This document provides information about algebraic formulae including variables, constants, writing formulae based on situations, finding the subject of a formula, and solving for variable values. It includes examples and practice problems with solutions related to these concepts. The document is divided into sections covering variables and constants, formulae, the subject of a formula, and finding the value of a variable. Practice questions with answers are provided throughout for additional examples.

1. MODULE 15 - ALGEBRAIC FORMULAE
ALGEBRAIC FORMULAE
The concept of Variables and Constants The Concept of Formulae
Write a Formula Subject of a Formula Find the Value of a Variable
A. Variables and Constants
A variable is a quantity whose value is not fixed.
A constant is a quantity whose value is fixed.
Practice 1
Determine whether each of the following is a variable or constant.
1. The length of the Penang Bridge. (_____________)
2. The mass of new born babies in a hospital. (_____________)
3. The number of sides of a hexagon. (_____________)
4. The monthly consumption of electricity for Ahmad`s family. (_________)
B. Formulae
A formula is an equation which shows the relationship between two or
more variables.
Example: 1 kg of beef costs RM p and 1 kg of fish costs RM q. Puan
Fauziah buys x kg of beef and y kg of fish. The total that
She has to pay is RM t.
Solution:
1 kg of beef = RM p,
x kg = x x RMp = RM px
1 kg of fish = RM q,
y kg = y x RM q = RM qy
Therefore, RM t = RM px + RM qy Ignore the unit
t = px + qy
Maths tip:
Use the same unit or measurement throughout the calculation.
Practice 2: Write a formula based on the given statement or situation.
1

2. 1. Hani bought 5 books of RMx each 2. Chef Radi bought 20 kg of flour.
and 3 books of Rmy each. The total He used x kg to make cake and y kg
cost of the books was RM t. to make cookies. The mass of flour
that remains is m kg.
3. The price of a packet of soya bean 4. The perimeter of a rectangular fish
drink is 90 sen and the price of a pond of length l cm and width w cm is
packet of chicken rice is RM 2.50. P cm.
Given that RM H represents the total
price for m packets of soya bean
drink and n packets of chicken rice.
C. Subject Of A Formula
The subject of a formula is a variable that is expressed in terms of
other variables in the formula.
If a variable is the subject of a formula, then the variable can only lie on
one side (usually left) of an equation and its coefficient must be 1.
Example 1: Given that 2k 2 = 4m 2 + n 2 , express m in terms of k and n .
Solution:
Note:“express m in terms of k and n ” means “express m as the
subject of the formula”.
2k 2 = 4m 2 + n 2
4m 2 + n 2 = 2k 2 Rewrite a = b as b = a
4m2 = 2k2 – n2 Isolate the term 4m2 from the other terms.
2k 2 − n 2
m =
2
Solve for m2 by dividing both sides of the equation by 4.
4
2k − n 2
2
m= Take square root on both sides of the equation.
4
2k 2 − n 2
m= 4=2
2
3w − r
Example 2: Given that = 2 , then w =
5
2

3. 3w − r
Solution: =2 3w − r = 4 × 5 = 20
5
2
 3w − r 
  = 22 3w = 20 + r
 5 
 
3w − r 20 + r
=4 w=
5 3
Practice 3: Express the specified variable as the subject of the given
formula.
1. a = 3b + 4c . Express c as the 2h
subject. 2. 4 g = 1 − . Express h as the
3
subject.
3w 4. e 2 = f 2 + g 2 . Express f as the
3. v = . Express w as the
8w − 9 subject.
subject.
5. p = 4 gr 2 + 1 . Express r as the q− p
6. s = . Express q as the
subject. 2r
subject.
D. Finding The Value Of A Variable
Example : Find the value of r if p = q (r − s ) when p = 5, q = 2 and s = 3.
3

4. Solution: p = q (r − s )
5 = 2(r – 3) Substitute the values of p,q,and s into the formula.
5 = 2r – 6 Expand the bracket.
2r = 5 + 6 Isolate the term 2r on the left side of the equation.
2r = 11
11
r= The value of r.
2
Practice 4: Solve each of the following.
1. Given that p = 3q − 4r 2 . If q = – 2 2. Given that T = a + (n − 1)d , find the
and r = 3, find the value of p. value of n when T = 20, a = 5 and
d = 3.
3t 2 − 4u 1 1 1
3. Given that s = , find the 4. Given = + , find the value of
5t f u v
value of s if t = – 2 and u = 5. v when f = 10 and u = 3.
5. Given p = 2 and q = – 3 , find the 1
6. Given r = 4 and p = − , find the
value of 4 p (2 p 2 − 3q ) . 2
value of 3r 2 − 5rp .
PMR FORMAT QUESTIONS
4

5. 2( p − 3)
1. Given that = 5 , express p in terms of k.
k
2. Given that r = – 1 and s = 4, thus (r 3 − 1) s 2 =
2(4 p − r )
3. Given that = 5 , thus r =
r
2p −3
4. Given that = 2 p − 5 , thus p =
3
5. Given that p = 3m 2 + 2 , express m in terms of p.
5p − k
6. Given that = 3 , thus p =
3
3n 2
7. Given that m = 7 + , express n in terms of m.
2
2
8. Given that k = , thus p =
2 p −1
3rs
9. Given that = t , express L as the subject of the formula.
L
p2
10. Given that p = – 2 and r = – 3, the value of (7 − p ) is
r
5

6. 11. Given that b = 4 and k = – 3, then bk − k 2 =
2b 2 − a
12. Given that a = 4 and b = – 2, then =
b
13. Sam buys p mangoes at 40 sen each and q oranges at 50 sen each.
He sells the mangoes at 55 sen each and the oranges at 70 sen each.
Write the expressions of total profits, T, in sen, from the sales of all the
mangoes and oranges.
14. Danial has RM 500. He spends all his money to buy x shirts and y
trousers. Given that the prices of a shirt and a pair of trousers are RM 20
and RM 120 respectively. Write the equation which involves x and y.
15. Table below shows the number of balls in a box.
Colour Number of Balls
x
Red
1
Blue x
2
x−4
White
If the total number of balls in the box is y , write the equation involving x
and y.
ALGEBRAIC FORMULAE
ANSWER:
6

7. Practice 1
1. constant 2. variable 3. constant 4. variable
Practice 2
1. t = 5x + 3y 2. m = 20 – x – y
3. H = 90m + 250n 4. P = 2l + 2w or 2(l + w)
Practice 3
a − 3b 3(1 − 4 g )
1. c = 2. h =
4 2
9v
3. w = 4. f = e 2 − g 2
8v − 3
p −1 1 p −1
5. r = or 6. q = 2rs + p
4g 2 g
Practice 4
1. p = – 42 2. n = 6
4 30 3
3. s = 4. v = − or − 4
5 7 7
5. 136 6. 58
PMR FORMAT QUESTIONS
7

8. 5k + 6 5k
1. p = or +3 2. – 32
2 2
8p
3. r = 4. p = 3
7
p−2 27 + k
5. m = 6. p =
3 5
2m − 14 2+k 1 1
7. n = 8. p = or +
3 2k k 2
9r 2 s 2
9. L = 10. – 12
t2
11. – 21 12. – 2
13. T = 15p + 20q 14. x + 6y = 25
5x
15. y = −4
2
8