Patterns and Algebra – AP Book 8, Part 2: Unit 4
6.
1.
7.
2.
3.
4.
a)
1
b)
–8
c)
–3
d)
–5
e)
2
f)
8
g)
–17
h)
–15
i)
26
j)
19
k)
–33
l)
39
a)
–6
b)
–12
c)
12
d)
–63
e)
–8
f)
5
g)
–6
h)
–2
i)
18
j)
8
k)
–33
l)
–7
b)
–3
c)
–18
d)
4
f)
6
g)
–2
h)
–19
i)
–14
j)
–51
k)
–9
l)
–26
m)
–19
n)
–15
o)
–29
p)
–6
h) and m), since:
3(–7) + 2 = 2 + 3(–7)
5.
a)
18°C, 27°C, 9n°C
b)
(–60 + 9n)°C
c)
–33°C
d)
n = –3; –87°C
V‐24 8.
9.
3
–2
a)
63 + 5w
–1
b)
w = –5; $38
0
0
–5
Teacher to check guess
and check.
a)
i)
–6, –6, –6, –6
1
–3
–8
a)
ii)
–8, –8, –8, –8
2
–6
–11
b)
2
c)
–3
d)
–6
iii)
–10, –10, –10,
–10
iv)
–20, –20, –20,
–20
b)
Circle:
all but 2a
b)
–35
c)
–12
d)
14
f)
–6
g)
–2
h)
23
i)
3
–9
–14
4
–12
–17
5
–15
–20
b)
2.
i)
–4
ii)
5
iii)
–2
iv)
2
a)
x
+ (–4)
–3
–15
5
1
0
–12
4
0
j)
49
–9
3
–1
l)
–6
–6
2
–2
m)
8
–3
1
–3
n)
9
0
0
–4
o)
25
x
3
–1
–5
BONUS 30
6
–2
–6
q)
–2
9
–3
–7
r)
–1
12
–4
–8
s)
–9
15
–5
–9
18
–6
–10
–5
u)
2
a)
b)
False;
Counter-example
will vary – teacher
to check.
Sample:
If a = –2, then
3a = –6 > –10 = 5a
b)
True
c)
True
d)
False;
3.
b)
÷ (–3)
c)
÷ (–5)
d)
+ (+5)
e)
– (–7)
f)
× (–9)
g)
× (–5)
h)
÷ (–7)
i)
÷8
Circle 3 expressions:
–6m ÷ (–6), m ÷ 6 × 6 and
6 + m + (–6)
If m = –5, values from left
to right are:
36
–36, –5, –5, –5 , –5, 17
3.
Teacher to check student
checks.
b)
x = 15
c)
x = –5
d)
x = –12
15
e)
x=3
ii)
–12
f)
x = –3
iii)
3
g)
x = 100
iv)
–15
h)
x = –8
i)
x = –21
j)
x = –35
–23, more than;
–9, –8, –10
34, less than 39;
39, equal to 39;
If a = 0, 3a = 0 = 5a
2.
b)
i)
–26, closer to –29
than –23;
c)
4.
Teacher to check
substitution checks.
Answers may also be
given in decimal form.
n = –9
d)
–2
AP Book PA8-18
page 110
1.
x
–3
t)
4.
23, more than 13;
a)
29
s= 3
AP Book PA8-17
page 109
25, further from 13
than 23;
b)
11
t= 2
1.
closer to –8 than –9,
next guess is –7
c)
8
y= 7
d)
21
x= 5
e)
9 3
a= 6 = 2
a)
n
(–3)n
(–3)n – 5
–5
15
10
–4
12
7
–3
9
4
–2
6
1
e)
–58, more than –100;
–64, closer to –100
than –58;
closer to –9 than –8,
next guess is –10
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
AP Book PA8-16
page 107
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
5.
6.
ii)
Teacher to check
substitution checks.
1.
a)
2.
5
6
5 × 2 = 10
10
k–1=t
t
5 × 3 = 15
15
9–1=8
8
2.
Circle: 1 and 3
7×s=t
t
2–1=1
1
3.
1 : 4 × FN = # Blocks
15 – 1 = 14
14
x = –22
7×1=7
7
g)
32
s= 3
7 × 2 = 14
14
7 × 3 = 21
21
6×s=t
t
h)
11
t= 3
i)
46
y= 5
6 × 2 = 12
12
6 × 3 = 18
18
j)
2 1
z= 6 = 3
6 × 4 = 24
24
c)
6
∆ = 4 or 1.5 circles
a)
No
b)
Yes; No; Method 1
c)
On the left side of
Line 2, the 3 should
subtract from x, not
add to it.
e)
f)
(–2)x + 6 = –4
2.
x–3=2
x=5
a)
Incorrect
b)
Incorrect
c)
Correct
d)
Incorrect
a)
Line 2 should read:
(–3)(x – 2) = –21
b)
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
5×1=5
f)
(–2)(x – 3) ÷ (–2) = –4 ÷ (–2)
3.
a)
c)
Correct
d)
Line 4 should read:
9×s=t
45
9 × 6 = 54
54
9 × 7 = 63
63
6×s=t
t
4.
ii)
a)
SB = 2 × FN
b)
B = 2 × FN + 2
iii)
a)
SB = 3 × FN
b)
B = 3 × FN + 2
iv)
a)
SB = 3 × FN
a)
s+9=t
b)
s+3=t
c)
s–1=t
d)
s–2=t
b)
0; 1; 2
c)
20; 35; 40
d)
9; 5; 14
e)
26; 27; 28
f)
28; 42; 63
g)
27; 15; 50
h)
12; 16; 20
i)
12; 21; 36
multiplication only;
a)
Multiply each input
number by 6.
multiplication and
addition.
b)
B = 3 × FN + 6
v)
a)
SB = 3 × FN
b)
B = 3 × FN + 1
vi)
a)
SB = 4 × FN
b)
B = 4 × FN + 3
c)
the number of
shaded blocks;
18
30
b)
Divide each input
number by 8.
7×s=t
t
c)
7 × 9 = 63
63
Subtract 3 from
each input number.
7 × 2 = 14
14
d)
x+3=y
7 × 15 = 105
105
e)
x+5=y
AP Book PA8-22
page 117
f)
TN × 7 = T
1.
10 × s = t
2×s=t
d)
15 × s = t
5.
b)
3 × FN ‒ 1
c)
5 × FN
Yes;
Sample explanation:
AP Book PA8-21
page 115
FN
Sudha
Nan
1
3+3+3=9
2+7=9
FN
# Blocks
2
4 + 4 + 3 = 11 4 + 7 = 11
s+7=t
t
1
3
3
5 + 5 + 3 = 13 6 + 7 = 13
1+7=8
8
2
6
2+7=9
9
2.
3
9
3 + 7 = 10
10
s+3=r
r
2+3=5
5
1.
b)
c)
FN
# Blocks
1
4
3+3=6
6
2
8
4+3=7
7
3
12
x = –26C
5+1=6
6
y = +1C
6+1=7
7
7+1=8
8
a)
(FN + 1) + (FN + 1) + 1
or 2 × FN + 3
Formula: 3 × FN
Formula: 4 × FN
d)
rd
3 : 7 × FN = # Blocks
6 × 5 = 30
t
AnswerKeysforAPBook8.2
6.
rd
st
6 × 3 = 18
s+1=t
z = –23C
5.
st
6
8×s=t
c)
4.
Formula: 2 × FN
6×1=6
c)
b)
f)
t
9 × 5 = 45
b)
Line 2 should read:
(–3)(x – 2) = –21
(–3)x ÷ (–3) = –27 ÷ (–3)
5.
4
3
x = 36
(–2)(x – 3) = –4
4.
2
7
e)
d)
3.
5
5+2=7
t = –7
10
∆ = 3 circles
2
3+2=5
d)
b)
# Blocks
t
5×s=t
s = –11
AP Book PA8-19
page 111
FN
3
c)
c)
d)
k
1+2=3
3; –36; –36 ÷ 4;
h = –9
b)
s+2=k
1
b)
d)
1.
e)
AP Book PA8-20
page 112
b)
(FN + 1) + (FN + 1) +
(FN + 1) + 1
a)
2
b)
three, 6;
or 3 × FN + 4
3.
four, 8;
(n – 1), 2 × (n – 1)
V‐25
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
8, 8 + 2 × (n – 1)
4.
When simplified,
8 + 2 × (n – 1)
= 8 + 2n – 2
= 2n + 6
d)
e)
66 blocks
f)
22 term
I would set the formula to
125 and then solve for TN.
5.
a)
i)
ii)
nd
iii)
0, 3, 6, 9, 12
b)
Gaps: –4
Start at 42, then
subtract 4 each time.
4, 9, 14, 19, 24
a)
b)
c)
3.
a)
6
3
8
FN
#B
Gaps: +3
1
3
Start at 3, then
add 3 each time.
2
5
3
7
FN
#B
1
5
2
9
3
13
6, 12, 18, 24, …
1.
a)
b)
n×G
T
1
2
2
3
3
4
4
5
c)
8
12
16
2
2
i)
2 × FN + 2
ii)
2 × FN + 1
iii)
4 × FN + 1
I×G
O
1×3=3
12
2×3=6
15
3×3=9
18
Formula: 4n – 3
d)
n×G
6
4
12
4
18
24
Formula: 6n + 3
e)
n×G
8
16
24
3
32
3
Formula: 8n – 5
f)
n×G
10
I×G
O
1×2=2
6
2×2=4
8
3×2=6
10
20
30
2
40
2
Formula: 10n – 1
g)
2n – 1
h)
9n
I×G
O
i)
8n – 1
4, 10, 16
4
3n + 6
Gap: +6
2 × 5 = 10
5
j)
9
k)
12n + 10
3 × 5 = 15
5
–8, –18, –28
14
l)
2n + 54
Subtract: 1
6.
5, 1; 5, 1
d)
I×G
1
4
1×5=5
Gap: –10
1
n×G
2
Add: 4
c)
1
Formula: n + 1
2
2, 4; 2, 4
7, 15, 23
The gap equals the
number in the formula that
is multiplied by the Input.
Substitute TN = 15
and solve.
b)
Gap: +8
c)
b)
3, 9; 3, 9
11, 15
It’s the number
multiplied by the
Term Number (TN).
c)
a)
Gap: +4
d)
Substitute TN = 1
and solve.
3.
In each case, the number
in the circles will equal
the gap.
Use the rule to
extend the sequence
until you get to the
th
15 term.
b)
b)
5.
Add: 9
Gaps: +6
It’s the starting
number.
iii)
Yes, they do.
–2, 4, 10, 16, 22
It’s the number
added or subtracted
each time.
ii)
3, 6, 9, 12, …
AP Book PA8-24
page 119
Start at –2, then add
6 each time.
2.
2
Stepwise rule:
The starting number
is equal to the gap
in the sequence.
Gaps: +5
e)
4
Formula:
There is no constant
added or subtracted
to the multiple of the
Term Number.
42, 38, 34, 30, 26
Start at 4, then add
5 each time.
#B
1
Start at 6, then
add 6 each time.
Start at 0, then add
3 each time.
d)
i)
FN
Gaps: +6
Gaps: +3
c)
a)
2, 4, 6, 8, …
Start at 2, then
add 2 each time.
Teacher to check that
charts are also completed
properly.
b)
Teacher to check if
students can predict the
gaps (prompt: how many
new blocks are added
each time?).
Gaps: +2
AP Book PA8-23
page 118
1.
2.
Sample explanation:
(n + 1) + (n + 1) + 4;
Yes, it gives same
number of blocks:
2n + 6
Answers may vary but
likely students will say the
formula is easier.
O
1 × 15 = 15 10
2 × 15 = 30 25
3 × 15 = 45 40
a)
6n + 2
b)
302 toothpicks
c)
Figure 13
15
15
Subtract: 5
15, 5; 15, 5
4.
V‐26 “Input × Gap”
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
c)
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
AP Book PA8-25
page 122
1.
a)
3
1
( 5, 3 )
5
3
( 7, 5 )
7
5
OP
1
( 1, 2 )
1
2
TN
T
( 3, 4 )
3
4
1
5
( 5, 6 )
5
6
2
7
3
9
4
11
5
13
6
15
nd
1
( 2, 3 )
2
3
( 4, 4 )
4
4
6
2
Note that T-tables
will vary depending
on the points
selected.
nd
OP
( 6, 5 )
5
Teacher to check the
marked grid points.
a)
st
1
( 0, 1 )
0
1
( 1, 3 )
1
3
( 2, 5 )
2
5
or
( 3, 7 )
3
7
b)
OP
1
( 0, 5 )
0
5
( 2, 6 )
2
6
( 4, 7 )
4
7
OP
1
( 0, 0 )
( 1, 3 )
( 2, 6 )
st
st
0
2
nd
OP
c)
2
2
Line B
5.
3
2
6
ii)
b)
c)
1.
c)
a)
TN
T
OP
1
5
( 1, 5 )
T = 2 × TN + 3
2
8
( 2, 8 )
3
11
( 3, 11 )
e)
There is no Term
Number “0” since
the term numbers
are based on the
counting numbers.
4
14
( 4, 14 )
a)
b)
( 5, 17 )
T
OP
1
4
( 1, 4 )
2
6
( 2, 6 )
C (¢)
1
20
3
8
( 3, 8 )
2
40
4
10
( 4, 10 )
3
60
5
12
( 5, 12 )
4
80
TN
T
OP
1
4
( 1, 4 )
C = 20L
c)
200¢ = $2.00
d)
5 minutes
st
17
L (min)
b)
1
5
TN
d)
c)
8.
a)
2
nd
TN
10¢
OP
2
1
( 2, 1 )
Line C
3
3
( 3, 3 )
i)
T
OP
2
8
( 2, 8 )
12
( 3, 12 )
1
5
( 1, 5 )
3
2
8
( 2, 8 )
4
16
( 4, 16 )
3
11
( 3, 11 )
5
20
( 5, 20 )
4
14
( 4, 14 )
2.
b)
ii)
I
O
I
O
4
5
( 4, 5 )
TN
T
OP
1
2
1
3
1
1
5
7
( 5, 7 )
1
1
( 1, 1 )
5
( 2, 5 )
2
5
2
4
3
3
2
3
8
3
5
5
5
3
9
( 3, 9 )
4
11
4
6
7
7
4
13
( 4, 13 )
b)
A: O = 3I – 1
b)
( 1, 3 ), ( 2, 5 ), ( 3, 7 ),
( 4, 9 ), ( 5, 11 )
O
a)
( 40, 157 )
x=6
I
4.
( 40, 122 )
ii)
d)
BONUS
6.
i)
AP Book PA8-26
page 125
nd
0
1
a)
c)
nd
a)
Line A
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
st
2
7.
Students should not
use the point ( 0, 3 ).
( 3, 1 )
st
2
nd
1
c)
3.
st
Teacher to check
the marked points.
OP
b)
2.
b)
c)
( 1, 41 ), ( 2, 38 ), ( 3, 35 ),
( 4, 32 ), ( 5, 29 )
d)
(1, 10), (2, 20), (3, 40),
(4, 80), (5, 160)
3.
a)
A: 0, 6, 12, 18, 24
B: 64, 32, 16, 8, 4
i)
C: 64, 32, 16, 32, 64
B: O = I + 2
D: 4, 9, 4, 9, 4, 9, 4, 9
C: O = I
E: 55, 46, 37, 28, 19
Teacher to check.
F: 3, 6, 11, 18, 27
b)
AnswerKeysforAPBook8.2
i)
C
ii)
A
iii)
D
iv)
B
v)
F
V‐27
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
vi)
c)
E
d)
A:
iii)
Teacher to check.
vi)
No
Sample descriptions:
2.
ii), iii) and v)
 Increasing
sequences go
up to the right;
decreasing
sequences go
down to the right.
( 5, 12 ), ( 6, 15 )
3.
a)
 If the gaps in a
sequence are
equal (ie. they
increase or
decrease by the
same amount
each time), the
line is straight.
B:
( 1, 0 ), ( 2, 3 ),
( 3, 6 ), ( 4, 9 ),
( 1, 3 ), ( 2, 2 ), ( 3, 1 ),
( 4, 0 ), ( 5, –1 )
Yes, it is linear.
iv)
( 1, 1 ), ( 2, 5 ),
b)
( 3, 9 ), ( 4, 12 ),
( 1, 1 ), ( 2, –2 ), ( 3, 3 ),
( 4, –4 ), ( 5, 5 ), ( 6, –6 )
( 5, 14 ), ( 6, 15 )
 A repeating
sequence makes
a zig-zag pattern.
C:
AP Book PA8-27
page 126
1.
a)
i)
No, it isn’t linear.
( 1, 0 ), ( 2, 1 ),
c)
( 3, 3 ), ( 4, 6 ),
( 5, 10 ), ( 6, 15 )
v)
( 1, –8 ), ( 2, –5 ),
( 3, –2 ), ( 4, 1 ), ( 5, 4 )
( 1, 4 ), ( 2, 2 ),
( 3, 0 ), ( 4, –2 ),
( 5, –4 ), ( 6, –6 )
D:
Yes, it is linear.
INVESTIGATION 1
ii)
E:
A.
( 1, 1 ), ( 2, 3 ),
ii)
1, 3, 5, 7, 9, 11
Gaps: +2
( 3, 5 ), ( 4, 7 ),
( 5, 9 ), ( 6, 11 )
vi)
iii)
( 1, 14 ), ( 2, 10 ),
0, 3, 6, 9, 12, 15
Gaps: +3
( 3, 8 ), ( 4, 4 ),
v)
( 5, 3 ), ( 6, 0 )
4, 2, 0, –2, –4, –6
Gaps: –2
B.
i)
0, 1, 3, 6, 10, 15
Gaps: +1, +2, +3,
+4, +5
iv)
1, 5, 9, 12, 14, 15
Gaps: +4, +4, +3,
+2, +1
vi)
b)
V‐28 i)
Gaps: –4, –2, –4,
–1, –3
No
ii)
Yes
iii)
Yes
iv)
No
v)
Yes
14, 10, 8, 4, 3, 0
C.
If all the gaps are equal,
the sequence is linear.
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
F:
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
4.
a)
b)
b)
B is linear because
the gaps are all
equal to –3.
A:
c)
B:
2.
INVESTIGATION 2
A.
i)
3, 5, 7, 9
Gaps: +2
ii)
a)
3n – 2
1
1
n
2n
(n, 2n)
n
4n – 2
(n, 4n – 2)
2
4
1
2
( 1, 2 )
1
2
( 1, 2 )
3
7
2
4
( 2, 4 )
2
6
( 2, 6 )
4
10
3
6
( 3, 6 )
3
10
( 3, 10 )
5
13
4
8
( 4, 8 )
4
14
( 4, 14 )
n
2n – 1
5
10
( 5, 10 )
5
18
( 5, 18 )
1
1
2
3
3
5
4
7
5
9
n
4n – 3
(n, 4n – 3)
1
1
( 1, 1 )
2
5
( 2, 5 )
3
9
( 3, 9 )
4
13
( 4, 13 )
5
17
( 5, 17 )
n
5n – 1
(n, 5n – 1)
1
4
( 1, 4 )
2
9
( 2, 9 )
3
14
( 3, 14 )
4
19
( 4, 19 )
5
24
( 5, 24 )
i)
n
2n + 3
(n, 2n + 3)
1
5
( 1, 5 )
2
7
( 2, 7 )
3
9
( 3, 9 )
4
11
( 4, 11 )
5
13
( 5, 13 )
b)
6, 7, 8, 9
c)
Gaps: +1
iii)
1, 4, 9, 16
1, 4, 7, 10
Gaps: +3
v)
0.25, 0.5, 0.75, 1
Gaps: +0.25
vi)
Gaps: +2, +4, +6
B.
C.
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
D.
ii)
0, 2, 6, 12
i), ii), iv) and v)
Choices will vary –
teacher to check.
The two non-linear
sequences both multiply
the Term Number with
itself, rather than just
with a constant.
In each case,
n has the same
coefficient (ie. the
number in front of
it: 2) and only the
“added number”
changes (3, –2, 0).
Teacher to check
combined graph.
c)
No, the lines will
never intersect:
they are parallel.
In the formula,
this is shown by
the common n
coefficient.
Gaps: +3, +5, +7
iv)
b)
iii)
n
3.
n
2n – 2
(n, 2n – 2)
1
0
( 1, 0 )
2
2
( 2, 2 )
3
4
( 3, 4 )
4
6
( 4, 6 )
5
8
( 5, 8 )
a)
n
3n + 2
(n, 3n + 2)
1
5
( 1, 5 )
2
8
( 2, 8 )
3
11
( 3, 11 )
4
14
( 4, 14 )
5
17
( 5, 17 )
d)
AP Book PA8-28
page 128
1.
a)
n
2n + 3
1
5
2
7
3
9
4
11
5
13
AnswerKeysforAPBook8.2
V‐29
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
e)
n
10n
(n, 10n)
1
10
( 1, 10 )
2
20
( 2, 20 )
3
30
( 3, 30 )
4
40
( 4, 40 )
5
50
( 5, 50 )
c)
( 1, 60 ), ( 2, 120 ),
( 3, 180 )
Gaps: +1, +6, +2
d)
60(4.5) = 270 km
b)
The term values
e)
c)
Vertical
Teacher to check
extended line (as
shown above).
th
th
 7 term = 19
iii)
 8 term = 25
3n + 1
n
Term
1
2
2
9
b)
+4
3
16
c)
+2
4
23
d)
–1
e)
+5
3(8) + 1 = 25 
5.
The gap equals the
coefficient of n.
6.
AP Book PA8-29
page 129
th
 7 term = 44
The following values are
missing:
a)
b)
7
Input:
Output: 5
b)
2.
3.
ii)
Term Number: 4, 7
iii)
Teacher to check lines.
a)
18
b)
55
a)
i)
Term
1
2
2
6
3
10
4
14
7.
+4
ii)
+2
3n – 2
iii)
+3
8.
7n – 5
a)
4.
a)
Yes: 12 term
ii)
Yes: 16 term
iii)
No, since 7
is not a factor
of 51.
th
n
# TP
1
5
2
9
3
13
f)
a)
i)
6
ii)
4
AP Book PA8-30
page 132
iii)
1
1.
b)
A: ( 1, 1 ), iii)
B: ( 1, 4 ), ii)
Formula: 3n – 2
 17 times
2.
i)
C: ( 1, 6 ), i)
9.
B, C, A
10.
Answers will vary –
teacher to check.
11.
a)
60h
b)
h
D (km)
1
60
2
120
3
180
ii)
V‐30 4, 6, 8
b)
2n + 2
c)
Figure 49
a)
4, 9, 14
b)
5n – 1
c)
No such figure.
No multiple of 5
equals 101.
iii)
th
n
a)
Sample:
 7 term = 26
ii)
60h = 360
h=6
iii), i), ii)
th
i)
6 hours
b)
7(7) – 5 = 44 
c)
n
i)
4(7) – 2 = 26 
4n – 2
3(7) – 2 = 19 
6, 18
Term Value:
i)
3, 4, 10, 12
a)
9, 12, 15
b)
3n + 6
c)
No such figure.
Term
Sample:
No multiple of 3
equals 94.
th
1
1
 8 term = 33
2
4
4n + 1
3
7
4(8) + 1 = 33 
4
10
b)
n
# TP
1
4
2
7
3
10
iv)
3.
a)
a)
4, 8, 12
b)
4n
c)
Figure 25
“The cost (in $) of
renting a pair of
skates for n hours”
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
1.
a)
Patterns and Algebra – AP Book 8, Part 2: Unit 4 (continued)
Sample explanation:
Formula for graph
is 4n + 2, which fits
the skate scenario
($2 flat fee + $4/hr)
whereas, for volume
of a cube, outputs
3
would be 1 = 1,
3
3
2 = 8, 3 = 27, etc.
b)
4.
Answers will vary –
teacher to check.
a)
a×b=b×a
6.
b)
3 × (–5 + 3)
3 × (–5) + 3 × 3
= 3 × (–2)
= –15 + 9
= –6
= –6
2
= 3 × (–1)
= –15 + 12
2
2
= –3
= –3
2
2
2
2
3
4 – 3 = 16 – 9 = 7
4
5 – 4 = 25 – 16 = 9
5
6 – 5 = 36 – 25 = 11
7 – 6 = 49 – 36 = 13
2
(n + 1) – n
2
3 × (–5 + 5)
3 × (–5) + 3 × 5
=3×0
= –15 + 15
=0
=0
3 × (–5) + 3 × (–6)
Sample explanation:
= –15 + (–18)
= –33
= –33
b)
Figure 33
7.
8.
9.
10.
d)
x=4
e)
3x + 2x – 4x = 13
b)
4x – 2x = 8 + 2
c)
–3x – 2x = –14 + 4
Teacher to check
underlining.
d)
5x – 2x = 9 + 3
a)
b)
2x = 8
 there are 16 red
and 4 yellow beads
x = 13
4.
8.
green: x
yellow: x + 3
x=4
c)
b)
3x = 9
red: x + 4
green: 5x
x=3
yellow: x
x=2
b)
4, 4
f)
x=1
c)
(–7) × 0 + (–7) × 3
g)
x=7
red: x + 5
d)
0×3+0×4
red: x
green: x; yellow: 3x
yellow: x ‒ 3
 x + 5 + x = 3x
x=5
a)
5.
red: 4x
e)
6.
b)
8, 8
f)
9–m+m=9
a)
Tara is correct
for all values of a
except a = 0.
AnswerKeysforAPBook8.2
x
so x + 4 = 20
5, 1
The two answers are
equal, so we know that:
1
3 ÷ 12 = 4 = 0.25
x
yellow: x ÷ 4 or 4
x + 3x = –16
4x = –16
3.
red: x
a)
m–5+5=m
f)
e)
x=4
x = –4
e)
3 × 12 = 36
 there are 11 red
and 9 yellow beads
8x – 3x = 20
d)
–1, –1
c)
d)
e)
a)
5, 3
b)
8, 5, –3;
c)
d)
7, –2, 9;
7.
b)
yellow: x + 3
9.
30 beads per box
Sample approach:
10.
3 hours
Sample approach:
7 + 3n = 10 + 2n
n=3
red: x
red: x
red: x
7 × 9 + (–2) × 9
yellow: x + 6
–3, 2, 0;
so x + x + 6 = 20
(–3) × 0 + 2 × 0
green: x
yellow: 4x
x
yellow: x ÷ 4 or 4
–3, –3
red: x – 2
red: x
yellow: x + 2
f)
c)
red: x
yellow: x ‒ 5
(a + b) × c = a × c + b × c
11.
so x + x ‒ 2 = 20
x = 11
–4, 12, 83
–70, –70
e)
yellow: x ‒ 2
c)
b)
12 ÷ 3 = 4
5x – 2x = 9
red: x
x = –3
c)
d)
b)
d)
c)
m÷2×2=m
12 × 3 = 36
+ 2x
3, 2, –5
b)
7+m–m=7
c)
f)
x=2
5, –2, 7
d)
15
+ 3x
b)
a)
c)
b)
e)
 there are 5 red
and 15 yellow beads
x=3
3, –5, 5;
m+3–3=m
Answers will vary –
teacher to check.
– 4x
3, –5, –6
b)
b)
d)
x = –4
= 3 × (–11)
2n + 10
+ 4x
d)
3 × (–5 – 6)
a)
c)
5x = 20
2009 row
4 019
so x + 3x = 20
x=5
– 3x
c)
red: x
yellow: 3x
3x = 9
e)
th
c)
b)
c)
AP Book PA8-31
page 133
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
2.
2
Teacher to check
student checks.
3.
Yes, though
answers will vary –
teacher to check.
3 × (–5) + 3 × 4
st
2.
c)
3 × (–5 + 4)
Using the 1 formula
above, n + 1 = 2010
and n = 2009.
1.
1.
5.
or 2n + 1
5.
Yes: –15
2
3 –2 =9–4=5
d)
b)
2
2
c)
Yes: –28
2
2 –1 =4–1=3
b)
a)
AP Book PA8-32
page 135
Circle: c), e)
2
1
6
4.
11.
32 units
Sample approach:
3x + 10 = 5x + 4 + x
3x + 10 = 6x + 4
–3x = –6
x=2
x=7
 there are 7 red
and 13 yellow beads
V‐31