Here's the way I solved it, although I discovered in doing so that two of the answer choices are correct, so that's annoying. Begin with the given inequality and isolate the terms a bit:
\(ax + 2 < 3x + b\)
\(ax - 3x + 2 - 2 < 3x - 3x + b - 2\)
\((a - 3)x < b - 2\)
Next we'll consider two cases. If \(a > 3\) then \(a - 3 > 0\) and dividing by that quantity will not change the direction of the inequality:
\(\displaystyle x < \frac{b - 2}{a - 3}\)
Now let's make a sign argument. We're given that \(x < 0\) so we really have:
\(\displaystyle \text{(negative)} < \frac{b - 2}{\text{(positive)}}\)
The only way for this to be true for every negative number is if \(b - 2 \ge 0\) (why?), so that gives us one possible solution \(a > 3, \: b \ge 2\). This doesn't exactly match one of the given answer choices, but it does imply that answer choice (a) is valid (why?).
Going back a step, we can consider another case. If \(a < 3\) then \(a - 3 < 0\) and dividing by that quantity will change the direction of the inequality:
\(\displaystyle x > \frac{b - 2}{a - 3}\)
Making our sign argument again:
\(\displaystyle \text{(negative)} > \frac{b - 2}{\text{(negative)}}\)
From this we can see that \(b > 2\) is a requirement, but we can also see that the right-hand side of the inequality will always be some fixed number, and there's no possible way we can guarantee that every negative number will be greater than some fixed negative number (why?). Therefore, we can conclude that if \(a < 3\) there is no solution.
Finally, we can consider the last case. If \(a = 3\) then \(a - 3 = 0\) and we cannot divide by that quantity. However, we can see that the inequality then becomes:
\(0 < b - 2\)
And this is true so long as \(b - 2 > 0 \implies b > 2\). This gives us another possible solution \(a = 3, \: b > 2\) which is answer choice (e).
Edit: Whoops! I made a sign error of my own. I've fixed this now, and also inserted a caveat to make my meaning more explicit. Sorry about any confusion this may have caused.