2019 - 12MM Headstart Booklet

x=2 1=2
b k = −3 5b
b = 10
2
a A unique solution for b ∈ R\{10}
9 x + 5y = 4 . . . (1)
2x + by = c . . . (2) b If b = 10 and c = 8 the corresponding
lines coincide and there are infinitely
Gradient of (1) = −1 many solutions.
5
c If b = 10 and c 8 the corresponding
Gradient of (2) = −2 lines are parallel and there are no
b solutions

For the lines to be parallel
or coincide:

66

Solutions to Exercise 2G c x + y = 5 . . . (1)
y + z = 7 . . . (2)
1 a 2x + 3y − z = 12 . . . (1) z + x = 12 . . . (3)
2y + z = 7 . . . (2)
2y − z = 5 . . . (3) Subtract (2) from (3)
x − y = 5 . . . (4)
Add (2) and (3)
4y = 12 Add (1) and(4)
y=3 ∴x=5
∴z=1 ∴y=0
∴z=7
Substitute in (1) to find x
x=2 d x − y − z = 0 . . . (1)
5x + 20z = 50 . . . (2)
b x + 2y + 3z = 13 . . . (1) 10y − 20z = 30 . . . (3)
−x − y + 2z = 2 . . . (2)
−x + 3y + 4z = 26 . . . (3) Simplify (2) and (3)
x + 4z = 10 . . . (4)
Add (1) and (2) y − 2z = 3 . . . (5)
y + 5z = 15 . . . (4)
Subtract (5) from (4)
Subtract (2) from(3) (x − y) + 6z = 7 . . . (6)
4y + 2z = 24
2y + z = 12 . . . (5) Subtract (1) from (6)
7z = 7
Multiply (4) by 2 and ∴z=1
subtract from (4)
−9z = −18 ∴x=6
z=2 ∴y=5
∴y=5
∴ x = −3 2 a y − 4z = −2
y = 4z − 2

b z=λ
y = 4λ − 2
∴ x + 8λ − 4 − 3λ = 4
x = 8 − 5λ

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3 a −y + 5z = 15 (2) + (1) 5 x + y + z + w = 4 . . . (1)
−y + 5z = 15 (3) − (2)
x + 3y + 3z = 2 . . . (2)

x + y + 2z − w = 6 . . . (3)

b They are the same (3) − (1)

c z=λ ⇒ z − 2w = 2
−y + 5λ = 15
y = 5λ − 15 Let z = t, t ∈ R
t−2 1
w = = t − 1
22

d x + 10λ − 30 + 3λ = 13 (2) − (3) gives
x = 43 − 13λ
2y + z + w = −4

2y + t + 1 − 1 = −4
t
2
3
4 a (1) +(2) 2y = −3 − t
⇒ 2z = 10 2
z=5 −3 3
Substitute into (1) y = − t
⇒ x−y+5=4 24
let y = λ Substitute into (1)
x=λ−1
x − 3 − 3 + t + 1 − 1 = 4
b Let z = λ t t
Substitute in (2) 24 2
⇒ x=3+λ 1 3 26 − 3t
Substitute into (1) x = 6 − t =
⇒ 6 + 2λ − y + λ = 6 24 4
y = 3λ when w = 6, t − 2 = 6 so t = z = 14
2
c (1) + 2 × (2) y = −3(14 + 2) = −12
6x + 3z = 14 4
Let z = λ x = 26 − 42 = −4
6x = 14 − 3λ 4
x = 14 − 3λ
6 6 a 3x − y + z = 4 . . . (1)
Substitute into (2) x + 2y − z = 2 . . . (2)
14 − 3λ + y + λ = 4 −x + y − z = −2 . . . (3)
6 (2) − (3) ⇒ 2x + y = 4
y = 4 − 14 + 3λ (3) + (1) ⇒ 2x = 2
6 x=1
y = 10 − 3λ y=2
6 Substitute into (3)
⇒ −1 + 2 − z = −2
z=3

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b x − y − z = 0 . . . (1) c 12x − y + z = 0
3y + 3z = −5 . . . (2) 2y + 2z = 2
3 × (1) + (2) Let z = λ
⇒ 3x = −5 y = 2 − 2λ
x = −5 2x − 2 + 2λ + λ = 0
3 2x = 2 − 3λ
3y = −5 − 3z x = 2 − 3λ
Let z = λ 2
y = −5 − 3λ
3

69

Solutions to technology-free questions

1 a 3x − 2 = 4x + 6 b x−y =2 ...1
4x − 3x = −2 − 6 43 ...2
x = −8 y−x=5

b x+1 = 4 Multiply 1 by 12:
2x − 1 3
3x − 4y = 24 . . . 3
3(x + 1) = 4(2x − 1)
3x + 3 = 8x − 4 Multiply 2 by 3:
8x − 3x = 3 + 4
5x = 7 3y − 3x = 15 . . . 4
x= 7
5 3 + 4 gives −y = 39, so y = −39

c 3x − 7 = 11 Substitute into 2:
5
3x = 18 −39 − x = 5, so x = −44
5
3x = 90 n+m
x = 30 3a

d 2x + 1 = x − 1 b
52
b
2(2x + 1) = 5(x − 1) b c+b
cd

6
d

q− p
m+n
e
m−n

a2
f

a−1

4x + 2 = 5x − 5 4 a intercepts 5, 0 , 0, 5
x=7 23

2 a y= x+4 ... 1

5y + 2x = 6 . . . 2

Substitute 1 into 2:

5(x + 4) + 2x = 6

5x + 20 + 2x = 6

7x = −14 b intercepts (6, 0), (0, −6)

x = −2
Substitute into 1:

y = −2 + 4 = 2

70

6 distance = (2 − (−1))2 + (4 − 6)2

= 32 + (−2)2
√

= 9+4
√

= 13

c intercepts (2, 0), (0, 3) 7 Midpoint = 4 + (−2), 6 + 8 = (1, 7)
22

8 a Let (x, y) be the coordinates of Y.

x + (−6), y + 2 = (8, 3)
22

∴ x − 6) = 8 and y + 2 = 3
22

∴ x = 22 and y = 4

5 a y − 3 = −2(x − 1) b Let (x, y) be the coordinates of Y.
y = −2x + 5 x + (−1), y + (−4) = (2, −8)
22
b m= 8−4 =2
3−1 ∴ x − 1) = 2 and y − 4 = −8
22
y − 4 = 2(x − 1)
∴ x = 5 and y = −12
y = 2x + 2
9a 2 1+4 0=6 1
c y = −2x + 6 has gradient m1 = −2; 32 12 44

for the gradient m2 of a perpendicular b 2 1 4 0= 9 2
3 2 1 2 14 4
line: m1m2 = −1
−1 1 c 2 1 −2 = −1
m2 = −2 = 2 32 3 0

y − 1 = 1(x − 1) d 4 0 −2 = −8
2 12 3 4

y = 1x+ 1 e 3 −2 = −6
22 39

d y = 6 − 2x has gradient −2; a parallel f 4 0 2 1=8 4
line has the same gradient: 1232 85
y − 1 = −2(x − 1)
y = −2x + 3

71

g 21 4 0 = −2 1 a Checking back in the equations there
− are infinitely many solutions when
32 12 20 m = −2.
Equation (1) becomes −2x − 4y = 1
2 1 = 2k k Equation (2) becomes 4x + 8y = −2
hk
3 2 3k 2k

i b There is a unique solution for
2 2 1 + 3 4 0 = 4 2 + 12 0 m ∈ R\{−2, −8}
32 12 64 36

= 16 2 12 a 2x − 3y + z = 6 . . . (1)
9 10
−2x + 3y + z = 8 . . . (2)
j 2 14 0=2 18 0
−2 − Add (1) and (2)
32 12 32 24
2z = 14
= −6 1
1 −2 z=7
Substitute in (1)
10 (10 − 5)2 + (y − 12)2 = 13 2x − 3y = −1
25 + (y − 12)2 = 169
(y − 12)2 = 144 ∴ y = 2x + 1
y − 12 = ±12 3
y = 0 or y = 24
Let x = λ
The solution is (λ, 2λ + 1, 7) where

3
λ∈R

b x − z + y = 6 . . . (1)

11 mx − 4y = m + 3 . . . (1) 2x + z = 4 . . . (2)

4x + (m + 10)y = −2 . . . (2) Let z = λ

Gradient of (1) = m Then x = 4 − λ
4 2

Gradient of (2) = − m 4 10 Substitute in (1)
+ 4−λ −λ+y = 6
Infinitely many or no solutions
2
when the gradients are the same. ∴ y = 8 + 3λ
2
−m = 4
4 m + 10 The solution is (4 − λ, 3λ + 8, λ)
22
m2 + 10m + 16 = 0
where λ ∈ R

(m + 8)(m + 2) = 0

m = −8 or m = −2

72

Solutions to multiple-choice questions

1 E y = mx + c 6 A Line passes through the points
(3, −2) and (−1, 10)
1
The m (gradient) value is − , ∴ y2 − y1,
x2 − x1
2 = 10 − −2
It passes through the point (1, 4) −1 − 3
= 12
4 = 1 + c −4
−1 = −3
2
∴ c=9 7 A Eqn 1: y = 2x + 3
2 Eqn 2: y = ax + 4
1 9 33
∴ y = − x + To be parallel gradients must be the
22 same.

2 E y = −2x + 4 ∴ a=2
Point (a, 3) 3
3 = −2a + 4
a= 1 ∴ a=6
2
8 C y = mx + c
3 D Line passes through the points m = 10 − −2
3 − −1
(−2, 0) and (0, −1) m=3
∴ y2 − y1 Passes through the point (3, 10)
x2 − x1
= −1 − 0 ∴ 10 = 9 + c
0 − −2 ∴ c=1
m (gradient) = −1 ∴ y = 3x + 1
2
Perpendicular line = − 1 9 B Distance between x points
m = |x2 − x1|
1
∴ − 1 =2 = |5 − 1|

− 2 =4
Distance between y points
4 C Midpoint at x2 + x1 , y2 + y1 = y2 − y1
22
= | − 2 − 4|
= −2, 17
22 =6
U√sing Pythagoras
= (−1, 8.5)
42 + 62
5 B 2ax − 10by = 22 √
+++ = 52
4ax + 10by = 2 √
= 2 13
∴ 6ax = 24
∴ x=4

a
Do not need to solve for y as there is

only one possible option.

73

10 C y = mx + c 12 A
Passes through points (4, 0) and (a − 1)x + 5y = 7 . . . (1)

(0, −3) 3x + (a − 3)y = 0 . . . (2)
m = −3 − 0
Gradient of (1) = a − 1
0−4 −
5
m= 3
4 Gradient of (2) = − 3
a−3
Y intercept = −3
∴ c = −3 Infinitely many or no solutions
∴ f (x) = 3 − 3
when the gradients are the same.
4 a−1 = 3
5 a−3
11 D bx + 3y = 0 . . . (1)
a2 − 4a − 12 = 0

4x + (b + 1)y = 0 . . . (2) (a − 6)(a + 2) = 0

Gradient of (1) = −b a = −2 or a = 6
3

Gradient of (2) = − b 4 1 13 D 0 + 4, d − 6 = 2, d − 6
+
Infinitely many solutions when the 22 2

gradients are the same. 14 C Gradient of line segment joining
(3, 0) and (0, −6) is 6 = 2 Gradient
b = b 4 1 3
3 + 1
of line perpendivular to this is -
b2 + b − 12 = 0 2

(b + 4)(b − 3) = 0

b = −4 or b = 3

74

Solutions to extended-response questions

1 a Graph is a straight line passing through (100, 50) and (50, 75).

Note that extending it back to the P axis shows that the intercept is (0, 100); this is
confirmed in part b below.

b Relationship is linear: P = aN + b
P = 50, N = 100: 50 = 100a + b . . . 1

P = 75, N = 50: 75 = 50a + b . . . 3

1 − 2: 50a = −25

a = 1
−
2
which implies b = 100

Hence P = 1 + 100.
−N
2

c i N = 88: P = 1 × 88 + 100
−
2

= 56

So the price would be $56.

ii P = 60: 60 = 1 N + 100
−
2

1 = 40
N
2

N = 80
So the number of jackets would be 80.

2 a The rule is of the form p = at + b
When t = 3, p = 12000 and when t = 8, p = 19240
Therefore the equations
12000 = 3a + b1
and 19240 = 8a + b2 are satisfied.
Subtract 1 from 2 to give 5a = 7240.
Hence a = 1448

75

Substitute in 1 to find that b = 7656
Therefore p = 1448t + 7656
b

The p axis intercept gives the initial population.
c When t = 10,

p = 14480 + 7656
= 22136

d The average rate of growth is the gradient. The growth rate is 1448 people per year.
3

a Midpoint of AB = 7 + 2, 5 + 2 = 9, 7
22 22
Gradient of AB = 5 − 2 = −3
2−7 5

Therefore equation of perpendicular bisector of AB is

y−7 = 5 x−9
23 2
5
Therefore y = x−4
3

76

b Solving the equations y = 4x − 26 and y = 5 −4 simultaneously for x and y will
x
3
give the coordinates of D

Consider 4x − 26 = 5 x − 4
3

7x = 22
3
x = 66

7
Substitute x = 66 in the equation y = 4x − 26 to give y = 82

77
Coordinates of D are 66, 82

77

c 5
Line BC is perpendicular to line AB. Therefore gradient of BC is
3

d B(2, 5) and C(8, c). The gradient of BC can also be written as 5 − c
−6

Therefore 5 − c = 5
−6 3

Solving for c gives c = 15

e The area will be found by calculating the area of triangle DXA and trapezium
BCDX. Let X be the midpoint of AB. From the above the coordinates of X are 9, 7
22

Distance XD = 66 − 9 2 + 82 7 2
72 −
72

= 8993 5 2+ 3 2
98 22
√

= 23 34
14

Distance XA = distance XB =

= 17
2

Area = area of triangle DXA + area of trapezium BCDX.

= 1XA × XD + 1 + XD)
BX(BC
22

= 1 + 2XD)
AX(BC
2

= 629
14

77

4

a Midpoint of BC = 4, 9
2
8 − 1 7
Gradient of AB = = −
2−6 4
Therefore gradient of perpendicular bisector = 4
7
The equation of the perpendicular bisector is

y − 9 = 4 − 4)
(x
27
Therefore y = 4 x + 31
7 14

b The perpendicular bisector passes through C as the triangle is isosceles.
When x = 3.5, y = 4 × 3.5 + 31 = 59
7 14 14
The coordinates of C are 7, 59
2 14

√
c The length of AB = (6 − 2)2 + (1 − 8)2 = 65

d The area of the triangle = 1 × √ × XC where X is the midpoint of AB
65
2√
XC = 7 − 4 2 + 59 − 9 2 = 65
2 14 2 √ 14
√
Therefore area = 1 × 65 × 65 = 65 square units.

2 14 28

5 A(−4, 6) and B(6, −7)

a Midpoint = −4 + 6, 6 + −2 = 1, −1
22 2

√
b/c The length AB = (−7 − 6)2 + (6 − −4)2 = 269 = the distance between A and B

78

d gradient of AB = 6 − −7
−4 − 6

= −13
10
13
The equation of AB is y − 6 = − (x + 4)
10
Rearranging gives

y = 13 x + 4
−
10 5

e The perpendicular bisector has gradient 10

1 10 13
(x − 1)
The equation is y+ =
2 13
10 33
Therefore y = x−
13 26

f

Triangles AXP and PY B are similar with scale factor 3.

AX : PY = 3 : 1
6 − b 3
Therefore b + 7 = 1

Therefore b = 15
−
4
Also XP : Y B = 3
a+4 =3
6−a
a= 7
2
coordinates of P are 7, −15
24

g

Triangles AXB and AY P are similar with scale factor 3.
Therefore

79

a+4 = 3
10 1

a = 26
Also b − 6 = 3

−7 − 6
b = −33
The coordinates of P are (26, −33)
6 a 25% of 500 = 125
125 litres of acid is required to produce 500 litres of a 25% acid solution.
b Let x denote the amount of 30% solution.
Let y denote the amount of 18% solution.

∴ x + y = 500 1
0.3x + 0.18y = 125 3
From 1 y = 500 − x. Substitute in 2
∴ 0.3x + 0.18(500 − x) = 125
∴ (0.3 − 0.18)x + 90 = 125

∴ 0.12x = 35
∴ x = 875
3

Substitute in 1 y = 500 − 875 = 625
33

875 625
litres of the 30% solution and litres of the 18% solution are required.

33
Graphical Calculator techniques for Question 6.
In a Calculator page use Algebra>Solve System of Equations>Solve System of
Equations.
For exact answers enter decimal inputs as fractions such as 30/100 or 30 as shown.

80