Algebra

Special Cases in the Solution of Simultaneous Equations

First, with two unknown quantities. 𝑎₁𝑥+𝑏₁𝑦=𝑐₁ 𝑎₂𝑥+𝑏₂𝑦=𝑐₂ } Then 𝑥=𝑐₁𝑏₂−𝑐₂𝑏₁𝑎₁𝑏₂−𝑎₂𝑏₁, 𝑦=𝑐₁𝑎₂−𝑐₂𝑎₁𝑏₁𝑎₂−𝑏₂𝑎₁ If the denominators vanish, we have 𝑎₁𝑎₂=𝑏₁𝑏₂, and 𝑥=∞, 𝑦=∞; unless at the same time the numerators vanish, for then 𝑎₁𝑎₂=𝑏₁𝑏₂=𝑐₁𝑐₂, and 𝑥=00, 𝑦=00; and the equations are not independent, one being produced by multiplying the other by some constant.211 Next, with three unknown quantities. See (60) for the equations. If 𝑑₁𝑑₂𝑑₃ all vanish, divide each equation by 𝑧, and we have three equations for finding the two ratios 𝑥𝑧 and 𝑦𝑧, two only of which equations are necessary, any one being deducible from the other two if the three be consistent.212

To solve simultaneous equations by Indeterminate Multipliers

Take the equations 𝑥+2𝑦+3𝑧+4𝑤=27, 3𝑥+5𝑦+7𝑧+𝑤=48, 5𝑥+8𝑦+10𝑧−2𝑤=65, 7𝑥+6𝑦+5𝑧+4𝑤=53. Multiply the first by 𝐴, the second by 𝐵, the third by 𝐶, leaving one equation unmultiplied; and then add the results. Thus (𝐴+3𝐵+5𝐶+7)𝑥+(2𝐴+5𝐵+8𝐶+6)𝑦+(3𝐴+7𝐵+10𝐶+5)𝑧+(4𝐴+𝐵−2𝐶+4)𝑤=27𝐴+48𝐵+65𝐶+53 To determine either of the unknowns, for instance 𝑥, equate the coefficients of the other three separately to zero, and from the three equations find 𝐴, 𝐵, 𝐶. Then 𝑥=27𝐴+48𝐵+65𝐶+53𝐴+3𝐵+5𝐶+7213

Miscellaneous Equations and Solutions

Example

𝑥⁶±1=0 Divide by 𝑥³, and throw into factors, by (2) or (3). See also (480)214

Example

𝑥³−7𝑥−6=0 𝑥=−1 is a root, by inspection; therefore 𝑥+1 is a factor. Divide by 𝑥+1, and solve the resulting quadratic.215

Example

𝑥3+16𝑥=455 𝑥4+16𝑥2=455𝑥=65×7𝑥 𝑥4+65𝑥2+6522=49𝑥2+65×7𝑥+6522 𝑥2+652=7𝑥+652 𝑥2=7𝑥 ∴ 𝑥=7 Rule: Divide the absolute term (here 455) into two factors, if possible, such that one of them, minus the square of the other, equals the coefficient of 𝑥. See (483) for general solution of a cubic equation.216

Example

𝑥⁴−𝑦⁴=14560 𝑥−𝑦=8 } Put 𝑥=𝑧+𝑣 and 𝑦=𝑧−𝑣 Eliminate 𝑣, and obtain a cubic in 𝑧, which solve as in (216).217

Example

𝑥⁵−𝑦⁵=3093 𝑥−𝑦=3} Divide the first equation by the second, and subtract from the result the fourth power of 𝑥−𝑦. Eliminate (𝑥²+𝑦²), and obtain a quadratic in 𝑥𝑦.218

On forming Symmetrical Expressions

Example

Take, for example, the equation (𝑦−𝑐)(𝑧−𝑏)=𝑎² To form the remaining equations symmetrical with this, write the corresponding letters in vertical columns, observing the circular order in which 𝑎 is followed by 𝑏, 𝑏 by 𝑐, and 𝑐 by 𝑎. So with 𝑥, 𝑦, and 𝑧. Thus the equations become (𝑦−𝑐)(𝑧−𝑏)=𝑎2 (𝑧−𝑎)(𝑥−𝑐)=𝑏2 (𝑥−𝑏)(𝑦−𝑎)=𝑐2 To solve these equations, substitute 𝑥=b+c+𝑥'; 𝑦=c+𝑎+𝑦'; 𝑧=𝑎+𝑏+𝑧'; and, multiplying out, and eliminating 𝑦 and 𝑧, we obtain 𝑥=𝑏𝑐(𝑏+𝑐)−𝑎(𝑏²+𝑐²)𝑏𝑐−𝑐𝑎−𝑎𝑏 and therefore, by symmetry, the values of 𝑦 and 𝑧, by the rule just given.219

Example

𝑦2+𝑧2+𝑦𝑧=𝑎21 𝑧2+𝑥2+𝑧𝑥=𝑏22 𝑧2+𝑥2+𝑧𝑥=𝑏23 ∴3(𝑦𝑧+𝑧𝑥+𝑥𝑦)2=2𝑏2𝑐2+2𝑐2𝑎2+2𝑎2𝑏2−𝑎4−𝑏4−𝑐44 Now add (1), (2), and (3), and we obtain 2(𝑥+𝑦+𝑧)2−3(𝑦𝑧+𝑧𝑥+𝑥𝑦)=𝑎2+𝑏2+𝑐25 From (4) and (5), (𝑥+𝑦+𝑧) is obtained, and then (1), (2), and (3) are readily solved.220

Example

𝑥2+𝑦𝑧=𝑎21 𝑦2+𝑧𝑥=𝑏22 𝑧2+𝑥𝑦=c23 Multiply (2) by (3), and subtract the square of (1). Result: 𝑥(3𝑥𝑦𝑧−𝑥3−𝑦3−𝑧3)=𝑏2c2−𝑎4 ∴𝑥𝑏2c2−𝑎4=𝑦c2𝑎2−𝑏4=𝑧𝑎2𝑏2−c4=𝜆4 Obtain 𝜆² by proportion as a fraction with numerator =𝑥²+𝑦𝑧=𝑎²221

Example

𝑥=c𝑦+𝑏𝑧1 𝑦=𝑎𝑧+c𝑥2 𝑧=𝑏𝑥+𝑎𝑦3 Eliminate 𝑎 between (2) and (3), and substitute the value of 𝑥 from equation (1). Result 𝑦²1−𝑏²=𝑧²1−c²=𝑥²1−𝑎²222

Sources and References

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