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Changing the Subject of a
Formula
(by grouping like terms and
factorizing)
Presented by Alona Hall
Simple Formulae Involving the Four Basic
Operations
The objectives of this lesson serves to accomplish the following:
1. Demonstrate the procedure related to ‘changing the subject’ in formulae
involving
 Grouping Like Terms and Factorizing
(A shorter method will be suggested: inspecting the formula then reversing the
process by using the inverse operation on the other side)
The Concept
Given the formula Area of a rectangle = length x width (written A = lw),
we say that A is the subject of the formula and this is so because:
 It is on the left hand side (the subject usually is)
 The coefficient is 1
 The power is 1
 It is in the numerator
If, however, we are interested in finding the length of the rectangle (l),
then we would get 𝑙 =
𝐴
𝑤
. We say that the subject of the formula has
been changed to l.
The Concept
The topic ‘changing the subject of a formula’ therefore implies that:
• A formula will be given
• It will have a subject
• The subject must be changed to something else
Procedure Related to “changing the subject”
In order to get back to the starting point, we had to ‘reverse’ which
involved ‘undoing’ that is ‘doing the opposite of what was done’.
In the context of mathematics, we will be doing the inverse of what
was done. Importantly, we must identify what is done to the subject in
order to determine what must be ‘undone’.
THUS: THE SAME RULES USED TO SOLVE AN EQUATION WILL BE
APPLIED
NB: The inverse operation eliminates the operation leaving the
number/symbol alone
Examples From Previous Lessons
Transpose the following formulae for the subject indicated in brackets:
1. 𝑣2= ‫ݑ‬2+2ܽ‫ݏ‬ (u)
2. T = 2𝜋
𝑒
𝑔
(e)
3. 𝑇2
=
4𝜋2 𝑅3
𝑔𝑟2 (𝑅)
Solutions
1. 𝑣2= ‫ݑ‬2+2ܽ‫ݏ‬ (u)
𝑢 𝑤𝑎𝑠 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 2 𝑎𝑠 𝑤𝑎𝑠 𝑎𝑑𝑑𝑒𝑑
𝑅𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑡𝑜 𝑔𝑒𝑡:
𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 2𝑎𝑠, 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒
𝑣2 − 2ܽ‫ݏ‬ = ‫ݑ‬2
(remember 2as will be eliminated from the RHS)
In order to eliminate the square, we square root, but show it on the other
side:
𝑣2 − 2𝑎𝑠 = 𝑢
(remember the square root eliminated the square leaving u on the RHS)
Rewrite to get: 𝑢 = 𝑣2 − 2𝑎𝑠
Solution (cont’d)
NB : 𝐺𝑖𝑣𝑒𝑛 𝑢 = 𝑣2 − 2𝑎𝑠,
𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑑𝑜𝑒𝑠 𝑁𝑂𝑇 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑣𝑒𝑟 𝑣.
REASON: 𝐼𝑓 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑔𝑜𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎 𝑝𝑙𝑢𝑠 𝑠𝑖𝑔𝑛 𝑜𝑟 𝑎 𝑚𝑖𝑛𝑢𝑠 𝑠𝑖𝑔𝑛, 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑠
𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑.
𝑇ℎ𝑖𝑠 𝑎𝑝𝑝𝑙𝑖𝑒𝑠 𝑡𝑜 𝑎𝑛𝑦 𝑟𝑜𝑜𝑡 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑝𝑜𝑤𝑒𝑟.
𝐻𝑜𝑤𝑒𝑣𝑒𝑟, 𝑖𝑓 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑔𝑜𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛, 𝑡ℎ𝑒𝑛
𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑.
𝑇ℎ𝑒 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑡𝑟𝑢𝑒, 𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑝𝑜𝑤𝑒𝑟𝑠 𝑤𝑖𝑙𝑙 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑟𝑜𝑜𝑡𝑠
𝑎𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Solution
2. T = 2𝜋
𝑒
𝑔
(e)
𝑒 𝑤𝑎𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 𝑔 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑤𝑎𝑠 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡𝑒𝑑 𝑎𝑛𝑑
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑑 𝑏𝑦 2𝜋
𝑅𝑒𝑣𝑒𝑟𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠, 𝑤𝑒 𝑔𝑒𝑡:
𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 2𝜋, 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒
𝑇
2𝜋
=
𝑒
𝑔
𝑡ℎ𝑒𝑛 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 (𝑟𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡)
(
𝑇
2𝜋
)2 =
𝑒
𝑔
Solution (cont’d)
(
𝑇
2𝜋
)2 =
𝑒
𝑔
(
𝑇2
22 𝜋2
) =
𝑒
𝑔
𝑇2
4𝜋2
=
𝑒
𝑔
𝑁𝑜𝑤, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑔 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒
𝑔 ×
𝑇2
4𝜋2
= 𝑒
𝑒 =
𝑔𝑇2
4𝜋2
Examine the next example carefully!
It will give you an idea about the inspection and transposing method
Solution
3. 𝑇2 =
4𝜋2 𝑅3
𝑔𝑟2 (𝑅)
𝐹𝑜𝑟 𝑡ℎ𝑖𝑠 𝑒𝑥𝑎𝑚𝑝𝑙𝑒, 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑖𝑛𝑠𝑝𝑒𝑐𝑡 𝑡ℎ𝑒𝑛 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑏𝑦 𝑑𝑒𝑚𝑜𝑛𝑠𝑡𝑟𝑎𝑡𝑖𝑛𝑔
𝑡ℎ𝑒 𝑠𝑡𝑒𝑝𝑠.
𝑅 𝑤𝑎𝑠 𝑐𝑢𝑏𝑒𝑑, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 4𝜋2
, 𝑡ℎ𝑒𝑛 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 𝑔𝑟2
𝑅𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑎𝑛𝑑 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑡ℎ𝑒𝑚 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒:
𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑔𝑟2
, 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 4𝜋2
𝑡ℎ𝑒𝑛 𝑐𝑢𝑏𝑒 𝑟𝑜𝑜𝑡
𝑅𝐸𝑀𝐸𝑀𝐵𝐸𝑅: 𝑇ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛. 𝐷𝑜 𝑡ℎ𝑒 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛:
𝑔𝑟2
× 𝑇2
= 4𝜋2
𝑅3
𝑔𝑟2
𝑇2
= 4𝜋2
𝑅3
𝑔𝑟2
𝑇2
4𝜋2
= 𝑅3
3 𝑔𝑟2 𝑇2
4𝜋2 = 𝑅
𝑅 =
3 𝑔𝑟2 𝑇2
4𝜋2
Examples
1. 𝑀𝑎𝑘𝑒 𝑣 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
1
𝑣
+
1
𝑢
=
2
𝑟
𝑅𝑒𝑚𝑜𝑣𝑒 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑏𝑦 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑒𝑎𝑐ℎ 𝑡𝑒𝑟𝑚 𝑏𝑦 𝑡ℎ𝑒 𝐿𝐶𝑀 = 𝑣𝑢𝑟
𝑣𝑢𝑟 × (
1
𝑣
) + 𝑣𝑢𝑟 × (
1
𝑢
) = 𝑣𝑢𝑟 × (
2
𝑟
)
𝑢𝑟 + 𝑣𝑟 = 2𝑣𝑢
𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑣 𝑡𝑒𝑟𝑚𝑠
𝑢𝑟 = 2𝑣𝑢 − 𝑣𝑟
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑣
𝑢𝑟 = 𝑣 2𝑢 − 𝑟
𝑣 =
𝑢𝑟
2𝑢 − 𝑟
Examples
2. 𝑇𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑒 𝑓𝑜𝑟 𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑥 =
𝑦 −4
𝑦 −7
(multiply by y -7 but show on the other side)
𝑥 𝑦 − 7 = 𝑦 − 4
𝑥𝑦 − 7𝑥 = 𝑦 − 4
𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑜𝑛 − 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒
(subtract y; add 7x but show it on the opposite side)
𝑥𝑦 − 𝑦 = −4 + 7𝑥
𝑥𝑦 − 𝑦 = 7𝑥 − 4
𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑦
𝑦 𝑥 − 1 = 7𝑥 − 4
𝑦 =
7𝑥 − 4
𝑥 − 1
Examples
3. 𝑀𝑎𝑘𝑒 𝑝 𝑡ℎ𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎:
𝐷
𝑑
=
𝑓+𝑝
𝑓 − 𝑝
𝑆ℎ𝑜𝑤 𝑠𝑞𝑎𝑢𝑟𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆, 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑖𝑠 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆
𝐷
𝑑
2
=
𝑓 + 𝑝
𝑓 − 𝑝
𝐷
𝑑2
2
=
𝑓 + 𝑝
𝑓 − 𝑝
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑑2
𝑜𝑛 𝑅𝐻𝑆; 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑓 − 𝑝 𝑜𝑛 𝐿𝐻𝑆; 𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠
𝐷2
𝑓 − 𝑝 = 𝑑2
(𝑓 + 𝑝)
𝑅𝑒𝑚𝑜𝑣𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑝 𝑡𝑒𝑟𝑚𝑠
𝐷2
𝑓 −𝐷2
𝑝 = 𝑑2
𝑓 + 𝑑2
𝑝
𝐷2
𝑓 −𝑑2
𝑓 = 𝐷2
𝑝 + 𝑑2
𝑝
3 cont’d
𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑝
𝐷2 𝑓 −𝑑2 𝑓 = 𝑝 (𝐷2 + 𝑑2)
𝐷2 𝑓 − 𝑑2 𝑓
𝐷2+ 𝑑2
= 𝑝
𝑝 =
𝐷2
𝑓 − 𝑑2
𝑓
𝐷2+ 𝑑2

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Changing the subject of a formula (grouping like terms and factorizing)

  • 1. Changing the Subject of a Formula (by grouping like terms and factorizing) Presented by Alona Hall
  • 2. Simple Formulae Involving the Four Basic Operations The objectives of this lesson serves to accomplish the following: 1. Demonstrate the procedure related to ‘changing the subject’ in formulae involving  Grouping Like Terms and Factorizing (A shorter method will be suggested: inspecting the formula then reversing the process by using the inverse operation on the other side)
  • 3. The Concept Given the formula Area of a rectangle = length x width (written A = lw), we say that A is the subject of the formula and this is so because:  It is on the left hand side (the subject usually is)  The coefficient is 1  The power is 1  It is in the numerator If, however, we are interested in finding the length of the rectangle (l), then we would get 𝑙 = 𝐴 𝑤 . We say that the subject of the formula has been changed to l.
  • 4. The Concept The topic ‘changing the subject of a formula’ therefore implies that: • A formula will be given • It will have a subject • The subject must be changed to something else
  • 5. Procedure Related to “changing the subject” In order to get back to the starting point, we had to ‘reverse’ which involved ‘undoing’ that is ‘doing the opposite of what was done’. In the context of mathematics, we will be doing the inverse of what was done. Importantly, we must identify what is done to the subject in order to determine what must be ‘undone’. THUS: THE SAME RULES USED TO SOLVE AN EQUATION WILL BE APPLIED NB: The inverse operation eliminates the operation leaving the number/symbol alone
  • 6. Examples From Previous Lessons Transpose the following formulae for the subject indicated in brackets: 1. 𝑣2= ‫ݑ‬2+2ܽ‫ݏ‬ (u) 2. T = 2𝜋 𝑒 𝑔 (e) 3. 𝑇2 = 4𝜋2 𝑅3 𝑔𝑟2 (𝑅)
  • 7. Solutions 1. 𝑣2= ‫ݑ‬2+2ܽ‫ݏ‬ (u) 𝑢 𝑤𝑎𝑠 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 2 𝑎𝑠 𝑤𝑎𝑠 𝑎𝑑𝑑𝑒𝑑 𝑅𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑡𝑜 𝑔𝑒𝑡: 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 2𝑎𝑠, 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 𝑣2 − 2ܽ‫ݏ‬ = ‫ݑ‬2 (remember 2as will be eliminated from the RHS) In order to eliminate the square, we square root, but show it on the other side: 𝑣2 − 2𝑎𝑠 = 𝑢 (remember the square root eliminated the square leaving u on the RHS) Rewrite to get: 𝑢 = 𝑣2 − 2𝑎𝑠
  • 8. Solution (cont’d) NB : 𝐺𝑖𝑣𝑒𝑛 𝑢 = 𝑣2 − 2𝑎𝑠, 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑑𝑜𝑒𝑠 𝑁𝑂𝑇 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑣𝑒𝑟 𝑣. REASON: 𝐼𝑓 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑔𝑜𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎 𝑝𝑙𝑢𝑠 𝑠𝑖𝑔𝑛 𝑜𝑟 𝑎 𝑚𝑖𝑛𝑢𝑠 𝑠𝑖𝑔𝑛, 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑. 𝑇ℎ𝑖𝑠 𝑎𝑝𝑝𝑙𝑖𝑒𝑠 𝑡𝑜 𝑎𝑛𝑦 𝑟𝑜𝑜𝑡 𝑎𝑛𝑑 𝑖𝑡𝑠 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑝𝑜𝑤𝑒𝑟. 𝐻𝑜𝑤𝑒𝑣𝑒𝑟, 𝑖𝑓 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑔𝑜𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑. 𝑇ℎ𝑒 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑡𝑟𝑢𝑒, 𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑝𝑜𝑤𝑒𝑟𝑠 𝑤𝑖𝑙𝑙 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑟𝑜𝑜𝑡𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛
  • 9. Solution 2. T = 2𝜋 𝑒 𝑔 (e) 𝑒 𝑤𝑎𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 𝑔 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑤𝑎𝑠 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡𝑒𝑑 𝑎𝑛𝑑 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑑 𝑏𝑦 2𝜋 𝑅𝑒𝑣𝑒𝑟𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠, 𝑤𝑒 𝑔𝑒𝑡: 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 2𝜋, 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 𝑇 2𝜋 = 𝑒 𝑔 𝑡ℎ𝑒𝑛 𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 (𝑟𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡) ( 𝑇 2𝜋 )2 = 𝑒 𝑔
  • 10. Solution (cont’d) ( 𝑇 2𝜋 )2 = 𝑒 𝑔 ( 𝑇2 22 𝜋2 ) = 𝑒 𝑔 𝑇2 4𝜋2 = 𝑒 𝑔 𝑁𝑜𝑤, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑔 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑖𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 𝑔 × 𝑇2 4𝜋2 = 𝑒 𝑒 = 𝑔𝑇2 4𝜋2
  • 11. Examine the next example carefully! It will give you an idea about the inspection and transposing method
  • 12. Solution 3. 𝑇2 = 4𝜋2 𝑅3 𝑔𝑟2 (𝑅) 𝐹𝑜𝑟 𝑡ℎ𝑖𝑠 𝑒𝑥𝑎𝑚𝑝𝑙𝑒, 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑖𝑛𝑠𝑝𝑒𝑐𝑡 𝑡ℎ𝑒𝑛 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑏𝑦 𝑑𝑒𝑚𝑜𝑛𝑠𝑡𝑟𝑎𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑡𝑒𝑝𝑠. 𝑅 𝑤𝑎𝑠 𝑐𝑢𝑏𝑒𝑑, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 4𝜋2 , 𝑡ℎ𝑒𝑛 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 𝑔𝑟2 𝑅𝑒𝑣𝑒𝑟𝑠𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑎𝑛𝑑 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑏𝑢𝑡 𝑠ℎ𝑜𝑤 𝑡ℎ𝑒𝑚 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒: 𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑔𝑟2 , 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 4𝜋2 𝑡ℎ𝑒𝑛 𝑐𝑢𝑏𝑒 𝑟𝑜𝑜𝑡 𝑅𝐸𝑀𝐸𝑀𝐵𝐸𝑅: 𝑇ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛. 𝐷𝑜 𝑡ℎ𝑒 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛: 𝑔𝑟2 × 𝑇2 = 4𝜋2 𝑅3 𝑔𝑟2 𝑇2 = 4𝜋2 𝑅3 𝑔𝑟2 𝑇2 4𝜋2 = 𝑅3 3 𝑔𝑟2 𝑇2 4𝜋2 = 𝑅 𝑅 = 3 𝑔𝑟2 𝑇2 4𝜋2
  • 13. Examples 1. 𝑀𝑎𝑘𝑒 𝑣 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 1 𝑣 + 1 𝑢 = 2 𝑟 𝑅𝑒𝑚𝑜𝑣𝑒 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑏𝑦 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑒𝑎𝑐ℎ 𝑡𝑒𝑟𝑚 𝑏𝑦 𝑡ℎ𝑒 𝐿𝐶𝑀 = 𝑣𝑢𝑟 𝑣𝑢𝑟 × ( 1 𝑣 ) + 𝑣𝑢𝑟 × ( 1 𝑢 ) = 𝑣𝑢𝑟 × ( 2 𝑟 ) 𝑢𝑟 + 𝑣𝑟 = 2𝑣𝑢 𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑣 𝑡𝑒𝑟𝑚𝑠 𝑢𝑟 = 2𝑣𝑢 − 𝑣𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑣 𝑢𝑟 = 𝑣 2𝑢 − 𝑟 𝑣 = 𝑢𝑟 2𝑢 − 𝑟
  • 14. Examples 2. 𝑇𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑒 𝑓𝑜𝑟 𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑥 = 𝑦 −4 𝑦 −7 (multiply by y -7 but show on the other side) 𝑥 𝑦 − 7 = 𝑦 − 4 𝑥𝑦 − 7𝑥 = 𝑦 − 4 𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑜𝑛 − 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 (subtract y; add 7x but show it on the opposite side) 𝑥𝑦 − 𝑦 = −4 + 7𝑥 𝑥𝑦 − 𝑦 = 7𝑥 − 4 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑦 𝑦 𝑥 − 1 = 7𝑥 − 4 𝑦 = 7𝑥 − 4 𝑥 − 1
  • 15. Examples 3. 𝑀𝑎𝑘𝑒 𝑝 𝑡ℎ𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝐷 𝑑 = 𝑓+𝑝 𝑓 − 𝑝 𝑆ℎ𝑜𝑤 𝑠𝑞𝑎𝑢𝑟𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆, 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑖𝑠 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆 𝐷 𝑑 2 = 𝑓 + 𝑝 𝑓 − 𝑝 𝐷 𝑑2 2 = 𝑓 + 𝑝 𝑓 − 𝑝 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑑2 𝑜𝑛 𝑅𝐻𝑆; 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑓 − 𝑝 𝑜𝑛 𝐿𝐻𝑆; 𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠 𝐷2 𝑓 − 𝑝 = 𝑑2 (𝑓 + 𝑝) 𝑅𝑒𝑚𝑜𝑣𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠, 𝑡ℎ𝑒𝑛 𝑔𝑟𝑜𝑢𝑝 𝑡ℎ𝑒 𝑝 𝑡𝑒𝑟𝑚𝑠 𝐷2 𝑓 −𝐷2 𝑝 = 𝑑2 𝑓 + 𝑑2 𝑝 𝐷2 𝑓 −𝑑2 𝑓 = 𝐷2 𝑝 + 𝑑2 𝑝
  • 16. 3 cont’d 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑝 𝐷2 𝑓 −𝑑2 𝑓 = 𝑝 (𝐷2 + 𝑑2) 𝐷2 𝑓 − 𝑑2 𝑓 𝐷2+ 𝑑2 = 𝑝 𝑝 = 𝐷2 𝑓 − 𝑑2 𝑓 𝐷2+ 𝑑2