Question 335260
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From the fact that x is negative and the quotient of x and y is z, you can deduce that not all three numbers are negative, hence y has to be positive because it is greater than each of the other two, and z must be negative because a negative (x) divided by a positive (y) must be negative.


From the fact that z/2 and z/3 are integers, and that z is negative, you can deduce that z must be -6 or an integer multiple of -6.


So let z = -6n where n is an natural number.  Then from x/z = -z, you can deduce that x = -36n^2. And then from x/y = z you can determine that y = 6n.


Solution set: *[tex \Large \{x,\,y,\,z\ |\ x\ =\ -36n^2,\,y\ =\ 6n,\,z\ =\ -6n,\ \forall\,n\,\in\,\mathbb{N}\}]


Check:


*[tex \Large 6n\ >\ -36n^2].


*[tex \Large -6n\ <\ 6n].


*[tex \Large -36n^2\ <\ 0].


*[tex \Large \frac{-36n^2}{-6n}\ =\ 6n].


*[tex \Large \frac{-36n^2}{6n}\ =\ -6n].


*[tex \Large \frac{-6n}{2}\ =\ -3n\ \in\ \mathbb{Z}].


*[tex \Large \frac{-6n}{3}\ =\ -2n\ \in\ \mathbb{Z}]


All statements true, answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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