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+2
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6

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
+2

Jul 30, 2019
#2
+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
+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   CPhill  Jul 30, 2019
#4
+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
+2

Thanks, Alan.....that makes sense.....!!!!   CPhill  Jul 30, 2019
#6
+1

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

Jul 31, 2019