1
A5 Formulae
A5.1 Literal Equation
In the equation 3x + 5 = 7, we can find one variable and three numerals. If we replace the numerals by letters, we have a literal equation.
For example, the equation 3x + 5 = 7 can be written as ax + b = c
The equation ax + b = c is a literal equation.
To solve the above literal equation is to express the variable x in terms of the letters a, b and c.
(1) Solve equations involving like terms.
Simple equation Literal equation
3 9 27 27 ) 7 2 (
27 7 2
=
=
= +
= +
x x x
x x
b a x c
c b a x
c bx ax
= +
= +
= +
) (
(2) Solve equations involving brackets.
Simple equation Literal equation
6 2 12 12 2
6 6 3
6 6
3
6 )
2 ( 3
=
=
= +
=
−
=
−
−
=
−
−
x x x x x
x x
x x
1 )
1 (
) (
−
= + +
=
−
+
=
−
=
−
−
=
−
−
a c x ab
c ab a
x
c ab x ax
c x ab ax
c x b x a
(3) Solve equations involving fractions.
Simple equation Literal equation
3 2 5 5 2 5 1
2
=
−
=
= + + =
x x x x
a bc x
bc a x
b c a x
−
=
= + + =
Assume a + b≠0
Assume a – 1≠0
Assume b≠0
Example 1
Solve the literal equation a(x−b)=c(x+d)+e for x.
Solution
c a
e cd x ab
e cd ab c a x
ab e cd cx ax
e cd cx ab ax
e d x c b x a
− +
= +
+ +
=
−
+ +
=
−
+ +
=
−
+ +
=
−
) (
) ( ) (
Example 2
Solve the literal equation c b x a
x − = , a≠ b0, ≠0 for x.
Solution
a b x abc
abc a
b x
ab c a b x
ab c ax bx
b c x a x
= −
=
−
− =
− =
=
−
) (
) (
Example 3
Solve the literal equation b(x c) a
x = − , a≠0 , for x.
Solution
1 )
1 (
) (
) (
= −
=
−
−
=
−
=
−
=
−
=
ab x abc
abc ab
x
x abx abc
abc abx x
c x ab x
c x a b x
Assume a – c≠0
Assume b – a≠0
Assume ab – 1≠0
3 Checkpoint 1
Solve the following literal equations for x. Suppose that a, b, x represent non-zero numbers.
(a) ab(x− )c +ac=c+xd
(b) c
b ax−c =
(c) (x c)
b c a a
x+ = −
(d) a d
x c bx+ = +
A5.2 Formulae and Substitution
We learnt that a formula is a rule or relation between two or more quantities and these quantities are usually represented by letters.
Some examples of formulae:
Area of rectangle: A=l×b
Perimeter of rectangle: P=2(l+b)
Speed:
T S = D
S:
D:
T:
speed distance time
Pythagoras’ Theorem: c2 =a2 +b2
In the formula, the value of a certain letter or variable can be obtained by the method of substitution if the values of the other letters are known.
Example 4
Find the value of P from the formula
P D T +2
= , P≠0, when D = 5 and T = 3.
Solution
1 5 5
2 5 3
2
=
=
= +
= +
P P
P P D T
b l
b
a c
5 Checkpoint 2
In each of the following, a formula and the values of some of the variables are given. Find the value of the unknown.
(a) 2
2 1 ft ut
s= + If u = 16, t = 2 and f = 10, find s.
(b) R nr
C nE
= + If C = 3.5, E = 17, R = 6 and r = 4, find n.
(c)
−
=t u
d 1
1 If d = 3 and t = 12, find u.
A5.3 Change of Subject of a Formula
In the formulaA=lb, A is called the subject of the formula. For l ≠0, this formula can be written as
l b= A
Now, b is isolated on the left side of the equal sign and it becomes the subject of the formula.
Example 5
Make x the subject of the formula y= x3 +4.
Solution
3 4 4 3
4 3
= −
−
= +
=
x y y x
x y
Example 6
Change the subject of the formula A (a b)h 2
1 +
= to a.
Solution
h hb a A
hb A ha
hb ha A
h b a A
h b a A
= −
−
= +
= +
= +
=
2 2 2
) ( 2
) 2(
1
Example 7
Change the subject of the formula T =2rt−4rs to r.
Solution
s t r T
s t r T
rs rt T
4 2
) 4 2 (
4 2
= −
−
=
−
=
Assume h≠0
7 Checkpoint 3
In each of the following, make the letter in bracket the subject of the formula.
(a) A=πr2 +2πrh [h]
(b) t(R−2r)=3R [R]
(c) 2
2 1 ft ut
s= + [f]
(d) a
c Mb Ma− =
, c≠0 [M]
Example 8
Change the subject of the formula
ax y a
= −
1 , 1− ax ≠0, to a.
Solution
xy a y
y xy a
y axy a
a yax y
a ax y
ax y a
= +
= +
= +
=
−
=
−
= −
1 ) 1 (
) 1 (
1
Example 9
Change the subject of the formula
k n h nk
) 1 ( 1
2 +
= + , 01+(n+1)k ≠ , to (a) n.
(b) k.
Solution (a)
hk k
hk n h
hk h hk k n
nhk nk hk h
nk hk nhk h
nk k
nk h
nk k
n h
k n h nk
−
= + +
=
−
−
= +
= + +
= + +
= + +
+
= +
2 ) 2
(
2 2 2 ) 1
(
2 ] ) 1 ( 1 [
) 1 ( 1
2
(b)
) 1 ( 2 )]
1 ( 2 [
) 1 ( 2
2 ) 1 (
2 ] ) 1 ( 1 [
) 1 ( 1
2
+
= −
= +
−
= +
−
= + +
= + +
+
= +
n h n k h
h n
h n k
h n
hk nk
nk n
hk h
nk k
n h
k n h nk
Assume 1 + xy ≠0
Assume 2k – hk ≠0
Assume 2n – h(n + 1) ≠0
9 Example 10
Change the subject of the formula
1 2 +
= R
A PI to R.
Solution
2 2
2 1 1
1 2
1 2
1 2
) 1 2 (
1 2
−
=
−
=
−
=
= +
= +
= +
A R PI
A R PI
A R PI
A R PI
PI R
A
R A PI
Checkpoint 4
Change the subject of the formula
np m=1−mnp
, n,p≠0, to n.
Checkpoint 5
In each of the following, make the letter in bracket the subject of the formula.
(a) b
bc c a
a −
=
+ , b≠0 [b]
(b) n k
k n mk m
− +
= + 2
) 1
( , 2n− k ≠0 [n]
(c) 2u =vw [u]
(d) b a
c b
a = +
+ , b+ c ≠0 [c]
11 Example 11
Change the subject of the formula 2 2 n
b = m , n≠0 to n.
Solution
b m b n m
b n m
n b m
±
=
±
=
=
=
2 2 2
2 2
Example 12
Change the subject of the formula 2 22 n m m mn
= +
+ , 0m+ n2 ≠ , to n.
Solution
2 2 2 2
2 2
) )(
2 (
2
2 2 2
2 2 2 2
2 2
2 2
m n m
m m n
mn n mn m m
mn n
m m
n m m mn
−
± −
=
−
−
=
= + + +
= + +
= + +
Checkpoint 6
Change the subject of the formula ak2 −b=ck2 to k.
A5.4 Applications of Formulae
Example 13
Celsius (C) scale and Fahrenheit (F) scale are used to measure temperature. The relationship between F and C is given by the formula
5 32
= C9 +
F .
(a) Express C in terms of F.
(b) Find the reading on Celsius scale if the reading on Fahrenheit scale is
(i) –4; (ii) 32.
(c) At what temperature will the reading on Celsius scale the same as the reading on Fahrenheit scale?
Solution (a)
) 32 9(
5 5 32 9
5 32 9
−
=
−
= +
=
F C
F C
C F
(b) (i)
20
) 32 4 9( 5
) 32 9(
5
−
=
−
−
=
−
= C
F C
(ii)
0
) 32 32 9( 5
) 32 9(
5
=
−
=
−
= C
F C
(c)
4 40 9 9 160
9 160 9
4
9 160 9
5
9 32 5 9 5
) 32 9(
5
−
=
×
−
=
−
=
−
=
−
×
−
=
−
=
C C C C
C C
C C
13 Example 14
The volume (V) of a cylinder with base radius (r) and height (h) is given by the formula
h r V =π 2 . (a) Make r the subject of the formula.
(b) Hence find r if V =88cm3, h = 7 and
7
= 22 π .
Solution (a)
(rejected)
or
2 2
h V h
r V h r V
h r V
π π
π π
−
=
=
=
(b)
cm 2
4 7 7 22
88
=
=
×
=
= h r V
π
Checkpoint 7
The measure of an interior angle (θ) of a rectangular polygon with n sides is given by the formula
n n 2) ( 180° −
θ = .
(a) Express n in terms of θ.
(b) If the measure of an interior angle of a regular polygon is 156o, find the number of sides of the polygon.
h
r
Exercise A5 Formulae
A5.1
1. Solve the following equations for x. All letters represent non-zero numbers.
(a) 3x−b=d (b) 4mx 8= n
(c) ax+b+c=0 (d) 4(x−3t)=7s (e) 5(x+2b)=3(x−2c) (f) mx−nx= pq (g) ax+9=3x+14 (h) rx+h=sx−k (i) 3px=2q(r−5x) (j) m(x−a)=n(x−b)
(k) t
s r
x 3
2 − = (l)
p n x m
x + = 1
(m) d
c b ax 2 4
3 + =
(n) 0
3 6
5 − + =
r x q px
A5.2
2. Given that A bh 2
= 1 , find A when b = 9 and h = 8.
3. Given that ( 1) 2
1 +
= n n
S , find S when n = 100.
4. Given that A=πr2, find A when
7
= 22
π and r = 14.
5. Given that 2
2 1gt ut
s = − , find s when u = 40, t = 5 and g = 10.
6. Given that
f v u
1 1
1+ = , find f when u = 15 and v = 30.
7. Given that 32
59 +
= C
F , find C if F = 50.
8. Given that V =blh, find h if V = 120, b = 4 and l = 10.
9. Given that A=P(1+rt), find t if A = 1240, P = 1000 and r = 0.06.
10. Given that g = ab, find b if a = 9 and g = 12.
15
11. Given that
x y x
−
= − 4
3
1 , find x if y = 10.
12. Given that
g
T =2π l , find l if T =10π and g = 10.
A5.3
13. In each of the following, make the letter in the brackets the subject of the formula.
(a) V Ah
3
=1 [h] (b)
V
D= M [V]
(c) t
P= Fd [F] (d) P=2(l+b) [b]
(e) ( 32)
9
5 −
= F
C [F] (f) 3u+7v=10V [v]
(g) v=u+at [t] (h) H =ms(T −t) [T]
(i) x
y x
−
= + 3
3 [x] (j)
n m t ms
= + [m]
(k) t
u v F m( − )
= [v] (l)
1 5 2
+ + −
= x
y x [x]
(m) W =IR2 [R] (n) 2
d Gmn
F = [d]
(o) v2 =u2 +2as [u] (p)
g
w= m [m]
(q) T =4 m [m] (r)
t r 5s
2
= 1 [t]
(s) D= b2 −4ac [b] (t) x=a+ b2 +c2 [c]
(u) a a 5b
1 +2
= [b] (v) A=πr2 +r h2 +r2 [h]
(w) R
u mv
= 2 [v] (x) A=b+ B2 −C2 [B]
A5.4
14. Celsius (C) scale and Fahrenheit (F) scale are used to measure temperature. The relationship between F and C is given by the formula
) 32 9(
5 −
= F
C .
(a) Express F in terms of C.
(b) Find the reading on Fahrenheit scale if the reading on Celsius scale is
(i) –5; (ii) 0; (iii) 5.
15. The total surface area (A) of a cylinder is given by the formula rh
r A=2π 2 +2π
where h is the height and r is the base radius of the cylinder.
(a) Make h the subject of the formula.
(b) Find the value of h if A = 550 cm3, r = 5 cm and
7
= 22 π .
16. The sum of the squares of first n positive integers is
6 ) 1 2 )(
1
( + +
= n n n
S .
(i.e.
6 ) 1 2 )(
1 ... (
3 2
12 2 2 2 + +
=
= + + +
+ n n n
S
n )
(a) Find the sum of the squares of the first 50 positive integers.
(b) Find the sum of the squares of the positive integers from 51 to 100 inclusive (包括在 內).
h
r
