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(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<0 and 1-x<0, so x>1. Not sufficient.

(2) a-b>0 --> a+ax>0 --> a(1+x)>0 --> either a>0 and 1+x>0, so x>-1ORa<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, so we have first range from (1): x<1 and first case from (2): x>-1 --> -1<x<1, so x may or may not be negative. Not sufficient.

x=-b/a; x>0 means a and b have to be different sign. I am usually not good at picking numbers, but in this case if you see a+b>0 and a-b>0 it is clear that 5+2>0 5-2>0 and 5+(-2)>0 5-(-2)>0 hence we cannot say that a and b will have the same sign. Hence E.
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(1) a+b>0 --> a-ax>0 --> a(1-x)>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 --> a+ax>0 --> a(1+x)>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, so we have first range from (1): x<1 and first case from (2): x>-1 --> -1<x<1, so x may or may not be negative. Not sufficient.

Answer: E.

you've made things easier bunuel. +1
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awesome explanations both bunuel and mainhoon --these two approaches would probably serve two different types of students really well. nice work!

there's also a potential blend of the two-- mainhoon's rephrase (are the signs of a and b different) and the stacking/adding of inequalities when evaluating choice c for that same rephrased question (if your head gets turned around when too many numbers are involved)
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(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<0 and 1-x<0, so x>1. Not sufficient.

(2) a-b>0 --> a+ax>0 --> a(1+x)>0 --> either a>0 and 1+x>0, so x>-1ORa<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, so we have first range from (1): x<1 and first case from (2): x>-1 --> -1<x<1, so x may or may not be negative. Not sufficient.

Answer: E.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

(1) a+b>0 --> a-ax>0 --> a(1-x)>0 --> either a>0 and 1-x>0, so x<1ORa<0 and 1-x<0, so x>1. Not sufficient.

(2) a-b>0 --> a+ax>0 --> a(1+x)>0 --> either a>0 and 1+x>0, so x>-1ORa<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, so we have first range from (1): x<1 and first case from (2): x>-1 --> -1<x<1, so x may or may not be negative. Not sufficient.

Answer: E.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

Why does a+b>0 imply that either a or b is negative? Why cannot both be positive?
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