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`-=[]โŸจโŸฉ\;',./~!@#$%^&*()_+{}|:"<>? ๐‘Ž๐‘๐‘๐‘‘๐‘’๐‘“๐‘”โ„Ž๐‘–๐‘—๐‘˜๐‘™๐‘š๐‘›๐‘œ๐‘๐‘ž๐‘Ÿ๐‘ ๐‘ก๐‘ข๐‘ฃ๐‘ค๐‘ฅ๐‘ฆ๐‘ง ร…โ€‰โˆ’โ€‚ร—โ€ƒโ‹…โˆ“ยฑโˆ˜๊žŠ๏นฆโˆ—โˆ™ โ„ฏ ๐”ธ๐”นโ„‚๐”ป๐”ผ๐”ฝ๐”พโ„๐•€๐•๐•‚๐•ƒ๐•„โ„•๐•†โ„™โ„šโ„๐•Š๐•‹๐•Œ๐•๐•Ž๐•๐•โ„ค๐ด๐ต๐ถ๐ท๐ธ๐น๐บ๐ป๐ผ๐ฝ๐พ๐ฟ๐‘€๐‘๐‘‚๐‘ƒ๐‘„๐‘…๐‘†๐‘‡๐‘ˆ๐‘‰๐‘Š๐‘‹๐‘Œ๐‘ โˆผโˆฝโˆพโ‰โ‰‚โ‰ƒโ‰„โ‰…โ‰†โ‰‡โ‰ˆโ‰‰โ‰Œโ‰โ‰ โ‰ก โ‰คโ‰ฅโ‰ฆโ‰งโ‰จโ‰ฉโ‰ชโ‰ซ โˆˆโˆ‰โˆŠโˆ‹โˆŒโˆ โŠ‚โŠƒโŠ„โŠ…โІโЇ ๐›ผ๐›ฝ๐›พ๐›ฟ๐œ€๐œ๐œ‚๐œƒ๐œ„๐œ…๐œ†๐œ‡๐œˆ๐œ‰๐œŠ๐œ‹๐œŒ๐œŽ๐œ๐œ๐œ‘๐œ’๐œ“๐œ” โˆ€โˆ‚โˆƒโˆ…โฆฐโˆ†โˆ‡โˆŽโˆžโˆโˆดโˆต โˆโˆโˆ‘โ‹€โ‹โ‹‚โ‹ƒ โˆงโˆจโˆฉโˆช โˆซโˆฌโˆญโˆฎโˆฏโˆฐโˆฑโˆฒโˆณ โˆฅโ‹ฎโ‹ฏโ‹ฐโ‹ฑ โ€– โ€ฒ โ€ณ โ€ด โ„ โ— สน สบ โ€ต โ€ถ โ€ท ๏น ๏น‚ ๏นƒ ๏น„ ๏ธน ๏ธบ ๏ธป ๏ธผ ๏ธ— ๏ธ˜ ๏ธฟ ๏น€ ๏ธฝ ๏ธพ ๏น‡ ๏นˆ ๏ธท ๏ธธ โœ   โ   โŽด  โŽต  โž   โŸ   โ    โก โ†โ†‘โ†’โ†“โ†คโ†ฆโ†ฅโ†งโ†”โ†•โ†–โ†—โ†˜โ†™โ–ฒโ–ผโ—€โ–ถโ†บโ†ปโŸฒโŸณ โ†ผโ†ฝโ†พโ†ฟโ‡€โ‡โ‡‚โ‡ƒโ‡„โ‡…โ‡†โ‡‡ โ‡โ‡‘โ‡’โ‡“โ‡”โ‡Œโ‡โ‡โ‡•โ‡–โ‡—โ‡˜โ‡™โ‡™โ‡ณโฅขโฅฃโฅคโฅฅโฅฆโฅงโฅจโฅฉโฅชโฅซโฅฌโฅญโฅฎโฅฏ
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Content

Algebra
โ€ƒSimultaneous Equations
โ€ƒโ€ƒGeneral Solution with Two Unknown Quantities
โ€ƒโ€ƒGeneral Solution with Three Unknown Quantities
โ€ƒMethods of Solving simultaneous Equations between Two Unknown Quantities
โ€ƒโ€ƒBy substitution
โ€ƒโ€ƒโ€ƒExamples
โ€ƒโ€ƒBy the Method of Multipliers
โ€ƒโ€ƒโ€ƒExamples
โ€ƒโ€ƒBy changing the quantities sought
โ€ƒโ€ƒโ€ƒExamples
โ€ƒโ€ƒโ€ƒExamples
โ€ƒโ€ƒBy Substituting ๐‘ฆ=๐‘ก๐‘ฅ
โ€ƒโ€ƒโ€ƒExamples
โ€ƒโ€ƒโ€ƒExamples
โ€ƒSources and References

Algebra

Simultaneous Equations

General Solution with Two Unknown Quantities

Given ๐‘Ž1๐‘ฅ+๐‘1๐‘ฆ=๐‘1๐‘Ž2๐‘ฅ+๐‘2๐‘ฆ=๐‘2}, ๐‘ฅ=๐‘1๐‘2โˆ’๐‘2๐‘1๐‘Ž1๐‘2โˆ’๐‘Ž2๐‘1 ๐‘ฆ=๐‘1๐‘Ž2โˆ’๐‘2๐‘Ž1๐‘1๐‘Ž2โˆ’๐‘2๐‘Ž1

General Solution with Three Unknown Quantities

Given ๐‘Ž1๐‘ฅ+๐‘1๐‘ฆ+๐‘1๐‘ง=๐‘‘1๐‘Ž2๐‘ฅ+๐‘2๐‘ฆ+๐‘2๐‘ง=๐‘‘2๐‘Ž3๐‘ฅ+๐‘3๐‘ฆ+๐‘3๐‘ง=๐‘‘3}, ๐‘ฅ=๐‘‘1(๐‘2๐‘3โˆ’๐‘3๐‘2)+๐‘‘2(๐‘3๐‘1โˆ’๐‘1๐‘3)+๐‘‘3(๐‘1๐‘2โˆ’๐‘2๐‘1)๐‘Ž1(๐‘2๐‘3โˆ’๐‘3๐‘2)+๐‘Ž2(๐‘3๐‘1โˆ’๐‘1๐‘3)+๐‘Ž3(๐‘1๐‘2โˆ’๐‘2๐‘1) and symmetrical forms for ๐‘ฆ and ๐‘ง.

Methods of Solving simultaneous Equations between Two Unknown Quantities

By substitution

Find one unknown in terms of the other from one of the two equations, and substitute this value in the remaining equation. Then solve the resulting equation.

Examples

๐‘ฅ+๐‘ฆ=237๐‘ฆ=28} From (2), ๐‘ฆ=4, substitute in (1); thus ๐‘ฅ+20=23, ๐‘ฅ=3

By the Method of Multipliers

Examples

3๐‘ฅ+5๐‘ฆ=362๐‘ฅโˆ’3๐‘ฆ=5} Eliminate ๐‘ฅ by multiplying (1) by 2 and (2) by 3; thus 6๐‘ฅ+10๐‘ฆ=726๐‘ฅโˆ’9๐‘ฆ=15} By subtraction 19๐‘ฆ=57 ๐‘ฆ=3 By substitution in (2) ๐‘ฅ=7

By changing the quantities sought

๐‘ฅโˆ’๐‘ฆ=2๐‘ฅ2โˆ’๐‘ฆ2+๐‘ฅ+๐‘ฆ=30} Let ๐‘ฅ+๐‘ฆ=๐‘ข, ๐‘ฅโˆ’๐‘ฆ=๐‘ฃ, and substitute in equations: ๐‘ฃ=2๐‘ข๐‘ฃ+๐‘ข=30} โˆด 2๐‘ข+๐‘ข=30 ๐‘ข=10 โˆด ๐‘ฅ+๐‘ฆ=10 ๐‘ฅโˆ’๐‘ฆ=2 From which ๐‘ฅ=6, and ๐‘ฆ=4

Examples

2๐‘ฅ+๐‘ฆ๐‘ฅโˆ’๐‘ฆ+10๐‘ฅโˆ’๐‘ฆ๐‘ฅ+๐‘ฆ=9๐‘ฅ2+7๐‘ฆ2=64} Substitute ๐‘ง for ๐‘ฅ+๐‘ฆ๐‘ฅโˆ’๐‘ฆ in (1): โˆด 2๐‘ง+10๐‘ง=9 2๐‘ง2โˆ’9๐‘ง+10=0 From which ๐‘ง=52 or 2, ๐‘ฅ+๐‘ฆ๐‘ฅโˆ’๐‘ฆ=2 or 52. From which ๐‘ฅ=3๐‘ฆ or 73๐‘ฆ Substitute in (2), thus ๐‘ฆ=2 and ๐‘ฅ=6, or ๐‘ฆ=67 and ๐‘ฅ=27

Examples

3๐‘ฅ+5๐‘ฆ=๐‘ฅ๐‘ฆ2๐‘ฅ+7๐‘ฆ=3๐‘ฅ๐‘ฆ} Divide each quantity by ๐‘ฅ๐‘ฆ 3๐‘ฆ+5๐‘ฅ=12๐‘ฆ+7๐‘ฅ=3} Multiply (3) by 2, and (1) by 3, and by subtraction ๐‘ฆ is eliminated.

By Substituting ๐‘ฆ=๐‘ก๐‘ฅ

By Substituting ๐‘ฆ=๐‘ก๐‘ฅ, when the equations are homogeneous in the terms which contain ๐‘ฅ and ๐‘ฆ.

Examples

52๐‘ฅ2+7๐‘ฅ๐‘ฆ=5๐‘ฆ215๐‘ฅโˆ’3๐‘ฆ=172} From (1) and (2), 52๐‘ฅ2+7๐‘ก๐‘ฅ2=5๐‘ก2๐‘ฅ235๐‘ฅโˆ’3๐‘ก๐‘ฅ=174} (3) gives 52+7๐‘ก=5๐‘ก2 A quadratic equation from which ๐‘ก must be found, and its value substituted in (4). ๐‘ฅ is thus determined; and then ๐‘ฆ from ๐‘ฆ=๐‘ก๐‘ฅ.

Examples

2๐‘ฅ2+๐‘ฅ๐‘ฆ+3๐‘ฆ2=1613๐‘ฆโˆ’2๐‘ฅ=42} From (1) and (2), by putting ๐‘ฆ=๐‘ก๐‘ฅ, ๐‘ฅ2(2+๐‘ก+3๐‘ก2=163๐‘ฅ(3๐‘กโˆ’2)=44} Squaring (4), ๐‘ฅ2(9๐‘ก2โˆ’12๐‘ก+4)=16 โˆด 9๐‘ก2โˆ’12๐‘ก+4=2+๐‘ก+3๐‘ก2, a quadratic equation for ๐‘ก. ๐‘ก being found from this, equation (4) will determine ๐‘ฅ; and finally ๐‘ฆ=๐‘ก๐‘ฅ.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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ID: 210600003 Last Updated: 6/3/2021 Revision: 0 Ref:

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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