SOLUTION: If x is an integer greater than 1 and if y = x + (1/x), which of the following must be true ? I y ≠ x II y is an integer III xy > x^2 (A) I

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: If x is an integer greater than 1 and if y = x + (1/x), which of the following must be true ? I y ≠ x II y is an integer III xy > x^2 (A) I       Log On


   

Question 296658: If x is an integer greater than 1 and if y = x + (1/x), which of the following must be true ?
I y ≠ x
II y is an integer
III xy > x^2
(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I , II ,and III only
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, we need to see if statement I is true or not. Assume that y=x, if this is the case, then x%2B1%2Fx=x. Subtract 'x' from both sides to get 1%2Fx=0. Since this equation is NEVER true for any 'x' values (let alone positive 'x' values), we can say that y=x is false where y=x%2B1%2Fx. So this means that y%3C%3Ex and that statement I is true.
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If x%3E1, then 1%2Fx is NOT an integer. So this consequently means that x%2B1%2Fx is also NOT an integer. So statement II is false meaning that choices C and E can be ignored.
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Finally, because 'x' is an integer and 'x' is positive, we also know that x%3Cx%2B1%2Fx or that x%3Cy (do a simple substitution here). Multiply both sides by 'x' to get x%5E2%3Cxy (note: the sign doesn't flip because 'x' is positive).

So we've shown that if 'x' is an integer and 'x' is positive, then x%5E2%3Cxy where y=x%2B1%2Fx. So statement III is true.

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So because statements I and III are both true, we can say that the answer is choice D.