alex.gellatly wrote:If xy does not equal 0 and x^2y^2-xy=6, which of the following could be y in terms of x?
I. 1/2x
II. -2/x
III. 3/x
I only
II only
I and II only
I and III only
II and III only
Let x=1.
Plugging x=1 into x^2y^2-xy=6, we get:
(1²)y² - 1*y = 6.
y² - y - 6 = 0.
(y-3)(y+2) = 0.
y=3 or y=-2. These our are targets.
Now we plug x=1 into the answers to which yield our targets 3 and -2.
I. 1/2x = 1/2.
II. -2/x = -2/1 = -2.
III. 3/x = 3/1 = 3.
II and III work.
The correct answer is
E.
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