Solve the inequality and you'll know :)

15x - \frac2x > 1

Since you don't know whether x is positive or not, bring all to the same side.

15x - \frac2x -1> 0

Now combine to form a big fraction.

\frac{15x^2 -x - 2}{x} > 0

\frac{(5x - 2)(3x + 1)}{x} > 0

Now try in some values of x. Usually, you do this in a table, but unfortunately, it is rather difficult and time consuming to do it here... can you try by yourself and post what you get? :)

The critical values are 2/5, 0 and -1/3. So you try in:

x < -1/3, -1/3 < x < 0, 0 < x < 2/5 and x > 2/5

And to elaborate on what Unknown008 did, once you have the problem set up as a big fraction on the left, and zero on the right, you'll want to determine all the possible values of x where the left side could potentially change between a positive value and a negative. There are only two things that allow that to happen: (1) Either the numerator of the function equals zero (because for a continuous function, you can't go from positive to negative, or vice-versa, without crossing through zero) or (2) the function "blows up" (i.e. goes to +/- infinity) at a point because the denominator goes to zero. For what it's worth, the values where the numerator goes to zero are called the "zeros" of the function (which is self-explanatory), and the values where the denominator goes to zero are called the "poles" of the function.

So the trick is to find all of the poles and zeros for the big fraction, and then do like Unknown008 said: make a table. For example, since he already factored it for you and everything, it should be pretty obvious that the zeros are at x = -1/3, and x = 2/5. And the one pole is at x=0. So you'd make a table with entries for x < -1/3, -1/3 < x < 0, 0 < x < 2/5 and x > 2/5. Then plug in a test value from each of those ranges into your original problem to find out which one(s) make the inequality a true statement.

\begin{matrix}
\text{Range } & \text{ Test value } & 15x-\frac{2}{x} & \text{ Is it > 1?}\\
x<-\frac{1}{3} & -1 & -13 & \text{no}\\
-\frac{1}{3} < x < 0 & -0.1 & 18.5 & \text{yes}\\
0 < x < \frac{2}{5} & 0.1 & -18.5 & \text{no}\\
x > \frac{2}{5} & 1 & 13 & \text{yes}
\end{matrix}

Lol, I had to spread the rep jcaron. Taking what you posted, here is what I usually do, using array so that I can add the separators.

\begin{array}{|c|c|c|c|c|} \hline
\text{Range } & (5x-2) & (3x+1) & \frac1x & \frac{(5x-2)(3x+1)}{x}\\ \hline
x<-\frac{1}{3} & -ve & -ve & -ve & -ve\\ \hline
-\frac{1}{3} < x < 0 & -ve & +ve & -ve & +ve\\ \hline
0 < x < \frac{2}{5} &-ve & +ve & +ve& -ve\\ \hline
x > \frac{2}{5} & +ve & +ve & +ve & +ve\\ \hline
\end{array}

Vardhan7, to understand my table, pick a value in the given range, and put it in the expression heading. If it's a negative value, put -ve and if positive, +ve. When done with the 3, the last column becomes easy, since (---) = -, (-+-) = +, (-++) = - and (+++) = +

And using \frac{(5x-2)(3x+1)}{x}, we know that it should be positive, that is greater than zero. Look for +ve in the right column, and look for the range corresponding to that.

It's a little different to that of jcaron, but it's basically the same principle.

So... can you tell us the final answers you get? :)

Cool! Definitely much clearer with the separators.