SOLUTION: (Revision)solve for x, y and z making each statement true: y > x z < y x < 0 x / z = -z x / y = z z / 2 and z / 3 are integers

Question 335260: (Revision)solve for x, y and z making each statement true:
y > x
z < y
x < 0
x / z = -z
x / y = z
z / 2 and z / 3 are integers
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

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:

Check:

All statements true, answer checks.

John

My calculator said it, I believe it, that settles it

RELATED QUESTIONS
Solve for x, y and z making each statement true: y > x z < y x < 0 x / z = -Z (answered by greenestamps)

If y>x z (answered by Alan3354)
x + y + z = 2 x - y + z = 1 x - y - z = 1 solve for x, y and... (answered by sowmya)
Solve the system for x, y and z: x+y+z=8 x-y+z=2... (answered by Alan3354,TigerAlgebra)
(y + z)/x - z/x (answered by Fombitz)
Solve xyz=x + y + z for... (answered by stanbon)
solve for X, y and Z 2x+y+z=1 +3y-z=0 x+ +z+1 please and thank you (answered by Alan3354)
Y=X+Z/3 for... (answered by solver91311)
solve for x,y and z 1. x+y-2z=5 2. x+z=4 3.... (answered by Alan3354)