KSSM_Form_4_Additional_Mathematics_Notes

CHAPTER 3 SYSTEMS OF EQUATIONS

Chapter 3 Systems of Equations

3.1 Systems of Linear Equations in Three Variables

1) x − 2y + z = 10 … ○1

2x + 2y + 3z = 13 … ○2

–2x + 5y + z = –7 … ○3

○1 ×2: 2x – 4y + 2z = 20 … ○4

○2 –○4 : 6y + z = –7

z = –7 – 6y … ○5

○2 +○3 : 7y + 4z = 6

4z = 6 – 7y … ○6

Substitute ○5 into ○6

4(–7 – 6y) = 6 – 7y

–28 – 24y = 6 – 7y

17y = –34

y = –2

Substitute y = –2 into ○5

z = –7 – 6(–2)

z=5

Substitute y = –2 and z = 5 into ○1

x – 2(–2) + 5 = 10

x=1

So, x + y + z = 1 + (–2) + 5

=4

2) x + 2y – z = 2 … ○1

2x – 3y + z = –1 … ○2

5x – y – 2z = –3 … ○3

○1 +○2 : 3x – y = 1 … ○4

○2 ×2: 4x – 6y + 2z = –2 … ○5

○3 +○5 : 9x – 7y = –5 … ○6

○4 ×7: 21x – 7y = 7 … ○7

○7 –○6 : 12x = 12

x=1

Substitute x = 1 into ○4

3(1) – y = 1

y=2

Substitute x = 1 and y = 2 into ○1

1 + 2(2) –z =2

z=3

97

CHAPTER 3 SYSTEMS OF EQUATIONS

3) x + y + z = 6 … ○1

y + 5z = −7 … ○2

2x + 5y − z = 27 … ○3

From ○2 : y = –7 – 5z …○4

○1 ×2: 2x + 2y + 2z = 12 … ○5

○3 –○5 : 3y – 3z = 15

y=5+z … ○6

Compare ○5 and ○6

–7 – 5z = 5 + z

6z = –12

z = –2

Substitute z = –2 into ○4

y = –7 – 5(–2)

y=3

Substitute y = 3 and z = –2 into ○1

x + 3 + (–2) = 6

x=5

4) Let the variables
Y: The number of yellow balls
P: The number of pink balls
B: The number of blue balls
There are 90 balls altogether and 25 are black, so the total number of yellow, pink and blue balls
are 90 – 25 = 65. The equation written as:
Y + P + B = 65 … ○1

60 of the balls are not pink and 25 are black, so the number of not pink ball excluded black balls:
60 – 25 = 35
35 balls are the total number of yellow balls and blue balls, the equation written as:
Y + B = 35 … ○2

67 are not yellow ball and 25 are black, so the number of not yellow balls excluded black balls:
67 – 25 = 42
42 balls are the total number of pink balls and blue balls, the equation written as:
P + B = 42 … ○3
From ○1 and ○2 we can deduce that
○1 –○2 : P = 30
Substitute P = 30 into ○3
30 + B = 42
B = 12
Substitute B = 12 into ○2
Y + 12 = 35
Y = 23
Thus, Adam has 23 yellow balls, 30 pink balls and 12 blue balls respectively.

98

CHAPTER 3 SYSTEMS OF EQUATIONS

5) Using the standard equation given: x2 +y2 + Ax + By + C = 0
Substitute 3 coordinates point (1,1), (2, –4) and (5,5) into the standard equation given to form

three new equations.

For (1,1): 1+1+A+B+C=0

A + B + C = –2 … ○1

For (2, –4): 22 + (–4)2 + A(2) + B(–4) + C = 0

20 + 2A – 4B + C = 0

2A – 4B + C = –20 … ○2

For (5,5): 52 + 52 + 5A + 5B + C = 0

50 + 5A + 5B + C = 0

5A + 5B + C = –50 … ○3

○1 ×2: 2A + 2B + 2C = –4 … ○4

○4 –○2 : 6B + C = 16

C = 16 – 6B … ○5

○1 ×5: 5A + 5B + 5C = –10 … ○6

○6 –○3 : 4C = 40

C = 10

Substitute C = 10 into ○5

10 = 16 – 6B

B=1

Substitute B = 1 and C = 10 into ○1

A + 1 + 10 = –2

A = –13
Hence, the equation of the circle that passes through the points (1,1), (2, –4) and (5,5) are

x2 + y2 – 13x + y + 10 = 0

6) Let the variables

x: The amount of money invested at 6%

y: The amount of money invested at 8%

z: The amount of money invested at 10%

The total amount of money invested at three different rate are RM50000

The equation can be written as:

x + y + z = 50000 … ○1

The yearly interest is RM3700. Simple interest in percentage form can be convert to decimal

form. For example, 6% simple interest is equivalent to 0.06 times the amount of money invested
at 6%.
0.06x + 0.08y + 0.10z = 3700 … ○2

Amount of money invested at 6% is twice as much as invested at 10%.

x = 2z … ○3
x – 2z = 0

From ○3 : x = 2z … ○4

○1 ×0.08: 0.08x + 0.08y + 0.08z = 4000 … ○5

○5 –○2 : 0.02x – 0.02z = 300 … ○6

Substitute ○4 into ○6

0.02(2z) – 0.02z = 300

99

CHAPTER 3 SYSTEMS OF EQUATIONS
z = 15000
Substitute z = 15000 into ○4
x = 2(15000)
= 30000
Substitute x = 30000 and z = 15000 into ○1
30000 + y + 15000 = 50000
y = 5000
Thus, RM30,000 is invested at 6%, RM5,000 is invested at 8% and RM15,000 is invested at
10%.

100

CHAPTER 3 SYSTEMS OF EQUATIONS

3.2 Simultaneous Equations Involving One Linear Equation
and One Non-Linear Equation

1) 2x + y = 1 … ○1

x2 + y2 = 1 … ○2

From ○1 : y = 1 – 2x … ○3

Substitute ○3 into ○2

x2 + (1 – 2x)2 = 1

x2 + (1 – 4x + 4x2) = 1

5x2 – 4x = 0

x(5x – 4) = 0

x = 0 or 4

5

Substitute x = 0 and x = 4 into ○3
5

y = 1 – 2(0) or y = 1 – 2( 4 )

5

=1 = – 3
5

When x = 0, y = 1

When x = 4 , y = – 3
5 5

2) x2 – y = 14 … ○1

2y – 4 = 12x … ○2

From ○2 : 2y = 12x + 4

y = 6x + 2 … ○3

Substitute ○3 into ○1

x2 – (6x + 2) = 14

x2 – 6x – 16 = 0

(x – 8)(x + 2) = 0

x = 8 or –2

Substitute x = 8 and x = –2 into ○3

y = 6(8) + 2 or y = 6(–2) + 2

= 50 = –10

When x = 8, y = 50

When x = –2, y = –10

3) x² − y² = 7 … ○1

2y = 2 + x … ○2

From ○2 : x = 2y – 2 … ○3

Substitute ○3 into ○1

(2y – 2)2 – y2 = 7

4y2 – 8y + 4 – y2 = 7

3y2 – 8y – 3 = 0

(y – 3)(3y + 1)= 0

y = 3 or – 1
3

101

CHAPTER 3 SYSTEMS OF EQUATIONS

Substitute y = 3 and y = – 1 into ○3
3

x = 2(3) – 2 or x = 2(– 31) – 2

=4 = – 8
3

When x = 4, y = 3

When x = – 8 , y = – 1
3 3

4) For equation x + 2y – 4 = 0

x –3 –2 –1 0 1 2 3
y 3.5 3 2.5 2 1.5 1 0.5

For equation x2 + y2 = 7 – xy 1 2 3
–3, 2 –3, 1 –2, –1
x –3 –2 –1 0

y 1, 2 –1, 3 –2, 3 ± √7

Based on the graph, there are two points of intersection which is (–2,3) and (2,1).
Thus, the solutions to the simultaneous equations are x = –2, y = 3 and x = 2, y = 1.

5) 4x + y – 5 = 0 … ○1

27x2 + 21xy = –2y2 … ○2

From ○1 : y = –4x + 5 … ○3

Substitute ○3 into ○2

27x2 + 21x(–4x + 5) = –2(–4x + 5)2

27x2 – 84x2 + 105x = –2(16x2 – 40x + 25)

–57x2 + 105x = –32x2 + 80x – 50

25x2 – 25x – 50 = 0

x2 – x – 2 = 0

(x – 2)(x + 1) = 0

102

CHAPTER 3 SYSTEMS OF EQUATIONS

x=2 or x = –1

Substitute x = 2 and x = –1 into ○3

y = –4(2) + 5 or y = –4(–1) + 5

= –3 =9

When x = 2, y= –3

When x = –1, y = 9
Therefore, the coordinates of point A is (2, –3) and point B is (–1,9)
[Alternative answer: Coordinate of point A is (–1,9) and point B is (2, –3)]

6) Area of the rectangular plot: 30 × x = 30x

Area of the fish pond: (x – 10) × (30 – y) = 30x – xy – 300 + 10y

The equation of area of the region grew with bananas can be written as:

30x – (30x + 10y – xy – 300) = 460

–10y + xy + 300 = 460
xy – 10y = 160 … ○1

The perimeter of the rectangular fish pond: 2(x – 10) + 2(30 – y) = 2x – 2y + 40

The equation of the perimeter of the fish pond can be written as :
2x – 2y + 40 = 48 … ○2

We obtain two equation
From ○2 : x – y + 20 = 24

x=y+4 … ○3

Substitute ○3 into ○1

y(y + 4) – 10y = 160

y2 + 4y – 10y – 160 = 0

y2 – 6y – 160 = 0

(y – 16)(y + 10) = 0

y = 16 or y = –10 (rejected since area cannot be negative)
Substitute y = 16 into ○3

x = 16 + 4

= 20

Thus, x = 20 and y = 16.

103

Chapter 4 Indices, Surds and Logarithms

Chapter 4 Indices, Surds and Logarithms

4.1 Law of Indices

13 13

(4 2) 43( 2)

1. (i) 2√ = 1

2 2
3

= 64 2 Simplify the expression by
using the law of indices
1

2 2

= 32 23−21

= 32

(ii) 3 +1+9 = 3 31+9 Notice 3 + 3 is the common factor
3 −1+1 3 3−1+1 for the numerator and denominator.
3(3 +3) We can factor and cancel it out.
= 3−1(3 +3)

=9

2. (i) 3 + 3 +1 + 3 +3 = 3 + 3(3 ) + 33(3 ) Factor out the common factor

= 3 (1 + 3 + 33) 3 . We find 31 is one of the

= 3 (31) factors of the expression.

Since 3 + 3 +1 + 3 +3 can be written as 3 (31), which means it has a factor of 31. Thus, the

expression is divisible by 31.

(ii) 7 + 7 +2 + 72 + 1 = 7 + 72(7 ) + 49 + 1
= 7 (1 + 72) + 50
= 7 (50) + 50
= 50(7 + 1)

Since 7 + 7 +2 + 72 + 1 can be written as 50(7 + 1), which means it has a factor of 50. Thus, the

expression is divisible by 50.

3. (i) 3 −3 = 9 Make the bases for both sides
3 −3 = (32) equal by using the law of indices.
3 −3 = 32

By comparison,
− 3 = 2

= −3

(ii) 25(5 +1) = 52 −1
52(5 +1) = 52 −1
5 +1+2 = 52 −1
5 +3 = 52 −1

By comparison,

+ 3 = 2 − 1

= 4

104

Chapter 4 Indices, Surds and Logarithms

4.2 Law of Surds

1. (i) Let = 0. 1̅̅̅4̅

100 = 14. ̅1̅̅4̅ Form two equations and solve for .

100 − = 14. ̅1̅̅4̅ − 0. ̅1̅̅4̅ Take note that the recurring decimal does not
start directly after the decimal point. Therefore,
99 = 14 forming two new equations of 10 and 1000
is necessary to solve for the fraction.
= 14
99
14
∴ 0. 1̅̅̅4̅ = 99

(ii) Let = 0. 1̅

10 = 1. 1̅

10 − = 1. 1̅ − 0. 1̅

9 = 1

= 1
9
∴ 0. 1̅ = 1
9

(iii) Let = 0.0̅5̅̅6̅
10 = 0. ̅5̅̅6̅

1000 = 56. ̅5̅̅6̅
1000 − 10 = 56. 5̅̅̅6̅ − 0. 5̅̅̅6̅

990 = 56
= 56

990

= 28

495

2. (i) 3 = 3 × √2
2√2 2√2 √2

= 3√2

4

(ii) 1 = 1 × 3√7+2√2 Multiply the conjugate for the surds in the
3√7−2√2 3√7−2√2 3√7+2√2 denominator in order to rationalise the fractions.

= 3√7+2√2
(3√7)2−(2√2)2

= 3√7+2√2

63−8

= 3√7+2√2
55

(iii) 2√3−3√5 = 2√3−3√5 × 2√3−3√5
2√3+3√5 2√3+3√5 2√3−3√5

= (2√3−3√5)2
(2√3)2−(3√5)2

= (2√3)2−2(2√3)(3√5)+(3√5)2

12−45

= 12−12√15+45

−33

= − 57−12√15

33

= 4√15−19

11

105

Chapter 4 Indices, Surds and Logarithms

4.3 Law of Logarithms

1. (i) log2(log3(2 + 1)) = 2 Apply the law of logarithm by converting
22 = log3(2 + 1) logarithmic expression to indices form.
34 = 2 + 1
81 = 2 + 1 Changing the base of logarithms to make
= 40 both sides have the same base.

(ii) log64(log2(3 − 2)) = log25 3√5 106

log64(log2(3 − 2)) = log5 3√5
log5 25
1

log64(log2(3 − 2)) = log5 53
log5 52

log64(log2(3 − 2)) = 1 log5 5
3 5

2 log5

log64(log2(3 − 2)) = 1
6
1
log2(3 − 2) = 646

log2(3 − 2) = 2
3 − 2 = 22

= 2

(iii) 3log3(2 −3) = 15
log3 15 = log3(2 − 3)

By comparison,
2 − 3 = 15
= 9

2. (i) log9(2 + 12) = log3( + 2)

log3(2 +12) = log3( + 2)
log3 9

log3(2 +12) = log3( + 2)
log3 32

log3(2 +12) = log3( + 2)
2 log3 3

log3(2 + 12) = 2 log3( + 2)
log3(2 + 12) = log3( + 2)2

By comparison,

2 + 12 = ( + 2)2

2 + 12 = 2 + 4 + 4

2 + 2 − 8 = 0

( + 4)( − 2) = 0

= −4 2

(ii) 2 2 = log2(4 − 1)
log7

2 7) = log2(4 − 1)
(log12

2 log2 7 = log2(4 − 1)
log2 72 = log2(4 − 1)

By comparison,

4 − 1 = 49

= 12.5

Chapter 4 Indices, Surds and Logarithms

3. 2 log3( − ) = 1 + log3 + log3
log3( − )2 = 1 + log3 + log3
log3( − )2 = log3 3 + log3 + log3
log3( − )2 = log3 3

By comparison,
( − )2 = 3

2 − 2 + 2 = 3
2 + 2 = 5 (shown)

4. (i) log2( + 2) + 5 log4 − 3 log2 √ Apply the law of logarithms
to simplify the expressions.
= log2( + 2) + 5 log2 − 3 log2 √
log2 4
3
= log2( + 2) + 5 log2 − log2
2 2

53

= log2( + 2) + log2 2 − log2 2

5
= log2 [( +23) 2]
2

= log2[( + 2) ]

(ii) log2( + 2) + 5 log4 − 3 log2 √ = 3
log2[( + 2) ] = 3
( + 2) = 23
2 + 2 − 8 = 0

( + 4)( − 2) = 0

= −4 2

5. log2 3 × log3 4 × log4 5 × log5 6 = log 3 × log 4 × log 5 × log 6 Notice the pattern of the expression
log 2 log 3 log 4 log 5 where terms can be cancelled out
after changing their bases.
= log 6
log 2

= 2.5850

107

Chapter 4 Indices, Surds and Logarithms

4.4 Application of Indices, Surds and Logarithms

1. Suppose AMON be a rectangle with triangle ABC inscribed in it as follows:

Let the radius of circle A, B, and C be , , and respectively, we have:

( + )2 = 2 + 2
( + 12)2 = 2 + 2
( + )2 = 2 + 2
( + 6)2 = 2 + 2
( + )2 = 2 + 2
(12 + 6)2 = 2 + 62

2 = 288

We know that = − = 12 − , = − = 6 − , and = +

From equation 1,
2 = ( + 12)2 − 2
2 = 2 + 24 + 144 − (144 − 24 + 2 )
2 = 48

From equation 2,
2 = ( + 6)2 − 2
2 = 2 + 12 + 36 − (36 − 12 + 2 )
2 = 24

Therefore from = + , we have

√288 = √48 + √24

√24√12 = √24√2 + √24√

√12 = √ (√2 + 1)

12 = (3 + 2√2)

= 12
3+2√2

= 2.059

108

Chapter 4 Indices, Surds and Logarithms

2. (i) = 5, Let the value of the car after 5 years be , we have

= 80000 (8)5

9

= 44394.32

(ii) 80000 (8) < 35000

9
(89)
< 35000
80000
log (8) < log 7

9 16

log 8 < log 7
9 16
log176
> log98

> 7.019

8 years will be needed for the car to drop below RM35000 for the first time.

109

Chapter 5 Progressions

5.1 Arithmetic Progression

1) a = 1038
= 964

l = 594

= a + (3 1)d
964 = 1038 + 2d

‫ﮠ‬ਤ

d=

= 37

= a + (n 1)d 37)
594 = 1038 + (n 1)(
594 = 1075 - 37n

ਤ

n=

= 13

Since the last term is the term, the arithmetic progression has 13 terms.

2) Money Harry saves every day follows the arithmetic sequence,
12, 15, 18, …

a = 12
d = 15 12

=3

Total amount of money saved till the tenth day, = [2(12) + (10 1)3]

= 5(51)
= RM255

x = 605 255
= 350

3) The middle two terms are the term and the term.
+ = 268

a + (8 1)d + a + (9 1)d = 268

2a + 15d = 268 ○1

= 101
a + (3 1)d = 101

a + 2d = 101 ○2

From ○2 ,

a = 101 2d ○3

Substitute ○3 into ○1 ,

2(101 2d) + 15d = 268

202 + 11d = 268

‫ﮠ‬

d=

=6

Substitute d = 6 into ○3 ,

The last term, a = 101 2(6)

= 89

‫ = ﮠ‬a + (16 1)d
= 89 + (15)6
= 179

4) The radius of the circles follows the arithmetic sequence,
3, 6, 9, ….

a=3
d=6 3

=3

Sum of the seven terms, = [2(3) + (7 1)3]

= (24)

= 84
Length of the rectangle = 84 x 2

= 168 cm
Radius of the seventh circle, = 3 + (7 1)3

= 21 cm
Height of the rectangle = 21 x 2

= 42 cm
Area of the rectangle = 168 cm x 42 cm

= 7056 cm

‫ﮠ‬

5) =

‫ﮠ‬

[2a + (5 1)d] =

2a + 4d = ‫ﮠ‬ x

‫ﮠ‬ਤ ○1

2a + 4d =

= ‫ﮠ‬
‫ﮠ‬
[2a + (8 1)d] = ‫ﮠ‬

2a + 7d = x

‫ﮠ‬ ○2

2a + 7d =

○2 ○1 , ‫ﮠ‬ ‫ﮠ‬ਤ

7d 4d =
3d = 13
d=

Substitute d = into ○1 ,
2a + 4(
) = ‫ﮠ‬ਤ
( )
2a = ‫ﮠ‬ਤ

‫ﮠ‬

a=

= 1080

= 1080 + (10 1)( )

= 1080 + ( 39)
= 1041

6) Given the following arithmetic sequence,
78, 82, 86, …

a = 78
d = 82 78

=4

(a) Sum of the 10 terms after the term =

= [2(78) + (18 1)4] [2(78) + (8 1)4]

= 9(224) 4(184)
= 2016 736
= 1280

(b) > 200

78 + (n 1)4 > 200

74 + 4n > 200

4n > 126

‫ﮠ‬

n> ਤ
n > 31.5

⸫ The smallest term which is more than 200 for this sequence is the term.

= 78 + (32 1)4
= 78 + 124
= 202

(c) First, we need to know the number of terms for the sequence.

= 158

78 + (n 1)4 = 158

74 + 4n = 158

4n = 84

ਤ

n= ਤ
= 21

Since 158 is the last second term, the sequence has a total of (21 + 1) terms.

Sum of the last five terms = [78 + (158 + 4)] [2(78) + (17 1)4]
= The last term

= 11(240) (220)

= 2640 1870
= 770

5.2 Geometric Progression

1) Given the geometric sequence.
20, 40, 80, …

a = 20

ਤ

r=

=2

(a) = 20
= 10240

(b) < 10000
20 < 10000
< 500
ln < ln 500
(n - 1) ln 2 < ln 500

ln

n 1 < ln
n < 8.97 + 1
n < 9.97

⸫ The last term of the sequence will be the term.

= 20
= 5120

(c) The sum of the term from term to term = a

o h

= 20

= 2520

2) The saving amount follows a geometric sequence,
5000, 1.03 x 5000, 1.03 x 1.03 x 5000, …

a = 5000
r = 1.03

> 6000

5000 > 6000

> 1.2

ln > ln 1.2

(n 1) ln 1.03 > ln 1.2
n > ln ln ln
n > 7.168

Since starting from the eighth term only the saving amount is more than RM6000, the minimum
years required for the fixed deposit is 7 years.

3) Area of the squared paper = 22 x 22
= 484 cm

When the paper is folded in half again and again, its area decreases in a geometric sequence,
484, 0.5 x 484, 0.5 x 0.5 x 484, …

a = 484
r = 0.5

Since the paper is folded in half for six times, we have to find the value of

= 484o h
= 7.5625

⸫ The area of the folded paper is 7.5625 cm .

4) a = 2000
‫ = ﮠ‬1180.98
n=6

‫ = ﮠ‬1180.98
2000o h‫ = ﮠ‬1180.98

= 0.59049
ln = ln 0.59049
5 ln r = ln 0.59049

ln r = 0.10536
r= ‫ﮠ‬

= 0.9

Sum of the first four terms = o ਤh

= 6878

5) The height of the ball follows the geometric progression,
21, 0.875 x 21, 0.875 x 0.875 x 21, …

a = 21
r = 0.875

=
= 168

Total distance travelled = 24 + 2 x168
= 24 + 336
= 360 cm

6) The radius and height of the cylinders follows geometric progression, G1 and G2 respectively.
G1: 1, 2 x 1, 2 x 2 x 1, …
G2: 5, 1.4 x 5, 1.4 x 1.4 x 5, …

For G1,

=1
= 64

For G2,

=5 ਤ
= 37.65

Volume of the last cylinder = π h
= πo‫ﮠ‬ਤh (37.65)
= 154214.4π cm

7) 5.3888… = 5.3 + (0.08 + 0.008 + 0.0008 + …)
= 5.3 +

= 5.3 +

ਤ

= 5.3 + ਤ

ਤ

= +ਤ

ਤ

=+

ਤ

=

=

REFERENCES

g

Elsaket, N. (2016, Mac 20). Real-life application of arithmetic and
geometric sequence. Retrieved from
https://prezi.com/jw4lnrb6waos/real-life-application-of-arithmetic-
and-geometric-sequence/

Geometric Sequences and Series. Retrieved from
https://courses.lumenlearning.com/boundless-
algebra/chapter/geometric-sequences-and-series/

Geometric Series and Three Applications. Retrieved from
http://www.math.niu.edu/~richard/Math302/ch1sec5.pdf

Wong, M. K., Zaini bin Musa, Azizah binti Kamar, Saripah binti Ahmad,
Nurbaiti binti Ahmad Zaki, & Burham@Borhan, Z. H. b. (2019).
KSSM Additional Mathematics Form 4 Textbook. Johor Bahru:
Ministry of Education Malaysia

Wong, P. W., & Wong, S. M. (2015). Success Additional Mathematics SPM.
Selangor: Oxford Fajar Sdn.Bhd.