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Q-161744-reposted-with-some-change-Solve-for-integer-numbers-x-y-5-x-y-5-5-y-x-y-5-5-y-5-x-




Question Number 161843 by Rasheed.Sindhi last updated on 23/Dec/21
Q#161744 reposted with some change.  Solve for integer numbers:  (x/y) + (5/x) + ((y - 5)/5) = ((y + x)/(y + 5)) + ((5 + y)/(5 + x))
You can't use 'macro parameter character #' in math modeSolveforintegernumbers:xy+5x+y55=y+xy+5+5+y5+x
Commented by malwan last updated on 23/Dec/21
x=0 or x=5  then we have one equation  and one variable y
x=0orx=5thenwehaveoneequationandonevariabley
Commented by Rasheed.Sindhi last updated on 23/Dec/21
x≠0 because it′s in denominator.  See Q#161744
x0becauseitsindenominator.You can't use 'macro parameter character #' in math mode
Commented by mr W last updated on 23/Dec/21
x≠0, x≠−5, y≠0, y≠−5  we can see x, y could (but not must)  be multiples of 5, then we can get  follwing solutions:  (5,5)  (5,10)  (−10,5)  but it′s not easy to prove that other  (not multiples of 5) solutions don′t  exist.
x0,x5,y0,y5wecanseex,ycould(butnotmust)bemultiplesof5,thenwecangetfollwingsolutions:(5,5)(5,10)(10,5)butitsnoteasytoprovethatother(notmultiplesof5)solutionsdontexist.

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