My Math Forum Hard Fraction problem

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

 November 9th, 2016, 05:19 AM #31 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,082 Thanks: 723 Amen.
 November 11th, 2016, 03:49 PM #32 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 371 Thanks: 68
 November 12th, 2016, 01:07 AM #33 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 As x and y are the zeros of u² - 4u + 5, 1/x and 1/y are the zeros of 5v² - 4v + 1, and so their sum is 4/5.
 November 15th, 2016, 11:00 PM #34 Banned Camp   Joined: Nov 2016 From: St. Louis, Missouri Posts: 28 Thanks: 4 Math Focus: arithmetic, fractions Although this problem is impossible to solve, you can make things easier in the future by making it so that the solution of x+y is larger than that for x times y.
 November 16th, 2016, 03:58 AM #35 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 That doesn't make sense.
November 16th, 2016, 05:12 AM   #36
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Quote:
 Originally Posted by toddhicks209 Although this problem is impossible to solve, you can make things easier in the future by making it so that the solution of x+y is larger than that for x times y.
Say what?

 December 1st, 2016, 01:43 AM #37 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $x+y=4...........(1)$ $xy=5.............(2)$ dividing eq(1) by $xy$ $\dfrac{x}{xy}+\dfrac{y}{xy}=\dfrac{4}{xy}$ $\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{4}{xy}$ since $xy=5$ therefore, $\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{4}{5}$

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