cm
Career profile
PETER RICHARDSON — Analyst Programmer
I use basic mathematical skills
throughout the day to calculate screen
heights and check whether all necessary
fields and labels will fit. More advanced
mathematics such as working with
formulas and other secondary school
mathematics are used in Excel
spreadsheets for statistics and data
manipulation.
During a typical day, all my work is
done on computer, usually using a
software package to write code in Java or
Cobol. I create screens for use by clients,
and the supporting code to ensure screens
react as expected.
Qualifications:
Bachelor of Applied Science (Computer
Science and Software Engineering)
I entered this field as a change of career and
find it to be interesting and diverse.
Questions
1. What computer language does Peter
use to write code?
2. Name one aspect of Peter’s job.
3. Find out what courses are available to
become an analyst programmer.
MQ10 VIC ch 02 Page 39 Tuesday, November 20, 2001 10:54 AM
Algebra and
equations
2
To raise money for a charity,
a Year 10 class has decided to
organise a school dance.
Tickets to the school dance
will cost $6 each. Expenses
have been calculated as $200
for the hire of the venue and
$250 for a DJ. The dance
organisers hope to raise more
than $1000 for charity. How
many people need to attend
the dance to achieve this
goal?
One way to come up with
an answer to this problem is
to solve an inequation. In this
chapter we will look at
equations and inequations
and consolidate and extend a
number of your algebraic
skills.
MQ10 VIC ch 02 Page 40 Tuesday, November 20, 2001 10:54 AM
40
Maths Quest 10 for Victoria
Operations with pronumerals
Like terms are terms that contain the same pronumerals and can be collected (added or
subtracted) in order to simplify an algebraic expression.
WORKED Example 1
Simplify the following.
a 4j − 5c + c + 3j
b d 2 + 9d − 12d − 15
THINK
a 1 Write the expression.
2 Identify the like terms and group
them together.
3 Simplify by collecting like terms.
WRITE
a 4j − 5c + c + 3j
= 4j + 3j − 5c + c
b
b d 2 + 9d − 12d − 15
= d 2 − 3d − 15
1
2
Write the expression.
Simplify by collecting like terms.
= 7j − 4c
When multiplying and dividing algebraic terms, it is not necessary to have like terms.
In fact, any terms can be multiplied or divided and the result is a single new term.
WORKED Example 2
Simplify the following.
a 5m × −6p
18mcd
b ----------------10mf
a2 b
c -----------26ab
THINK
a 1 Write the expression.
2 Rearrange, writing the coefficients
first.
3 Multiply the coefficients and
pronumerals separately.
WRITE
a 5m × −6p
= 5 × −6 × m × p
b
18mcd
b ----------------10mf
9cd
= --------5f
c
1
Write the expression.
2
Cancel 18 and 10 (common factor of
2) and cancel m from numerator and
denominator.
1
Write the expression.
2
Cancel a from numerator and
denominator. Cancel b from
numerator and denominator.
= −30mp
a2b
c -----------26ab
a
= -----6b
MQ10 VIC ch 02 Page 41 Tuesday, November 20, 2001 10:54 AM
Chapter 2 Algebra and equations
41
remember
1. Like terms contain the same pronumeral parts and can be collected (added or
subtracted) in order to simplify an algebraic expression.
2. When multiplying and dividing algebraic terms, it is not necessary to have like
terms. In fact, any terms can be multiplied or divided and the result is a single
new term.
2 Simplify the following.
a 10m − 7m + 5c − 3c
d 7t − 11t + 7 − 5
g 6p − 9 − 2p − 5
j 2j + 4c − 3j − c
m 5y2 + 3y + 2y2 − 7y
b 6m + 9m + 10f + 6f
e 7g + 2j + 5g + 11j
h 5y + 6h2 + h2 + 9y
b
e
h
k
n
14a − 6a + 11f − 8f
10r + 15 − 11r − 10
10w − 6w + 14 − 10
−2k + 14m − 5k − 10m
4x3 − 2x3 + 3x4 − 9x4
4 Simplify the following.
a x2 + 5x + 2x + 6
d a2 − 4ab + 2ab + b2
5 Simplify the following.
a 8f × 5h
2a
d 6p × −9hn
4d + 3c + 5d + 2c
12d + 8 + 3d + 5
4nv + 8u + 7nv + 2u
c
f
i
l
o
5k + 11 − 3k − 9
6v − 8v + 12 − 17
4c − 8 + 3c − 9
−d + 5c − 2c − 3d
c2 − 2c2 + 5 + 7
D −3q − 5p
E 5p − 7q
D −9r + 9y
E −r + 9y
D −2j 6
E 4j 2 − 2j
b d2 − 3d + 7d − 10
e u2 + 4u − 3u − 12
c
f
v2 − 2v − 8v − 6
5n4 − 12n2 + 6n2 − 25
b 10ab × 6c
e −2b × −6at × 7s
c
f
−4m × −7g
3ma × −6t × 4hs
6 Simplify the following.
24 jkl
2b, c
a ------------8 jl
16ab
d -----------12bc
x 2 yz
g ----------5xyz
12mnt
b ---------------6t
27kg
e --------------– 36kh
32mn 2 k 2
h – --------------------12m 2 n 2
c
– 42kgh
-----------------7kh
50a 2 b
--------------215ab
– 3abc
----------------– 2b 2 d 2
WORKED
Example
2.1
HEET
2.2
Math
WORKED
Example
HEET
cad
3 multiple choice
a 2q + 6p − 5q − p simplifies to:
A 7q + 5p
B 7q − 5p
C −3q + 5p
b −4r − 6y − 3y + 5r simplifies to:
A −9r − 9y
B r − 9y
C r + 9y
2
2
c j − 6j − 3j + 4j simplifies to:
A −2j 2 − 2j
B −2j 4 − 2j
C −3j 2 − 2j
c
f
i
SkillS
Example
SkillS
1 Simplify the following.
a 2k + 3k + 2c + 9c
1
d 2f + 3h + 2f + 5h
g 9n + 4 + 3n + 7
WORKED
f
i
Operations
with
pronumerals
HEET
SkillS
2A
Operations with
pronumerals
2.3
MQ10 VIC ch 02 Page 42 Tuesday, November 20, 2001 10:54 AM
42
Maths Quest 10 for Victoria
Substituting into expressions
When the numerical values of pronumerals are known, we can substitute them into an
algebraic expression and evaluate it. It can be useful to place any substituted values in
brackets when evaluating an expression.
WORKED Example 3
If a = 4, b = 2, and c = −7, evaluate the following expressions.
a a−b
b a3 + 9b − c
THINK
WRITE
a
a a−b
b
1
Write the expression.
2
Substitute a = 4 and b = 2 into the
expression.
=4−2
3
Simplify.
=2
1
Write the expression.
2
Substitute a = 4, b = 2 and c = −7
into the expression.
= (4)3 + 9(2) − ( −7)
3
Simplify.
= 64 + 18 + 7
= 89
b a3 + 9b − c
WORKED Example 4
If c =
a 2 + b 2 , find c if a = 12 and b = −5.
THINK
WRITE
c=
a2 + b2
1
Write the expression.
2
Substitute a = 12 and b = −5 into the
expression.
=
( 12 ) 2 + ( – 5 ) 2
3
Simplify.
=
144 + 25
=
169
= 13
remember
1. When the numerical values of pronumerals are known, we can substitute them
into an algebraic expression and evaluate it.
2. It is sometimes useful to place any substituted values in brackets when
evaluating an expression.
MQ10 VIC ch 02 Page 43 Tuesday, November 20, 2001 10:54 AM
Chapter 2 Algebra and equations
2B
43
Substituting into
expressions
1 If a = 2, b = 3 and c = 5, evaluate the following expressions.
a a+b
b c−b
c c−a−b
3
a b c
--- + --- + --d c − (a − b)
e 7a + 8b − 11c
f
2 3 5
g abc
h ab(c − b)
i a2 + b2 − c2
j c2 + a
k −a × b × −c
l 2.3a – 3.2b
WORKED
HEET
4
i
3k − 5d
c
xy
9x
-----2y
E
L Spread
XCE
L Spread
XCE
E
f
Substitution
game
4 Complete the following.
a If c =
a 2 + b 2 , calculate c if a = 8 and b = 15.
b If A = 1--2- bh, find the value of A if b = 12 and h = 5.
c
d
e
f
g
h
Substitution
The perimeter, P, of a rectangle is given by P = 2L + 2B. Find the perimeter, P, of a
rectangle given L = 1.6 and B = 2.4.
C
If T = ---- , find the value of T if C = 20.4 and L = 5.1.
L
n+1
If K = ------------ , find the value of K if n = 5.
n–1
9C
Given F = ------- + 32 , calculate F if C = 20.
5
If v = u + at, evaluate v if u = 16, a = 5, t = 6.
The area, A, of a circle is given by the formula A = πr 2. Calculate the area of a
circle, correct to 1 decimal place, if r = 6.
i
If E = 1--2- mv2, calculate E if m = 5, v = 4.
j
Given r =
A
--- , evaluate r to 1 decimal place if A = 628.
π
5 multiple choice
a If p = −5 and q = 4, then pq is equal to:
A 20
B 1
C −1
D −20
2
2
2
b If c = a + b , and a = 6 and b = 8, then c is equal to:
A 28
B 100
C 10
D 14
2
c Given h = 6 and k = 7, then kh is equal to:
A 294
B 252
C 1764
D 5776
E − 5--4E 44
E 85
sheet
Example
k−d
d2
Substituting
into
expressions
sheet
WORKED
c
f
Math
cad
2 If d = –6 and k = –5, evaluate the following.
a d+k
b d−k
d kd
e −d(k + 1)
k–1
g k3
h ----------d
1
1
3 If x = --3- and y = --4- , evaluate the following.
a x+y
b y−x
x
d -e x 2y 3
y
2.4
SkillS
Example
MQ10 VIC ch 02 Page 45 Tuesday, November 20, 2001 10:54 AM
Chapter 2 Algebra and equations
45
Expanding
Expanding brackets in an algebraic expression is achieved by multiplying the term outside the brackets by each of the terms inside.
For example,
2(x + 4y) = 2 × x + 2 × 4y
= 2x + 8y
Remember that rather than writing multiplication signs (×), we can write the pronumerals in brackets. In this case we could have written: 2(x + 4y) = 2(x) + 2(4y)
= 2x + 8y
When more than one set of brackets appears in an expression, we can often simplify
by collecting any like terms that result from expanding the brackets.
For example: 2(x + 5y) + 3(2x − y)
= 2(x) + 2(5y) + 3(2x) + 3(−y)
= 2x + 10y + 6x − 3y
= 8x + 7y
WORKED Example 5
Expand:
a 7(m − 4)
b −6(a − 3).
THINK
WRITE
a
Write the expression.
Multiply each term inside the
brackets by the term outside.
a 7(m − 4)
= 7(m) + 7(−4)
= 7m − 28
Write the expression.
Multiply each term inside the
brackets by the term outside.
b −6(a − 3)
= −6(a) − 6(−3)
= −6a + 18
1
2
b
1
2
WORKED Example 6
Expand and simplify 6(m − 4r) − 2(2m + 7r).
THINK
1
2
3
WRITE
Write the expression.
Multiply each term inside the brackets
by the term outside.
Simplify by collecting like terms.
6(m − 4r) − 2(2m + 7r)
= 6(m) + 6(−4r) − 2(2m) − 2(7r)
= 6m − 24r − 4m − 14r
= 2m − 38r
remember
1. Expanding brackets in an algebraic expression is achieved by multiplying each
term inside the brackets by the term outside.
2. When more than one set of brackets appears in an expression we can often
simplify by collecting any like terms that result from expanding the brackets.
MQ10 VIC ch 02 Page 46 Tuesday, November 20, 2001 10:54 AM
46
Maths Quest 10 for Victoria
2C
Mat
d
hca
Expanding
1 Expand the following.
a 5(k + 1)
5a
d 8(d − 9)
g 4(5m − 2)
j 5(m + n)
m b(c − d)
b
e
h
k
n
7(m + 4)
12(h − 5)
5(6t + 5)
8(4y − 3f )
k(i + ef )
c
f
i
l
o
4(y + 7)
2(k − 6)
8(2k − 11)
6(3v + 7w)
6p(2j − 3m)
2 Expand the following.
a −3(c + 1)
5b
d −8(c + d)
g −10(5 − y)
b −5(d + 2)
e −4(3k − 2m)
h −k(k + 2)
c
f
i
−6(m + 11)
−7(2 − 3x)
−x(x − 3)
12(k + 1) + 5(k + 6)
7(t + 2) − 5(t + 5)
6(d − 3) − 5(d − 2)
2(3y − 8) + (4y − 3)
c + 3 + 5(c + 7)
c
f
i
l
o
8(m + 2) + 5(m + 5)
9(m + 7) + 4(m − 6)
12(w − 4) − 8(w + 8)
5(2x + 3) − (x − 12)
−2(m − 9) − (3m − 4)
b w(w + 2) − 6(w + 2)
e f (3f + 2) − 8(3f + 2)
c
f
x(2x − 5) − 4(2x − 5)
2a(4a − 3) + 3(4a − 3)
WORKED
Expanding
Example
EXCE
et
reads
L Sp he
Expanding
WORKED
Example
GC pr
ogram
Expanding
3 Expand and simplify.
a 2(c + 4) + 3(c + 5)
6
d 2( j − 5) − 3( j + 2)
g 10(c + 4) + 6(c − 9)
j −4(2h + 7) − 10(h − 5)
m 10(h − 1) − (3h − 5)
WORKED
Example
SkillS
2.5
HEET
4 Expand and simplify.
a y(y − 6) + 2(y − 6)
d 3h(h + 1) + 5(h + 1)
GAM
me
E ti
Algebra and
equations
— 001
Work
ET
SHE
2.1
b
e
h
k
n
5 multiple choice
a y(3 − 2y) simplifies to:
A 3 − 2y2
B 6 − y2
C 3y − 6y2
b –5(k – 3) simplifies to:
A −5k − 15
B −5k + 15
C −5k + 8
c 4(b − 3) − 2(b − 2) simplifies to:
A 2b − 16
B 2b + 8
C 2b − 1
D 3y − 2y2
E y3 − 2y2
D −5k − 8
E −5k + 3
D 2b − 11
E 2b − 8
1
1 Simplify 5r − 2t + 3r − t.
64e
2 Simplify ----------2- .
40e
3 Simplify −3gh × 2gkm.
6 If c = a 2 + b 2 , evaluate c when a = 12
and b = 6. Leave your answer in
simplified surd form.
4 If l = 7 and m = − 1--2- , evaluate
2l – 6m
------------------ .
5m
5 If c = 0.05 and d = 0.4, evaluate
6c − 4d.
8 Expand −3q(p − 5q).
7 Expand 6(w − 2v).
9 Expand and simplify 3(2 + 5u) + 2(8u + 7).
10 Expand and simplify 7(12 − 3r) − (5r + 14).
MQ10 VIC ch 02 Page 47 Tuesday, November 20, 2001 10:54 AM
Chapter 2 Algebra and equations
47
Factorising using common factors
Factorising is the opposite process to expanding. In this section, we will look at factorising by taking out the highest common factor (HCF) of an algebraic expression. (Further factorising techniques will be covered in chapter 4, Quadratic equations.)
WORKED Example 7
Factorise the following.
a 6a – 15
b 20p6 + 15p4
THINK
WRITE
a
Write the expression.
Find the highest common factor
(HCF) of the terms.
Place the HCF in front of a set of
brackets and divide each term by the
HCF to complete the inside of the
brackets.
a 6a − 15
Write the expression.
Find the highest common factor
(HCF) of the terms.
Place the HCF in front of a set of
brackets and divide each term by the
HCF to complete the inside of the
brackets.
b 20p6 + 15p4
1
2
3
b
1
2
3
HCF = 3
= 3(2a − 5)
HCF = 5p4
= 5p4(4p2 + 3)
remember
1. To factorise an expression:
(a) Find the highest common factor (HCF) of the terms and place in front of
the brackets.
(b) Divide each term by the HCF and place inside the brackets.
2. If the first term is negative, take out a negative common factor.
HEET
2.6
SkillS
2D
Factorising using common
factors
1 Factorise the following.
a 4x + 12
7a
d 10y + 110
g 5b − 45
j 6l − 72
WORKED
Example
6y + 24
2f + 28h
6d − 6
12n − 36p
c
f
i
l
7m + 49
3a − 9
8e − 24
7f − 98d
Math
cad
b
e
h
k
Factorising
MQ10 VIC ch 02 Page 48 Tuesday, November 20, 2001 10:54 AM
48
Maths Quest 10 for Victoria
2 Factorise the following.
a 6t + 10
d 30m + 15
g 6c − 27
j 5c + dc
b
e
h
k
9m + 6
14m + 12n
100h − 15
6ak − 30am
c
f
i
l
12k + 28
10j − 25
20m − 2
4abc + bcd
3 Factorise the following (by taking out a negative common factor).
a −3c + 15
b −7m + 35
c −8k + 24j
d −5j − 20
e −4h − 28j
f −6p − 12s
g −9k + 15
h −16ac + 12ad
i −12bm − 20abc
4 Factorise the following.
a m2 + 5m
7b
d 8f + f 2
g 4q2 − 10q8
j 20m3n4 − mn5
WORKED
Example
b
e
h
k
d3 − 6d
y2 − y
15r5 − 5r4
6k3p2 + 8k2p
c
f
i
l
4x4 + 16x
7p3 + 21p5
12a2b + 15ab2
11x3y − 11xy2
5 multiple choice
QUEST
GE
S
EN
M AT H
a 10h + 12 factorises to:
A 2(5h + 6)
B 10(h + 12) C 12(h + 10) D 5(5h + 6)
E 5(2h + 6)
b −8 + 4r factorises to:
A −4(−2 − r) B −4(2 − r)
C −4(2 + r)
D −4(−2 + r) E 4(2 − r)
c When fully factorised, 8au − 10uh factorises to:
A 2(4au − 5uh)
B 8u(a − 1.25h)
C 2u(4a + 5h)
D 8(au − 1.25h)
E 2u(4a − 5h)
d 30a5n4 + 12a7n3 factorises to:
B 6an(5a4n3 + 2a6n2)
C 6a5n3 (5an + 2a2n)
A 6(5a5n4 + 2a7n3)
5 3
2
5 3
2
E 12a n (5 + a )
D 6a n (5n + 2a )
CH
AL
L
1 The lines in the diagram at right show A
paths between A and B. The distance
from A to B along an outside edge is 8
units. Starting at A and ending at B,
what is the longest distance you could
travel along the given paths without
travelling along any line more than
once?
2 Which numbers less than 100 have
exactly 3 factors (including one and
B
the number itself)?
3 Which 2 numbers less than 100 have exactly 5 factors?
4 Which number less than 100 has exactly 7 factors?
5 Which number or numbers less than 100 have the largest number of
factors?
MQ10 VIC ch 02 Page 49 Tuesday, November 20, 2001 10:54 AM
49
Chapter 2 Algebra and equations
This is predicted to occur in 2061.
Find common factors to
factorise the expressions given
to find the answer code.
F = 18 – 12x
=
2
A = 4x – 10
=
E = –6x – 8
=
C = ae + a
=
L = –2x – 1
=
2
2
2
E = x + 3x
=
N = 6x – 3x
=
F = 12c – 4
=
Y = –36e + 24e
=
2
2
H = 20x – 10x
=
T = 2ce + 4e
=
I = aec – ac
=
E = 10ex + 5e
=
T = 6x
=
3
+ 3x
2
E = –4ex + 2x
=
3
2
V = 8x – 4x + 12x
=
2
O=x +x
=
2
R = 12e + e
=
2
Y = 4x + 12
=
S = 2ae + 4ac
=
A = 3x – 12x
=
R = 6ce – 3c
=
E = –3x – 1
=
3
2
2
M = 4x – 8x
=
T = x – 2x – x
=
R = cx – c
=
E = 5x – 10x
=
2
2
U = 4e – 6e
=
L = 9e + 18
=
A = –3x – 9
=
S = –2x + 5
=
2
2
T = 15x – 3x
=
F = –3x – 6x + 3
2
O = –5x + 10x
=
E = 6x – 2x + 2
=
2
=
2
2
N = 12c – 2c
=
Y = 3x + 3
=
V = 21x – 14
=
E = x – 7x
=
2
2
H = –16 – 8x
=
2
S = 8a – a
=
T = 22x + 11
=
E = 7ae – 14
=
R = 2x – 6x
=
3
2
3x2(2x + 1) 10x (2x – 1) 5e (2x + 1) e (12e + 1) x (x + 3) x (x 2 – 2x – 1) 2e (2e – 3) c (x – 1) 3x (2 – x) x (x + 1) 4 (3c – 1)
–8 (x2 + 2) 2 (2x – 5) 9 (e + 2) – (2x2 + 1) x (x – 7) 4 (x2 + 3) – (2x – 5) a (e + 1) –5x (x – 2) 4x (x – 2) –2 (3x + 4) 3x (5x – 1)
–3 (x + 3) 6 (3 – 2x) 11 (2x + 1) – (3x + 1) 3c (2e – 1) a (8 – a) 5x (x – 2) 7 (3x – 2) 7 (ae – 2) 2c (6c – 1) 2e (c + 2) 3 (x 2 + 1)
–3 (x 2 + 2x – 1) ac (e – 1) 4x (2x 2 – x + 3) 2x (1 – 2e ) –12e (3e – 2) 2 (3x 2 – x + 1) 3x (x – 4) 2x 2 (x – 3) 2a (e + 2c)
answers
MQ 10 Answers part A Page 660 Monday, June 24, 2002 1:50 PM
660
Answers
Maths Quest challenge
q 1250
1 8, 18, 50 (other answers possible)
2 a 177 768 889
b 399 960 001
c 16
u 12 15
2 a E
3 a
Exercise 1D — Simplifying surds
1 a 2 5
f
5 2
b 2 2
c 3 2
d 7
g 2 7
h 6 3
i
k 10 5 l
p 7 2
2 a 4 2
f
m 2 13 n
12
f
90
4 a D
f E
5 a 15 3
e
30
12 2 j
4 3
55
o 2 21
78
t
q 11 3 r 7 7
s
b 15 3 c 48
d 35 2 e 20 6
10 3 g 4 42 h 72 2 i
3 a
f
9 2
b
175 c
g
32
b A
g D
b 19 5
f
b 3 3m
7 a 60 14 cm
e
80
384
h
720 i
600 j
338
c C
d C
e D
h B
c 16 7 d 17 2 e a c
d 2 b g hk j h
6 a 3 2m
4 10
27 5 j 132 2
108 d
f
c 12 – 3 3 ≈ 6.804 m
b 224 cm
Exercise 1E — Addition and subtraction
of surds
1 a 2 2
b –4 5
d 0
e
c –6 3
5 11
f
j
16 2 – 11 5 k 2 6
m 8 7 – 11
n 11 – 2
b C
5
d
e
15
f
3
g
5
h 4
i
–2 5
j
9 3
l
1
p ------6
3
m ------4
n 15
o −8
4
q 1 ----45
-----r 1 13
32
s
b C
b 1 --16-
c C
c 1 --12-
4 a B
5 a 1 --23-
6 a 3 2+3 5
3 2
---------5
d D
d 1 --34-
3
--4
t
b 5 6 – 5 2 c 6 5 + 6 11
d 8 2 + 24
e 4 7 – 20 f
g 42 + 7 7
h
j
6 + 15 i
42 – 8 14 k 5 + 2 5
l
10 – 2 2
2 5 + 2 10
6–5 6
m 12
n 60 – 18 5 o 12 14 + 8 35
p −15
q 50 – 15 2 r 8 6 – 60
7 a 2 10 – 15 + 2 6 – 3
b 3 35 – 14 + 3 10 – 2
d 2
8 10 + 7 3
l
5 5 + 15
i
42 – 24 3
o
7 – 13
e
7 + 10
g 8 – 2 15 h 5 + 2 6
b 4 2 cm
8 a 42
c
168 2 cm
e 42
d 504 2 cm
b 18 2 m2
9 a 3 2 m
c –5 3
d –4 7
10 Quick Questions 2
e 5 6+6 5
f 2 3–3 5
1 2.25
2 0.45
3
h 29 5 + 22 3
5 4.05
6 2
7 3 10
2
g 4 6 + 6 5 – 14 2
i
9 11 – 30
j 28 2 – 39 5
k 69 3 – 17 2
m 69 – 51 2
3 a D
b A
l
9 40 2
18 10 + 42 2
c ( 36 – 4 41 ) m
d 10.39 m
e
d
g −10
h 2 15
k 15 2
l
o −168
p 6 2
f 8
i
2 14
j
m −150
6 2
n 24 5
8 16 2
3
10 ------3
3
1 a ------3
c −5
– 66
99
b ( 31 + 4 41 ) m
b 5
e
4
Exercise 1G — Writing surd fractions
with a rational denominator
Exercise 1F — Multiplication and
division of surds
1 a 5 5
7
-----11
n 41 5 + 6 3 – 6 2
c C
d A
4 ( 40 + 5 2 + 2 13 ) cm
5 a 67 m
2
------2
k –2 2
b 0
2 a
2
c A
c
27 + 2 6
q 5 2–6 7r 2
28 7
2
2
p 3 6–4 3
t
b
c −1
2 7
s 1600 2
3
f
g 10 2 + 7 3 h 8 5 + 6 i
108 2
r
35
10
---------5
15
2 a ---------5
5
b ------5
6
c ------6
7
d ------7
f
15
g ---------5
30
h ---------5
6
c ------3
15
d ---------5
5
30
b ---------6
e
2 6
---------3
f
2 21
------------7
3 10
g ------------5
6
h ------2
i
30
j
6
30
k ---------2
l
24
70
---------2
661
Answers
2 15
3 a ------------b
7
2
4 a ------b
2
3 2
5 a ---------b
2
6 a 5(2 + 3)
2 42
------------3
2 3
---------3
2 6
c ---------5
2
c ------2
5
d ------4
15
d ---------2
2
c 4( 5 – 2)
b –2 ( 1 – 2 )
5
d 3(3 + 7)
6
e
3( 5 + 2) f
g
3 6(3 6 + 5 2)
2 ( 7 + 2 ) h ------------------------------------------- or 27 + 15 3
2
5 ( 5 – 3 ) or 5 −
15
Summary
1
3
5
7
9
11
decimal number
rational
multiple
surds
perfect
like
2
4
6
8
10
12
1
b − ----12
7
-----12
3
--4
c
1
-----12
c
0.285714 d 0.5̇
d
2 126
3 a 0.08
4 a C
5 a 4--5-
b 0.8125
b E
b 8--9c
6 16 = 4
7 a 7.9
b 13.7
c 0.5
d 25.7
b 5 7
c 24 2
d 12 10
b
c
d
8 a 3 11
9 A
10 a
150
83
--------100
180
5
--6
d
e
605
83
-----99
18
11 a 4 6 – 7
12 D
b 5 3
c 18 5 – 6 7
13 a 5 2
b 24 21
c 13
e 3 --12-
f
d –2 6
14 a 240
15 C
16 B
30
---------6
b 73 – 40 3
7
5 6
17 a ------- b ---------14
6
c
d
5–2
30 + 3 3
Exercise 2A — Operations with
pronumerals
b
e
h
b
e
15m + 16f
12g + 13j
7h2 + 14y
8a + 3f
−r + 5
c
f
i
c
f
9d + 5c
15d + 13
11nv + 10u
2k + 2
−2v − 5
e
4w + 4
i 7c − 17
−7k + 4m l −4d + 3c
2x3 − 6x4 o −c2 + 12
B
c A
b d 2 + 4d − 10
d a2 − 2ab + b2
f 5n4 − 6n2 − 25
60abc
c 28gm
84abst
f −72ahmst
2mn
c −6g
10a
3g
− -----f --------4h
3b
8k 2
– -------3m
i
3ac
-----------22bd
1 a
f
k
2 a
f
3 a
5
3
30
−11
36
f
4 a
f
5 a
48
17
68
D
7
-----12
b
g
l
b
g
b
2
30
−5
−1
−125
1
− ----12
b 30
g 46
b C
c 0
h 12
d 6
i −12
e −17
j 27
c 1
h 1
1
c ----12
d 30
i 15
d 1 --13-
e −24
c 8
d 4
h 113.1 i 40
c B
e
1
--------576
e 1.5
j 14.1
Exercise 2C — Expanding
1 a
e
i
m
2 a
d
g
3 a
d
g
j
m
4 a
c
e
5 a
5k + 5 b 7m + 28 c 4y + 28 d 8d − 72
12h − 60 f 2k − 12 g 20m − 8 h 30t + 25
16k − 88 j 5m + 5n k 32y − 24f l 18v + 42w
bc − bd n ki + kef o 12pj − 18mp
−3c − 3
b −5d − 10 c −6m − 66
−8c − 8d
e −12k + 8m f −14 + 21x
−50 + 10y
h −k2 − 2k i −x2 + 3x
5c + 23
b 17k + 42 c 13m + 41
−j − 16
e 2t − 11
f 13m + 39
16c − 14
h d−8
i 4w − 73
−18h + 22
k 10y − 19 l 9x + 27
7h − 5
n 6c + 38
o −5m + 22
y2 − 4y − 12
b w2 − 4w − 12
2x2 − 13x + 20
d 3h2 + 8h + 5
18f 2 − 12f − 16
f 8a2 + 6a − 9
D
b B
c E
10 Quick Questions 1
4 −6 4--57 6w − 12v
10 70 − 26r
8
2 -----5e
5 −1.3
8 −3pq + 15q2
3 −6g2hkm
6 6 5
9 20 + 31u
Exercise 2D — Factorising using
common factors
1 a
d
g
j
4(x + 3)
10(y + 11)
5(b − 9)
6(l − 12)
b
e
h
k
6(y + 4)
2( f + 14h)
6(d − 1)
12(n − 3p)
c
f
i
l
7(m + 7)
3(a − 3)
8(e − 3)
7( f − 14d)
1D
➔
5k + 11c
4f + 8h
12n + 11
3m + 2c
−4t + 2
b
e
b
h
1 8r − 3t
Chapter 2 Algebra and
equations
1 a
d
g
2 a
d
x
g --5
h
k
n
b
Exercise 2B — Substituting into
expressions
repeater
recurring
fractions
approximation
factor
multiplied
Chapter review
1 a
3
4
4p − 14
−j + 3c
7y2 − 4y
C
x2 + 7x + 6
v2 − 10v − 6
u2 + u − 12
40fh
−54hnp
3k
4a
d -----3c
g
j
m
a
a
c
e
a
d
a
answers
MQ 10 Answers part A Page 661 Tuesday, November 20, 2001 2:14 PM
2D
answers
MQ 10 Answers part A Page 662 Tuesday, November 20, 2001 2:14 PM
662
2 a
d
g
j
3 a
d
g
4 a
d
g
j
5 a
Answers
2(3t + 5)
b
15(2m + 1)
e
3(2c − 9)
h
c(5 + d)
k
−3(c − 5)
b
−5( j + 4)
e
−3(3k − 5)
h
m(m + 5)
b
f(8 + f )
e
2q2(2 − 5q6) h
mn4(20m2 − n) k
A
b B
3(3m + 2)
2(7m + 6n)
5(20h − 3)
6a(k − 5m)
−7(m − 5)
−4(h + 7j)
−4a(4c − 3d)
d(d2 − 6)
y(y − 1)
5r4(3r − 1)
2k2p(3kp + 4)
c E
c
f
i
l
c
f
i
c
f
i
l
4(3k + 7)
5(2j − 5)
2(10m − 1)
bc(4a + d)
−8(k − 3j)
−6(p + 2s)
−4b(3m + 5ac)
4x(x3 + 4)
7p3(1 + 3p2)
3ab(4a + 5b)
11xy(x2 − y)
d D
Maths Quest challenge
1 24 units
A
B
2 4, 9, 25, 49 (Square numbers have an odd number of
factors. Squares of prime numbers have 3 factors
only.)
3 16, 81
4 64
5 72 and 96 each have 12 factors.
Exercise 2E — Adding and subtracting
algebraic fractions
5y
1 a -----12
3w
e ------28
7x + 17
i -----------------10
5
2 a -----8x
7
e --------24x
1
i – -----6x
3y
b – -----40
y
f – --5
7x + 30
j -----------------12
5
b --------12x
9
f --------20x
3x 2 + 14x – 4
3 a ---------------------------------( x + 4)( x – 2)
13x
c --------12
89y
g --------35
2x – 11
k -----------------30
38
c --------21x
37
g -----------100x
14x
d --------9
37x
h --------15
19x + 7
l -----------------6
8
d -----3x
51
h --------10x
2x 2 + 3x + 25
b ---------------------------------( x + 5)( x – 1)
2x 2 + 6x – 10
c ------------------------------------( 2x + 1 ) ( x – 2 )
4x 2 – 17x – 3
d ------------------------------------( x + 1 ) ( 2x – 7 )
2x 2 + 10x – 2
e ---------------------------------( x + 4)( x – 1)
f
2x 2 + 6x + 7
g ---------------------------------( x + 1)( x + 4)
3x 2 + 6x + 4
h ---------------------------------( x + 2)( x + 6)
i
–x 2 + 7x + 15
---------------------------------( x + 1)( x + 2)
59x 2 + 62x – 7
k ------------------------------------( 8x + 9 ) ( x + 2 )
7x 2 + x
---------------------------------( x + 7)( x – 5)
j
x–7
---------------------------------( x + 3)( x – 2)
l
x 2 + 3x + 9
------------------------------------( x + 2 ) ( 3x – 1 )
Exercise 2F — Multiplying and dividing
algebraic fractions
4x
1 a -----y
5x
e -----4y
3x
i -----2y
2
2 a --------------3x – 2
2x
e ------------------2( x + 1)
3 a 3--5-
3x
b -----c
y
3w
f ------g
2x
5
j -----k
24
5
b ----------c
x–3
x+1
f ----------------------2 ( 2x – 3 )
b
e
1
-----25
f
i
8y 2
-------9
j
2
--9
c
4y
-----x
6z
-----7x
12z
-------x
9
-------------------2( x – 6)
9x
d -----4y
2z
h -----7x
x
l ------6w
1
d -----------x+3
1
--3
d 3
4y 2
2y 2
h -------25
or 5 5--6- g -------7
32xy
2
-----------k --35
35
-----6
9
4 a ------------------------------------( 3x – 7 ) ( x + 3 )
4x
c ------------------------------------( x – 1 ) ( 2x + 1 )
1
e -------------------2( x + 1)
l
y2
1
b ---------------------------------( x + 2)( x – 9)
4( x – 1)
d ---------------------------------( x + 1)( x – 5)
28
f ------------------------------------( 2x – 3 ) ( x – 7 )
Exercise 2G — Solving basic equations
1 a
e
i
2 a
e
a = 24
h = 0.26
x=0
f = 12
w = −5 1--3-
b k = 121 c g = 2.9
f i = −2
g t=5
b i = −60
f k = 10
d r=3
h q = --16-
c z = −7
d v=7
g a = 0.425 h m = 16 5--8-
i y = 21 1--23 a t = 100 b y = ±17 c q = 6.25
-------------e h = 16
f p = ± 3--8- g g = 225
484
49
d f = ±1.2
-----h j = ± 14
31
i a = ±1 --234 a a=4
b b=6
e q = 1 1--8- f r = 5 2--5-
d f=9
h t = 9 4--5-
i
5 a
e
6 a
a = −7 1--2f = 40
b g = 30 c r = −10
n = 28 f p = 62.4
x = 1 --13- b y = 9
c m = 4 --25-
e n = 5 --237 a k = 25
e
8 a
9 a
e
i
10 a
e
i
c i=3
g s = 4 5--6-
8
x = ----11
B
x = −5
h = −2
g = −0.8
x = −1
t=3
f = −12 1--4-
f
d m = 18
d k = 1 --12-
c = 1 --13-
b m = 16
c p = −11 3--7-
d u = −4 1--8-
c C
c p=7
g v = −20
d x = −11
h r = −3
b v=1
c l=2
f e = −23 --13- g j = −3 --38-
d g = −2
h k = −36
f
b
b
f
v=3
E
d = −1
t=5
663
Answers
11 a
e
i
12 a
x=2
t=3
a=0
x = −1
b b=5
f f = 2 1--3-
c w=2
g g = −1 1--3-
d f=7
h h = −2 1--5-
b c=2
c r = 2 2--3-
d k=1
g w=1
h m=
e y = −1 --18- f
i
g=7
b u<1
1
--5
d u = −2 5--7-
e f = 12 1--2i x=1
14 a A
h h = −12
f
r = 7 1--2-
b D
g d = −6
2 9
b x = 3 5--8-
8
- f
e x = −2 ----11
j
5
-----19
20
e x = 5 ----43
i x = 1.5
4 a B
x=
10
-----43
x = −2
b x = 15
e x = −1 1--2- f
3 a x=
x>1
f ≥ −6
y > 1 3--4x<3
d ≥ −1 2--5-
3 a h < 1 8--9e
i
4 a
e
i
5 a
6
7
8
9
10
11
12 13 g
j
x=3
h x = 5 3--4-
-----k x = 2 11
12
l
c x = −6 2--9-
-----d x = − 10
19
x = −6
h x = 12
4
--7
l
11
c x = 4 ----14
15
d x = −3 ----17
2
10
- g x = 1 -----x = −1 ----61
13
9
h x = −4 ----26
j x = −4 1--3- k x = 3
b A
x = 1 5--8-
x=1
l
b v < 10
f h > 45
c t ≤ −6
g k < 7.5
d p≥1
h j ≤ −0.16
b r≥8
f k > −2
c y>3
d e ≤ −3
b d ≥ –1
c y≤7
d p > 5 5--6-
g y ≥ −18
h k < 180
c u ≥ −1
g f < −28
d x ≤ −2
h w ≥ −4 1--2-
–8
–7
–6
–5
–4
–3
–2
3
4
5
6
7
8
9
11
12
13
14
15
16
17
18
19 m
11
12
13
14
15
16
17
18
19 j
–3
–2
–1
0
1
2
3
4
5
–1
0
1
2
3
4
5
6
7r
–4 –3
–2
–1
0
1
2
3
4
w
2
3
4
5
6
x
–4 –3
–2
–1
0
y
–1
i
x
–2
–1
0
1
–8
–7
–6
–5
y ≤ − 4 --14-
l
6 a C
b A
c A
Summary
1
3
5
7
9
11
13
pronumeral
2 multiplying
substitute
4 brackets
expanding
6 highest, divide
lowest, common, single 8 numerators
reciprocal
10 numerical
isolate
12 inequality
negative
Chapter review
1 a 7c − 13 b −7k + 3m c −5d − 5c d 7y2 − 5y
a
4cd
8a
2 a −21mp b -----c --------d -----20
25 f
3
3 B
4 35
5 D
6 a 6x − 18
b −8 + 4x
7 C
8 B
9 a 4a(p − 3g)
b −4(h + 18) c 3p4(4p2 + 5)
10 B
7y
7x + 18
22
11 a -----b ------------------ c --------6
10
15x
3x 2 + 6x – 9
---------------------------------( x + 3)( x + 2)
8y
12 a -----b
x
d
25z
-------4x
c
5
-----------x+3
2D
➔
–9
2
k x<2
d x = −7
29
-----36
k x = 52
u ≤ 5 5--8- f c > 6
x ≤ 20
x > −4 b h < −7
k > −4 4--5- f j ≤ −7 7--9a < 16
i < −5
w>
j
g x = −5
x = −192 g x =
31
b x = 1 ----58
f
c x=
y
1
− 1--2-
3 17
Exercise 2I — Solving inequations
1 a
e
i
2 a
e
5
r<3
i
Exercise 2H — Solving more complex
equations
x = 3 1--4-
4
h x≤1
6 t = −42
8 y=7
10 b = − --12-
1 51
i
3
g j ≥ 15
Maths Quest challenge
5
-----17
2
m > 15
f
2 −4(8 − 7g)
3x ( x + 1 )
4 ----------------------4
5 r = 12
7 v = −3
9 z=2
2 a x=
1
e y ≥ 4 1--2-
c B
1 4(4 − 9y)
8
3 --------15x
x = −2
0
d c > −16
10 Quick Questions 2
i
–1
–20 –19 –18 –17 –16 –15 –14 –13 –12 c
13 a x = −15 b y = −4 4--5- c t = 21
20
-----31
–2
c g<9
p = 1 2--3-
1 a x=
5
u
–3
answers
MQ 10 Answers part A Page 663 Tuesday, November 20, 2001 2:14 PM
2I
answers
MQ 10 Answers part A Page 664 Tuesday, November 20, 2001 2:14 PM
664
Answers
5
d --6
e
4
y2
-----50
2x
------------------------------------( x – 1 ) ( 9x + 1 )
13 a p = 88
b s = 3.01
d r = −35
e x = 144
c b = 16
-----f x = − 13
2
g y = 60
14 a b = 4
15 a x = 1--2-
h a = ±6
b t=2
b x = 6 1--5-
i k = 12
c p = −2
3
c x = − ----14
e x = 12 2--9-
f
b x = 22 --12-
c x=2
e x = 3 3--8-
f
b x<7
e u ≥ −1
c e ≤ −12
f x ≥ −14
f
d x=1
16 a x =
6
--7
d x=5
17 a x > 4
d y ≥ 30
18
–6
–5
–4
x
−10
−8
−6
−4
−2
0
2
4
y
23
19
15
11
7
3
−1
−5
y
20
–2
–1
0
1
–10
y = –2x + 3
0
–5
5
x
y
−4
−3
−2
−1
0
1
2
3
4
5
−25 −15
−5
−5
5
15
25
35
45
55
65
75
-----x = − 16
21
y
75
50
n
6
x
x
−10
−4 −2
0
2
4
6
8
10
y
−28 −22 −16 −10 −4
2
8
14
20
26
32
y
−5
−4
−3
−2
−1
0
1
2
3
4
5
−37 −32 −27 −22 −17 −12
−7
−2
3
8
13
y
20 y = 5x – 12
y
y = 3x + 2
30
10
20
–4 –2 0
–10
10
5
6x
4
2
–50
1
–10 –5 0
–10
y = 10x + 25
–6 –4 –2 0
–25
Exercise 3A — Plotting linear graphs
−6
10 x
5
–20
Chapter 3 Linear graphs
−8
10
–10
x = 1 1--6-
2
8
10
25
–3
6
−9 −13 −17
10 x
4
2
6x
–20
–30
–20
–30
2
7
x
−10
−8
−6
−4
−2
0
y
32
26
20
14
8
2
y
30
2
4
6
8
10
x
−4 −10 −16 −22 −28
y
−20 −16 −12
20
18
16
−8
−4
0
4
8
12
16
20
14
12
10
8
6
4
2
0
y
20
y = –0.5x + 10
10
y = –3x + 2
20
10
–10 –5 0
–10
5
10 x
–20
8
–30
x
3
x
−10
−8
−6
−4
−2
0
2
4
6
8
10
y
13
11
9
7
5
3
1
−1
−3
−5
−7
y
10 y = –x + 3
5
–10 –5 0
–5
–10
5
10 x
−10
−8
−6
–20 –10 0
–10
10 20 x
−4
0
−2
2
4
6
8
10
y −1240 −1040 −840 −640 −440 −240 −40 160 360 560 760
y
1000
500
–10 –5 0
–500
5 10 x
–1000
y = 100x – 240