271
Solve for ‘n’ (rounding to 2 decimal places):
= ln c 4, 285 m
81. nn =
4, 000
7, 200
m
4, 725
85. nn==
ln ^1.01h
ln c
82.
= ln c
nn =
6, 750
m
3, 200
83.
= ln c
nn =
86.
5, 120
m
2, 250
=
nn =
ln ^1.005h
87.
nn==
ln c
3, 645
m
2, 175
ln ^2.5h
ln ^1.03h
84. nn== ln c
=
88. nn =
75, 000
m
2, 200
ln ^3 h
ln ^1.02h
8.5 Rearranging Equations and Formulas
Introduction
Equations are mathematical statements formed by placing an equal (=) sign between two expressions
to indicate that the expression on the left side is equal to the expression on the right side of the equal sign.
5x + 3 = y – x
For example, Formulas are similar to equations. In formulas, the relationship among many variables is written as a
rule for performing calculations. Formulas are written so that a single variable, known as the subject
of the formula, is on the left side of the equation, and everything else is on the right side.
I = Prt
For example,
Isolating Variables
To isolate a particular variable in an equation or a formula, rearrange the terms and simplify, so that
the required variable is on the left side of the equation and all the other variables and numbers are
on the right side of the equation. Rearrangement can be performed by using the rules that you have
learned in the previous sections of this chapter and the following guidelines:
■■ Add or subtract the same quantity to or from both sides.
■■ Multiply or divide both sides by the same quantity.
■■ Take powers or roots on both sides.
■■ Expand the brackets and collect the like terms.
■■ Remove the fractions by multiplying both sides by the denominator or LCM.
For example, consider the formula for simple interest: I = Prt.
To solve for any of the variables, 'I', 'P', 'r', or 't' in this simple interest formula, we can rearrange the
variables as shown below:
I = Prt is the same as Prt = I
Solving for 'P':
Solving for 'r':
Solving for 't':
Prt
I
rt = rt I
P = rt
Prt
= I Pt
Pt
r= I
Pt
Prt
I
=
Pr
Pr
Dividing both sides by 'rt',
Dividing both sides by 'Pt',
Dividing both sides by 'Pr',
t = I
Pr
8.5 Rearranging Equations and Formulas
272
Example 8.5-a
Rearranging to Isolate Variables
Rearrange and isolate the variables indicated in the brackets:
Solution
(i)
S = C + M(M)
(ii)
C + E + P = S(P)
(iii)
P = RB(R)
(iv)
y = mx + b(m)
(v)
S = P(1 + rt)(P)
(i)
S = C + M
(M)
M+C=S
Subtracting ‘C’ from both sides,
M+C− C=S−C
M=S−C
(ii)
C + E + P = S
(P) Subtracting ‘C’ and ‘E’ from both sides,
C+E+P−C−E=S− C−E
P=S−C−E
(iii)
P = RB
(R)
RB = P
Dividing both sides by 'B',
RB = P
B
B
R= P
B
(iv) y = mx + b(m)
mx + b = y
Subtracting ‘b’ from both sides,
mx + b − b = y − b
mx = y − b
Dividing both sides by ‘x’,
y- b
mx
=
x
x
m=
(v)
y- b
x
S = P(1 + rt) (P)
P(1 + rt) = S
Dividing both sides by ‘(1 + rt)’,
P^1 + rt h
S
=
^1 + rt h
^1 + rt h
P=
Example 8.5-b
S
^1 + rt h
Solving for Variables Using the Rearranged Simple Interest Formula
In the simple interest formula I = Prt, find:
(i)
‘I’, when P = $1,000, r = 5% (= 0.05), t = 3 years
(ii)
‘P’, when I = $150, r = 3% (= 0.03), t = 1 year
(iii)
‘r’, when I = $500, P = $8,000, t = 2 years
(iv)
‘t’, when I = $40, P = $800, r = 5% (= 0.05)
(v) ‘P’ when I = $3,180, r = 3%, (= 0.03), t = 2 years
Round the answers to 2 decimal places, wherever applicable.
Chapter 8 | Basic Algebra
273
Solution
(i)
Substituting the values for ‘P’, ‘r’, and ‘t’ in the formula:
I = Prt
I = 1,000.00 × 0.05 × 3 = $150.00
(ii)
Substituting the values for ‘I’, ‘r’, and, ‘t’ in the rearranged formula:
I
P=
rt
P =
150.00
^0.03 # 1h
= $5,000.00
(iii) Substituting the values for ‘I’, ‘P’, and ‘t’ in the rearranged formula:
r = I
Pt
r =
500.00
= 0.03125 = 3.125% = 3.13%
8, 000.00 # 2
(iv) Substituting the values for ‘I’, ‘P’, and ‘r’ in the rearranged formula:
t = I
Pr
40.00
t =
= 1 year
800.00 # 0.05
(v)
Example 8.5-c
Substituting the values for ‘I’, ‘r’, and ‘t’ in the rearranged formula:
P= I
1 + rt
P = 3,180.00
(1 + 0.03 × 2)
= 3,180.00
1 + 0.06
= 3,180.00 = $3,000.00
1.06
Rearranging to Isolate Variables Involving Brackets and Fractions
Rearrange to isolate the variables indicated in the brackets.
Solution
(i)
S = P(1 + rt)(t)
(ii)
y = x + a (x)
x–a
(i)
S = P(1 +rt)(t)
P(1 + rt) = S
Expanding the bracket,
P + P . rt = S
Subtracting 'P' from both sides,
(ii)
P . r . t = S – P
t = S –. P
P r
Dividing both sides by 'P . r',
y = x + a (x)
x–a
y = x + a
Multiplying both sides by '(x – a)',
x–a
(x – a)y = x + a
Expanding the brackets,
xy – ay = x + a
Adding 'ay' to both sides,
xy = x + a + ay
Subtracting 'x' from both sides,
8.5 Rearranging Equations and Formulas
274
Solution
xy – x = a + ay
continued
x(y – 1) = a(1 + y)
Dividing both sides by '(y – 1)',
a(1 + y)
x=
(y – 1)
a(y + 1)
x=
y–1
Example 8.5-d
Rearranging to Isolate Variables Involving Powers and Roots
Factoring 'x' on the left side and 'a' on the right side,
Rearrange to isolate the variables indicated in the brackets.
(i)
F=
(ii)
y=
(i)
F=
Solution
mV 2
r
x–a mV 2
r
Multiplying both sides by 'r',
mV 2 = F . r
F.r
2
V =
m
F.r
V=
m
(ii)
y=
Dividing both sides by 'm',
Taking square root on both sides,
x–a (x)
x – a = y
Taking square on both sides,
2
x–a=y (x)
(V)
mV 2
= F
r
Example 8.5-e
(V)
x=a+y
2
Adding 'a' to both sides,
Rearranging to Isolate Variables Involving Factors
Rearrange to isolate the variables indicated in the brackets.
Solution
(i)
a(x – y) = b(x + y)(x)
(ii)
y = (x + y)(x – y)(x)
(i)
a(x – y) = b(x + y)(x)
a(x – y) = b(x + y)
Expanding both sides,
ax – ay = bx + by
Subtracting 'bx' from both sides,
ax – ay – bx = by
ax – bx = by + ay
Factoring 'x' on left side and 'y' on right side,
x(a – b) = y(b + a)
Dividing both sides by '(a – b)',
y(b + a)
a–b
y(a + b)
x =
(a – b)
x =
Chapter 8 | Basic Algebra
Adding 'ay' to both sides,
275
Solution
continued
(ii)
y = (x + y)(x – y)(x)
(x + y)(x – y) = y
2
x – y = y
x2 = y + y2
Example 8.5-f
Expanding left side,
2
x = y + y 2
x=
Adding 'y2 ' to both sides,
Taking square roots on both sides,
Taking 'y' as a factor within the square root,
y(1 + y)
Rearranging Formula and Evaluating to Find the Value of a Subject
The formula for the area 'A' of a rectangle of length 'l ' and width 'w' is A = l . w
Solution
Example 8.5-g
(i)
Rearrange the formula to find 'w' as the subject.
(ii)
Calculate the width (w) of a rectangle if A = 200 cm2 and l = 25 cm.
A=l.w
(i)
l . w = A
A
w=
l
(ii)
Substituting the value for 'A' and 'l ' in the rearranged formula,
w = 200 = 8 cm
25
Dividing both sides by 'l ',
Rearranging Formula and Evaluating to Find the Value of a Subject
2
The formula for the area 'A' of a circle is A = πr , where r is the radius of the circle.
(i)
(ii)
Solution
(i)
Rearrange the formula to find 'r' as the subject.
Find the radius of a circle with an area of 400 cm2. Use π = 3.14.
(Round the answer to two decimal places.)
A = πr2
πr2 = A
r2 = A π
r=
(ii)
Dividing both sides by π,
Taking square root on both sides,
A
π
Substituting the values for 'A' and π in the rearranged formula,
r=
400
3.14
= 11.286652... = 11.29 cm
8.5 Exercises
Answers to odd-numbered problems are available at the end of the textbook.
Rearrange to isolate the variables indicated in the brackets:
1. 4x + 5 = y(x)
2.
3. 3x – y = 7
4. 2y – x = 5
(x)
x + 6y = 15
(y)
(y)
5.
S = C + M(C)
6. L – N = d . L(d)
7.
N = L(1 – d)(L)
8.
C = 2πr(r)
8.5 Rearranging Equations and Formulas
276
C + E + P = S(E)
10. S – P – E = C(P)
11. 5x – 6 = 2x + y(x)
12. 3 – 5x = x + y(x)
13. b =
ac (a)
1+a
15. c = a – c (b)
b
c
17. b = + ac (a)
a–2
ac (c)
1–a
16. c = a + c (a)
b
c
18. b = – ac (c)
a+2
19. c = ab – b (b)
4+a
20. c = ab + b (a)
4–a
21. 6(a – x) = y(x)
22. 4(x + a) = y(x)
9.
23. 3(a + x) = x(x)
14. b =
24. 5(y + a) = y(y)
2
2
25. V = u + 2as(u)
26. V2 = u2 + 2as(s)
29. 5(x + 2) = 3(x – 7)
30. 4(9 – 5) = 5(y + 4)
2
2
2
27. x + y = r (y)
(x)
28. S = ut + at 2(u)
(y)
31. x – y = xy(x)
32. x – xy = y(y)
33. y = x + 5 (x)
x–5
34. x =
y+2
(y)
y–2
2x + 5 (x)
36. y =
7 – 3x (x)
x (x)
38. y =
x + 4
A (A)
4π
41. x + 5 = x(y)
y
40. C =
a2 + b2 (a)
43. A = (a + b)h (a)
2
44. A = (a + b)h (h)
2
45. y = (x – 4)(x + 4)
46. y = (3 – x)(3 + x)(x)
35. y =
37. y = 8 –
39. r =
(x)
(x)
42. 7 – x = x(y)
y
2
47. The formula for the area, 'A', of a square is A = S , where S is the length of the side.
a. Rearrange the formula to find 'S' as the subject.
b. Find the length of a square with an area of 441 cm2.
.
48. The formula for the area, 'A', of a triangle is A = b h , where 'b' is the base and 'h' is the height.
2
a. Rearrange the formula to find 'h' as the subject.
b. Find the height of a triangle with an area of 198 cm2 and a base length 18 cm.
49. The formula for the surface area, 'A', of a sphere is A = 4πr2, where r is the radius of the sphere.
a. Rearrange the formula to find r as the subject.
b. Find the radius of a sphere with an area of 800 cm2. (Use π = 3.14 and round the answer to 2 decimal places.)
50. The formula for the volume 'V' of a cone is given by the formula V = 1 πr2h, where r is the radius of the base and h
3
is the perpendicular height.
a. Rearrange the formula to find r as the subject.
b. Find the base radius of a cone with a volume of 900 cm3 and a perpendicular height of 10 cm. (Use π = 3.14 and
round the answer to 2 decimal places.)
Chapter 8 | Basic Algebra