GmatPrep DS

Hi,

On statement 1, your approach is right on.
You can plug in a positive number that meets x < 3, like x =1, which gives you 1/3 + 3 > 2, which is true.
You can then plug in a negative number (all of which will meet x < 3), like x = -1, which gives you -1/3 + -3 > 2, which is false.
So S1 is insufficient.

On Statement 2, we are told that x > 1. One thing this means is that x is positive. This may seem obvious, but that insight allows you to multiply through by x to get it out of the denominator without worrying about the possibility that x could be negative. (In S1, we could not do this, since x < 3 does not tell us the sign of x.)

So, we can multiply through by 3x to eliminate the numbers in the denominators. This gives us:

x/3 + 3/x > 2 ?
3x(x/3 + 3/x) > 3x(2)
x^2 + 9 > 6x

Now, the key here is to realize that this looks very similar to a perfect square. Specifically, if we bring 6x over to the other side, we get
x^2 + 9 > 6x
x^2 + 9 - 6x > 6x - 6x
x^2 - 6x + 9 > 0

Realizing that the expression x^2 - 6x + 9 is the perfect square of (x - 3) allows us to write this inequality as (x - 3)^2 > 0

From here, we know that if x - 3 = 0, the inequality is false (because we get 0 > 0), otherwise it is true. This allows us to see that x = 3 is the one value for which the inequality is not true when x > 1.