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I have a doubt in equality:
a>0 and 1−x>0, so x<1 OR a<0 and 1−x<0, so x>1. not able to get how did a>0 turned to a<0 and x<1 to 1-x<0
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Bunuel
If ax + b = 0, is x > 0?

    Given: \(b=-ax\).
    Question: is \(x>0\)?

(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):

    \((a+b)+(a-b)>0\);

    \(a>0\).

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 Bunuel
One 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
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Kimberly77
Bunuel
If ax + b = 0, is x > 0?

    Given: \(b=-ax\).
    Question: is \(x>0\)?

(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):

    \((a+b)+(a-b)>0\);

    \(a>0\).

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 Bunuel
One 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

If ax + b = 0, is x > 0?

(1) a + b = 0

Substituting b = -a in ax + b = 0, we get ax - a = 0, which simplifies to a(x - 1) = 0. Thus, either a = 0 (and x can be any number) or x = 1. Not sufficient.

(2) a - b = 0

Substituting b = a in ax + b = 0, we get ax + a = 0, which simplifies to a(x + 1) = 0. Thus, either a = 0 (and x can be any number) or x = -1. Not sufficient.

(1)+(2) Solving a + b = 0 and a - b = 0 gives a = b = 0. Since any x satisfies 0*x + 0 = 0, we still cannot determine whether x > 0. Not sufficient.

Answer: E.
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