My Math Forum Hard Fraction problem

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 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: 7 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
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Quote:
 Originally Posted by IamHere 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)
Are you sure you've got those two equations right?

Are you supposed to know about complex numbers yet?

October 2nd, 2016, 01:13 PM   #4
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Quote:
 Originally Posted by edrudathec There are no real solutions.
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: 7 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: 6,782 Thanks: 2197 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
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Quote:
 Originally Posted by GeoLifeScienceGuy 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.
I didn't mean that you did mean solve it that way.

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: 6,782 Thanks: 2197 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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