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Let \(x\) and \(y\) be real numbers such that

\(2 < \frac{x - y}{x + y} < 5.\)
If \(\frac{x}{y}\) is an integer, what is its value?

 Jul 30, 2019
 #1
avatar
+2

See this link:  https://www.wolframalpha.com/input/?i=Give+integer+solutions+++++%5B++2+%3C+%5Cfrac%7Bx+-+y%7D%7Bx+%2B+y%7D+%3C+5+%5D

 Jul 30, 2019
 #2
avatar+7711 
+2

\(\text{Let }\dfrac{x}{y}=z.\\ 2 < \dfrac{x - y}{x + y} < 5\\ 2 < \dfrac{z - 1}{z + 1} < 5\\ 2z + 2 \stackrel{\boxed{1}}{<} z - 1 \stackrel{\boxed{2}}{<} 5z + 5\\ \boxed{1}: 2z + 2 < z - 1\\ z < -3\\ \boxed{2}: z - 1 < 5z + 5\\ -6 < 4z\\ z > -\dfrac{3}{2}\\ \text{Combine }\boxed{1} \text{ and }\boxed{2}: -\dfrac{3}{2} < z < -3\\ \text{The only integer solution is } z = -2\\ \therefore \dfrac{x}{y} = -2\)

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 Jul 30, 2019
 #3
avatar+102913 
+3

One question, Max

 

If  the solution is  -3/2 < z < -3

 

What integer  is > -3/2   and < -3  ???.....it can't be  z =   -2

 

 

cool cool cool

CPhill  Jul 30, 2019
 #4
avatar+28125 
+4

Note that x-y must be larger than x+y if the ratio is to be between 2 and 5. This will be the case if y is negative.

 

So we have (x-y)/(x+y) > 2,  or. x-y > 2x+2y or x < -3y or x/y > -3 where the < becomes > because we are dividing by a negative number.

 

We also have (x-y)/(x+y)<5,  or x-y < 5x+5y or 4x > -6y or x > -(3/2)y or x/y < -3/2 where > becomes < because we are dividing by a negative number.

 

Hence -3 < x/y < -3/2, giving x/y = -2

 

Max got the right answer with incorrect reasoning! It happens!

 Jul 30, 2019
 #5
avatar+102913 
+2

Thanks, Alan.....that makes sense.....!!!!

 

cool cool cool

CPhill  Jul 30, 2019
 #6
avatar+133 
+1

Thank you to the numerous amount of people who answered this question

 Jul 31, 2019

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