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AlgebraSimultaneous EquationsGeneral Solution with Two Unknown QuantitiesGiven𝑎1𝑥+𝑏1𝑦=𝑐1𝑎2𝑥+𝑏2𝑦=𝑐2 }, 𝑥= 𝑐1𝑏2−𝑐2𝑏1𝑎1𝑏2−𝑎2𝑏1𝑦= 𝑐1𝑎2−𝑐2𝑎1𝑏1𝑎2−𝑏2𝑎1 General Solution with Three Unknown QuantitiesGiven𝑎1𝑥+𝑏1𝑦+𝑐1𝑧=𝑑1𝑎2𝑥+𝑏2𝑦+𝑐2𝑧=𝑑2𝑎3𝑥+𝑏3𝑦+𝑐3𝑧=𝑑3 }, 𝑥= 𝑑1(𝑏2𝑐3−𝑏3𝑐2)+𝑑2(𝑏3𝑐1−𝑏1𝑐3)+𝑑3(𝑏1𝑐2−𝑏2𝑐1)𝑎1(𝑏2𝑐3−𝑏3𝑐2)+𝑎2(𝑏3𝑐1−𝑏1𝑐3)+𝑎3(𝑏1𝑐2−𝑏2𝑐1)and symmetrical forms for 𝑦 and 𝑧. Methods of Solving simultaneous Equations between Two Unknown QuantitiesBy substitutionFind 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.Examples𝑥+𝑦=237𝑦=28 }From (2), 𝑦=4, substitute in (1); thus 𝑥+20=23, 𝑥=3 By the Method of MultipliersExamples3𝑥+5𝑦=362𝑥−3𝑦=5 }Eliminate 𝑥 by multiplying (1) by 2 and (2) by 3; thus 6𝑥+10𝑦=726𝑥−9𝑦=15 }By subtraction 19𝑦=57 𝑦=3 By substitution in (2) 𝑥=7 By changing the quantities sought𝑥−𝑦=2𝑥2−𝑦2+𝑥+𝑦=30 }Let 𝑥+𝑦=𝑢, 𝑥−𝑦=𝑣, and substitute in equations: 𝑣=2𝑢𝑣+𝑢=30 }∴ 2𝑢+𝑢=30 𝑢=10 ∴ 𝑥+𝑦=10 𝑥−𝑦=2 From which 𝑥=6, and 𝑦=4 Examples𝑥+𝑦𝑥−𝑦+10 𝑥−𝑦𝑥+𝑦=9𝑥2+7𝑦2=64 }Substitute 𝑧 for 𝑥+𝑦𝑥−𝑦in (1): ∴ 2𝑧+ 10𝑧=9 2𝑧2−9𝑧+10=0 From which 𝑧= 52or 2, 𝑥+𝑦𝑥−𝑦=2 or 52. From which 𝑥=3𝑦 or 73𝑦 Substitute in (2), thus 𝑦=2 and 𝑥=6, or 𝑦= 6and 𝑥=2 Examples}Divide each quantity by 𝑥𝑦 3𝑦+ 5𝑥=1 2𝑦+ 7𝑥=3 }Multiply (3) by 2, and (1) by 3, and by subtraction 𝑦 is eliminated. By Substituting 𝑦=𝑡𝑥By Substituting 𝑦=𝑡𝑥, when the equations are homogeneous in the terms which contain 𝑥 and 𝑦.Examples}From (1) and (2), }(3) gives 52+7𝑡=5𝑡2 A quadratic equation from which 𝑡 must be found, and its value substituted in (4). 𝑥 is thus determined; and then 𝑦 from 𝑦=𝑡𝑥. Examples}From (1) and (2), by putting 𝑦=𝑡𝑥, }Squaring (4), 𝑥2(9𝑡2−12𝑡+4)=16 ∴ 9𝑡2−12𝑡+4=2+𝑡+3𝑡2, a quadratic equation for 𝑡. 𝑡 being found from this, equation (4) will determine 𝑥; and finally 𝑦=𝑡𝑥. Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive©sideway ID: 210600003 Last Updated: 6/3/2021 Revision: 0 Ref: References
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