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gmattester
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Notice that the inequality will be false if x is negative, so 1) is not sufficient. From 2), we know x is positive, so we can multiply both sides by 3x (to get rid of the fractions), and since we know x is positive, we don't need to worry about whether to reverse the inequality when we do this:gmattester wrote:Is x/3 + 3/x >2
a) x<3
b) x>1
If x > 0,
x/3 + 3/x > 2
x^2 + 9 > 6x
x^2 - 6x + 9 > 0
(x-3)^2 > 0
The left side is squared so can't be negative; the inequality will be true for every value of x except x = 3. So Statement 2) is not sufficient on its own (x could be 3), but 1) and 2) together are sufficient (x cannot be 3). C.
