Bunuel
If ax + b = 0, is x > 0?Given: \(b=-ax\).
Question: is \(x>0\)?
(1) \(a+b>0\):
Either \(a>0\) and \(1-x>0\), so \(x<1\)
OR \(a<0\) and \(1-x<0\), so \(x>1\).
Not sufficient.
(2) \(a-b>0\):
Either \(a>0\) and \(1+x>0\), so \(x>-1\)
OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.
(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction):
\((a+b)+(a-b)>0\);
\(a>0\).
Thus, we have the first range from (1): \(x<1\) and the first case from (2): \(x>-1\), which means that \(-1<x<1\). Therefore, \(x\) may or may not be negative. Not sufficient.
Answer: E.
Always the best and the most efficient way, thanks
BunuelOne question regarding your explanations that include both +/- cases, is this due to statements 1 & 2 has inequality sign?
Do we still need to consider if statements 1 & 2 is (1) a + b = 0 & (2) a - b = 0 ?
As then we will get on substitution a (1-x) = 0
I presumed yes but could you kindly confirm? Thanks