October 2nd, 2016, 12:35 PM  #1 
Newbie Joined: Oct 2016 From: United States Posts: 2 Thanks: 0  Hard Fraction problem
I'm having quite a bit of trouble with this question. If x + y = 4 and xy = 5, then 1/x + 1/y = ? (Note: 1/x and 1/y are written were written in fraction form) 
October 2nd, 2016, 01:08 PM  #2 
Newbie Joined: Oct 2016 From: USA Posts: 8 Thanks: 0 
There are no real solutions. I'll try and figure out what the complex ones are. EDIT: This is false. It ends up with one real solution. I underestimated this problem. EDIT: No, wait, I overestimated it. Last edited by edrudathec; October 2nd, 2016 at 01:23 PM. 
October 2nd, 2016, 01:12 PM  #3 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,495 Thanks: 753  
October 2nd, 2016, 01:13 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra  This is incorrect. Complex numbers are not required for this problem, only fractions. Have you tried doing the sum $\frac1x + \frac1y$? Last edited by v8archie; October 2nd, 2016 at 01:17 PM. 
October 2nd, 2016, 01:25 PM  #5 
Newbie Joined: Sep 2016 From: USA Posts: 14 Thanks: 3 
Try thinking of it like x + y = 4 xy = 5 xy  1 = 4 x + y = xy  1 And then divide and substitute xy to solve from there. 
October 2nd, 2016, 01:30 PM  #6 
Newbie Joined: Oct 2016 From: USA Posts: 8 Thanks: 0 
Trying to solve for x and y (which is what I did) is not a good way to solve this, because they're both complex numbers.

October 2nd, 2016, 01:31 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra 
Given that it is described as a fractions problem, I am 95% certain that the OP is supposed to evaluate the sum he was given.

October 2nd, 2016, 01:44 PM  #8 
Newbie Joined: Sep 2016 From: USA Posts: 14 Thanks: 3 
I didn't mean solve it that way: solve for x=? and y=? Like if you divided and substituted xy then: x + y = xy  1 (Divide by xy) 1/y + 1/x = 1  1/xy (xy = 5) 1/x + 1/y = 1  1/5 Answer: 4/5. No complex numbers involved. I laid out the steps divide and substitute xy so that if u tried them you would see that the sum could easily be solved for. Last edited by GeoLifeScienceGuy; October 2nd, 2016 at 01:47 PM. 
October 2nd, 2016, 02:15 PM  #9  
Newbie Joined: Oct 2016 From: USA Posts: 8 Thanks: 0  Quote:
I wasn't sure what you meant by divide and substitute xy at first, so I advocated for v8archie's solution, which I think is the best out of the three. Sorry for the confusion.  
October 2nd, 2016, 02:16 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra 
It is even easier, and better practice for the OP to simply do the given sum.


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