What faf did was put the equation in terms of x= by manipulating it algebra style, like so:
4/x + 2/y = 1
[multiply everything by x, and y, to get rid of the x and y terms in the denominators]
4y + 2x = 1xy
[subract 2x to get x's on one side]
4y = 1xy - 2x
[take the common factor on the right side (which is x)]
4y = x(y - 2)
[divide by (y-2) to get x alone]
4y / (y-2) = x
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since we now know what X equals, we can write the ordered pair like this: (4y / (y-2), y)
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Now that I look at it, however, that answer is not sufficient, because you're asking how many INTEGER ordered pairs.
If it were simply ordered pairs, the solutions would be infinite, as the equation indicates. However, i'll have to ge tback to you because its INTEGER ordered pairs.
its probably something simple too, and were all just overlooking it