Sum (1) and (2), we can get a>0. Subtract (1) from (2), we can get b>0. In this setting, ax+b=0 means x is negative. Thus, I thought C is correct. What am I wrong?

I love this question. Probably, because one can solve it under 1 minute if you are clear on the rules of positive & negative (in inequality).

The question is saying the following: ax + b =0 => b=-ax or => -b/a = x

So the rephrased question is not to check x > 0 but is -b/a > 0. Now this has two implications as follows:-

1. Case I: If a > 0, b < 0 [Note: ',' indicates and]
2. Case II: If a <0, b > 0

So another deep down will give us the following phrasing:- Is Case I or Case II true (If either one of them is true, then you can see we will get our answer to the question)

Now, Statement A: a + b > 0 i.e. we don't the 'sign' of either a or b: both could be positive; one could be large positive, one could be small negative. So clearly, Not Sufficient

Moving on to statement B: a - b > 0 i.e. same as the above statement only with a minor tweak, hence not sufficient

Combining both the statement: Now inequalities can be added in the same direction; therefore, we get the following:-
2a > 0 => a > 0
Do you think this information is enough to answer the question?. The answer is a resounding 'No'. Why is that?
Well, the statement only gives us the sign of a; however, we don't know whether b is either > 0 or less < 0 or even equal to 0.

Kudos, if you liked the solution

Regards,
Abheek

Bunuel
On your solution of (1)+(2) in which you sum (1) and (2) as they have same sign>> (a+b)+(a−b)>0 and get a>0.
Question is whenever I encountered same sign, can I also subtract LHS terms >> (a+b)-(a−b)>0 and get b>0.
I was mistakenly did so with this Q and from (1)+(3); I got both a>0 (combine LHS) and b>0 (Subtract LHS), which results in X<0, if ax + b = 0, corresponding to ans. C

Please correct me on this point so that next time I can apply only the combination of LHS and not subtraction in context of equation with same sign.

Thank you!

Target question: Is x > 0

Given: ax + b = 0

Statement 1: a + b > 0
At this point, we have 1 equation, 1 inequality, and THREE variables.
Even if we had 2 EQUATIONS and 3 variables, we probably wouldn't be able to make any conclusions about whether x is positive or negative.
Given this, let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: a - b > 0
This is the same scenario as statement 1, so let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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