Is x > 0 ? (1) x + 1 > 0 (2) |x + 4| < |x - 2|

wminesh

My Two cents -

1) Whenever you're working with signs, you will have to consider the regions in which those signs hold true.

For example



In this case, you have assumed

x + 4 > 0 ⇒ x > -4
x - 2 < 0 ⇒ x < 2

From the working, you arrived at x < -1, not sure why did you indicate the net result as "Not Sufficient" because all it tells us that between -4 and -1 the equation holds true.

Hence, the values of x that satisfy the equation lie between -4 and -1.

So ideally this part is sufficient because all such values are NOT greater than 0.

Similarly -



In this case, you have assumed

x + 4 < 0 ⇒ x < -4
x + 2 > 0 ⇒ x > -2

The consideration itself is not valid as x cannot be less than -4 and greater than -2 at the same time.

2) You can form regions and evaluate the signs within that region -



Hope this clarifies.