SOLUTION: Please help! If x,y,z are integers greater than 1 and if xy =24 and yz=56,which of the following must be true? and how? A) x<y<z B) y<x<z C) z<x<y D) y<x z

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Question 1007123: Please help!
If x,y,z are integers greater than 1 and if xy =24 and yz=56,which of the following must be true? and how?
A) x B) y C) z D) y
Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Please help!
If x,y,z are integers greater than 1 and if xy =24 and yz=56,
which of the following must be true? and how?
A) x < y < z
B) y < x < z
C) z < x < y
D) y < x z
--------------------------------------------
We are given: xy =24 and yz=56.  Therefore,

y= 24/x  and y = 56/z, so

24/x = 56/z

24z = 56x

 3z = 7x

z/x = 7/3 = 14/6 = 21/9 = 28/12

So the potential solutions are

(z,x) = (7,3), (14,6), (21,9), (28/12)

y = 56/z so we must rule out (21,9) because 21 is not a 
factor of 56.

So the only possibilities for (z,x) are these three

(z,x) = (7,3), (14,6), (28/12) 

Which means, using y=56/z, the only possibilities for (x,y,z) are

(x,y,z) = (3,8,7), (6,4,14), and (12,2,28)



Therefore z is always greater than x,  

But something is wrong with the choices listed above:

In the case (x,y,z) = (3,8,7)
we have 3 < 7 < 8, or x < z < y which is none of those.

In the case (x,y,z) = (6,4,14)
we have 4 < 6 < 14, or y < x < z which is B.

In the case (x,y,z) = (12,2,28)
we have 2 < 12 < 28, or y < x < z which is B.

So, as this clearly proves, none of those choices are always true.
B is true for only 2 of the 3 solutions, and none of them is
true for (x,y,z)=(3,8,7) 

Edwin
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
xy = 24 yields y = 24/x
yz = 56 yields y = 56/z

y equal to both yields 24/x = 56/z
that leads to 24z = 56x
that leads to z = 56x/24

that means that z has to be greater than x and the ratio of z to x has to be 56/24.

since 56/24 is not an integer, then x has to be a multiple of whatever makes 56/24 * x an integer.

56/24 reduces to 7/3.

x can be equal to 3,6,9,12,15,18,.....

in other words x can be a multiple of 3.

but not all multiples of 3 will work.

that's because xy has to be equal to 24.

when x = 3, y is equal to 8.
when x = 6, y is equal to 4.
when x = 9, y is not an integer.
when x = 12, y is equal to 2.
when x = 15, y is not an integer.
when x = 18, y is not an integer.
hen x = 21, y is not an integer.
when x = 24, y is equal to 1 which is not greater than 1.

the only possible values of x are 3 or 6 or 12.

when x is 3, y is 8, and z is 7 because xy = 24 and yz = 56.
you get x < z < y
that is not one of the selections.

when x is 6, y is 4 and z is 14 because 6*4 = 24 and 4*14 = 56.
you get y < x < z
that is selection B.

when x is 12, y is 2 and z is 28 because 12*2 = 24 and 2*28 = 56.
you get y < x < z
this is selection B.

i would go with selection B.
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