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Solve-the-equation-x-y-5-i-x-x-y-y-31-ii-No-trial-and-error-

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Question Number 26649 by Mr eaay last updated on 27/Dec/17
Solve the equation:x+y=5−−−i x^x + y^y =31−−−ii No trial and error
Solvetheequation:x+y=5−−−ixx+yy=31−−−iiNotrialanderror
Commented by mrW1 last updated on 27/Dec/17
If you want to get an analytic formula, I think there is no chance. For concrete values you can find the solution via graphic method.
Ifyouwanttogetananalyticformula,Ithinkthereisnochance.Forconcretevaluesyoucanfindthesolutionviagraphicmethod.
Answered by Amstrongmazoka last updated on 28/Dec/17
from equ(i), y=(5−x) substituting for y in equ(ii), gives, x^x +(5−x)^((5−x)) =31 ⇒x^x +(5−x)^((5−x)) −31=0 Now, let f(x)=x^x +(5−x)^((5−x)) −31 ⇒((df(x))/dx)=f′(x)=x^x (1+lnx)−(5−x)^((5−x)) [1+ln(5−x)] Now, by the Newton′s iterative formula, x_2 =x_1 −((f(x_1 ))/(f′(x_1 ))), now for this particular f(x), x_2 =x_1 −((x_1 ^x_1 +(5−x_1 )^((5−x_1 )) −31)/(x_1 ^x_1 (1+lnx_1 )−(5−x_1 )^((5−x_1 )) [1+ln(5−x_1 )])) Now, choosing any arbitrary value of x_1 that makes lnx and ln(5−x) defined, then we can only choose from 0<x<5 Now, taking x_1 =1 for instance, gives x_2 =1.3706, x_3 =1.6905, x_4 =1.9049, x_5 =1.9888 x_6 =1.9998, x_7 =2.0000, x_8 =2.0000 Thus lim_(x→2) x_n =2, ∴ x=2 is a solution. Similarly, choosing another arbitrary value of say x_1 =4, gives x_2 =3.6294, x_3 =3.3095, x_4 =3.0952, x_5 =3.0112, x_6 =3.0002, x_7 =3.0000, x_8 =3.0000 Again, lim_(x→3) x_n =3, ∴ x=3 is also a solution to the equation. Any other arbitrarily chosen value of x_(1 ) only points to either 2 or 3. Thus these are the only solutions. ∴ either x=2 or x=3. But x+y=5, −⇒y=5−x ∴ when x=2, y=5−2=3, and when x=3, y=5−3=2 Finally, for the system of equations, x=2 and y=3, or x=3 and y=2.
fromequ(i),y=(5−x)substitutingforyinequ(ii),gives,xx+(5−x)(5−x)=31⇒xx+(5−x)(5−x)−31=0Now,letf(x)=xx+(5−x)(5−x)−31⇒df(x)dx=f′(x)=xx(1+lnx)−(5−x)(5−x)[1+ln(5−x)]Now,bytheNewton′siterativeformula,x2=x1−f(x1)f′(x1),nowforthisparticularf(x),x2=x1−x1x1+(5−x1)(5−x1)−31x1x1(1+lnx1)−(5−x1)(5−x1)[1+ln(5−x1)]Now,choosinganyarbitraryvalueofx1thatmakeslnxandln(5−x)defined,thenwecanonlychoosefrom0<x<5Now,takingx1=1forinstance,givesx2=1.3706,x3=1.6905,x4=1.9049,x5=1.9888x6=1.9998,x7=2.0000,x8=2.0000Thuslimx→2xn=2,∴x=2isasolution.Similarly,choosinganotherarbitraryvalueofsayx1=4,givesx2=3.6294,x3=3.3095,x4=3.0952,x5=3.0112,x6=3.0002,x7=3.0000,x8=3.0000Again,limx→3xn=3,∴x=3isalsoasolutiontotheequation.Anyotherarbitrarilychosenvalueofx1onlypointstoeither2or3.Thusthesearetheonlysolutions.∴eitherx=2orx=3.Butx+y=5,−⇒y=5−x∴whenx=2,y=5−2=3,andwhenx=3,y=5−3=2Finally,forthesystemofequations,x=2andy=3,orx=3andy=2.

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