F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
Chapter 4 Formulas
12/2002 P. 1
Name:_________( ) Class: F.2 (
Important Terms
algebraic fractions
polynomials
constant term
factor
variable
代數分式
多項式
常數項
因式
變量
numerator
denominator
change of subject
substitution
)
分子
分母
主項變換
代入法
Revision Notes:
1.
Algebraic Fractions
(a) A fraction
A
, where A and B are polynomials, is called an algebraic fraction if B contains at
B
least one non-constant term.
1 2a 2 x  1
e.g.
x, 3b, 2 x  1
(b) An algebraic fraction can be reduced to its simplest form by cancelling the common factor(s) in
the numerator and the denominator.
3x  6
3( x  2)
e.g.

the common factor
xy  2 y y ( x  2)
(x+2) is cancelled
3

y
(c) When adding or subtracting fractions with different denominators, we find the L.C.M. of the
denominators first.
e.g.
5
7
5 2 7 3



18 x 12 x 36 x 36 x

10  21
36 x

36x is the L.C.M.of
18x, 12x
11
36 x
(d) Multiplication and division of algebraic fractions
e.g.
4a 2 3ab ab 4a 2 3ab 3


 2 

2
3
2 ab
3b 2
3b

2.
6a 2
b2
Formulas
A formula is a relation between two or more quantities, and these quantities are usually represented
by letters.
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 2
(a) Change of subject
In the formula, A 
( a  b) h
, A is called the subject of the formula.
2
We can change the subject of a formula by rules used in solving equations.
e.g. Change the subject of the formula A 
A
( a  b) h
2
2 A  ah  bh
bh  2 A  ah
b
( a  b) h
to b.
2
b does not appear on
the left hand side.
2 A  ah
h
(b) Substitution
In a formula, we can obtain the value of a certain variable by substitution the other variables
into the formula.
e.g. Given the formula b 
b
2 A  ah
, if A=10, a = 3, h=5, then
h
2(10)  (3)(5)
5
加菲，你又偷懶啦，
5

5
成日睡覺，唔駛你做啊?!
=1
(c) Applications of formulas
e.g. The relation between Celsius scale (C 攝氏) and
Fahrenheit scale (F 華氏) can be expressed as C 
When F = 99, we have C 

5
(99  32)
9
5  67
9
= _______
5
( F  32) .
9
我好慘，發高燒
至 99 度呀!!
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 3
Exercise A
Level I Simplify
1.
2.
36 xy
6x
(a)
5x
20 y
(b)
–
(c)
15m  10n
5
(d)
ax  ay
a
(e)
2x  6 y
2y
(f)
c
ab
(b)
x
 6x
2y
(a) 5a 
20a
10a  5b
(c)
4a 2 b
ab


3
a
2
(d)
3x 2  12 x 15

5y
x
(e)
3y 2  3y 9 y

4
2
(f)
2x  1 2x  6

x3
4x
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 4
(a)
3
7

2a 2a
(b)
5
7

ab ab
(c)
x  y 2x  y

3xy
3xy
(d)
2x
2

x 1 x 1
4. (a)
1 t t  t 2 1 t
 2 
t
t
t  t2
(b)
2m  6 6n  12

3n  6 4m  12
5.
(a)
3w  2 w  4

w3 3 w
(b)
a
b

ba ab
(c)
14a
100 x

5x  7 7  5x
(d)
6x 2
18 x

3 x x 3
3.
Level II Simplify
F.2 Mathematics Supplementary Notes
6.
Chapter 4 Formulas
12/2002 P. 5
(a)
1
1

18a 12a
(b)
1
1

x 1 x  2
(c)
1
2

x 1 x
(d)
1
2

2a  4 3a  9
(e)
2x
2
x y
(f)
3
x
x2
(g)
2
3

2
3 x
( x  3)
(h)
1
1
 2
a  ab b  ab
2
Exercise B
Level I
1.
(a) C =
5
( F  32)
9
If F = –40, Find C
(b)
H=
1
( r  5)
2
If r = 10, Find H
F.2 Mathematics Supplementary Notes
(c) A =
n 1
n2
Chapter 4 Formulas
(d)
S = 2r2 + rL
If S = 2, r = –2, find L
If n = 8, Find A
2.
12/2002 P. 6
In each of the following, make the letter in the square brackets the subject of the formula
(a) K = 2C + 273
C
(b)
F = 6(M – N)
M
(c) V = U – gt2
g
(d)
l = mx – nx
x
a a
 k
h k
h
(b)
k=
PV
T S
P
a
1 R
R
(d)
k=
mn
mn
m
Level II
3.
(a)
(c) S =
F.2 Mathematics Supplementary Notes
(e) a =
bc
bc
Chapter 4 Formulas
C
(f)
m=
12/2002 P. 7
n2
 4L
3n
[n ]
2b
, express x in terms of a, b and c.
cx  b
4.
Given a  1 
5.
Given A  s(s  a)( s  b)( s  c) , where s 
Find the values of s and A if
(a) a = 3, b = 4, c = 5;
1
(a  b  c) .
2
b
a
Area A
c
(b) a = 10, b = 10, c = 16.
6.
n
Given S  [2a  (n  1)d ] .
2
(a) Express a in terms of S, n and d.
(b) Find a if S=5050, n=100 and d=1.
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 8
Exercise C
Multiple Choice
x y
1. If
= 4, then y =
x y
A.
5x
3
E.
3y  1
y3
B.
D.
4.
3y  1
y 3
3y  1
y3
1 b
, then b =
ab  1
A.
1 x
ax  1
B.
C.
1 x
ax  1
D.
1 x
ax  1
6.
1 x
ax  1
1 x
1  ax
If x =
y  3z
, express y in terms of x and z.
y
A. y =
x
–1
3z
B. y =
C. y =
x 1
3z
D. y =
E. y =
3z
x 1
If a =
x 1
3z
3z
x 1
bd
, then b =
b  2cd
A. 2c
3y  1
3 y
5. If x =
E.
3x
5
3x  1
, then x =
x3
3y 1
y3
C. –
E.
D.
3x
5
5
3x
3. If y =
A.
B. –
5
3x
C. –
2.
C.
2cd
1 d
E.
2acd
ad
If a = 1 +
B.
2ac
a 1
D.
2acd
ad
1
, then b =
1 b
A.
a2
a 1
B.
a 1
2a
C.
a 1
2a
D. –
E.
2a
a 1
a
a 1
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 9
**************************************Optional*************************************
Level III
Note: Division of polynomials
Method of long division (長除法)
3x 2  6 x  11
3
2
x  2 3x  0 x  x  5
3x 3  6 x 2
e.g. (3x 3  x  5)  ( x  2)
6x 2  x
quotient = 3x 2  6 x  11
remainder = 17
6 x 2  12 x
11x  5
i.e. 3x 3  x  5  ( x  2)(3x 2  6 x  11)  17
11x  22
17
Ex. (i)
1.
( x 2  4 x  5)  ( x  1)
(ii)
(2 x 3  4 x 2  3x  7)  ( x  2)
(iii)
( y 3  8)  ( y  2)
(a) Given s  r  t 
t
r t
. Express t in terms of s and r.
 100 
 100 
(b) Given a1 
  b 1 
 . Express x in terms of a and b .
x 
x 


F.2 Mathematics Supplementary Notes
2.
(a)
Chapter 4 Formulas
12/2002 P. 10
Simplify the following
1
1

1 x 1
1
x
(b)
x
y
 2
2
y
x
(c)
1
1
1

 2
2
x
xy y
(d)
1
1
a
b

1
1
b
a

ab
ab  1
(Ans. (b).0)
( x  1)( x  2)( x  3)  ( x  1)( x  2)  ( x  1)
( x  1)( x  2)  ( x  1)  2
(Ans. x – 1)
3. (a) Simplify
1
1

.
n n 1
(b) Using (a), find the value of
4.
1
1
1

 ...... 
.
2  3 3 4
99  100
Given that x  1  t 2 and t  1  y .
(a) Express t in terms of x.
(b) Express y in terms of t.
(c) Use the result of (a) and (b), express y in terms of x.
(Ans. (b). 49/100 )
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 11
5.
When x 2  3x  a is divided by (x+2) , the quotient is (x+1) and the remainder is 5.
Find the value of a.
(Hint : Dividend =Divisor  Quotient + Remainder )
(Ans: a =7)
6.
Let S denote the sum of the squares of the first n natural numbers, i.e. S  12  2 2  3 2    n 2 .
n(n  1)( 2n  1)
It is given that the value of S can be found by the formula S 
.
6
(a) (i)
Use the given formula to find the value of S when n=5.
(ii) Verify the result in (i) by adding of the squares of the first 5 natural numbers.
(b) Use the given formula to find the sum of the squares of each of the following:
(i)
7.
12  2 2  3 2    10 2 .
Suppose a 
(ii) 112  12 2  13 2    20 2 .
1
1
b
 2 , where a  1 , b  1 and a  b  2  0 .
a 1
b 1
Find the value of ab  a  b.
8.
If abc =1 , find the value of
(Ans: 2)
a
b
c


.
ab  a  1 bc  b  1 ca  c  1
(Ans: 1)
F.2 Mathematics Supplementary Notes
數學課外閱讀：＜＜幾何明珠：第六章
Chapter 4 Formulas
秦九韶公式＞＞。黃家禮
12/2002 P. 12
編著。九章出版社。
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 13
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
12/2002 P. 14
F.2 Mathematics Supplementary Notes
Chapter 4 Formulas
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