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Zitkin8
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 25 May 2023, 20:14
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uzzy12
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Hi Bunuel, why can't we add both equations and subtract both equations to get a>0 and b>0 simultaneously, and hence we will get x<0. Can you explain the error in my explanation?
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Bunuel
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 26 May 2023, 00:03
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Zitkin8
uzzy12
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Hi Bunuel, why can't we add both equations and subtract both equations to get a>0 and b>0 simultaneously, and hence we will get x<0. Can you explain the error in my explanation?

You can add the inequalities to get a > 0 because the sign are in the same direction. However, you cannot subtract. You can subtract one inequality from another, when their signs are in the opposite directions.

Check Manipulating Inequalities for more.

Hope it helps.
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 28 May 2023, 08:04
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Zitkin8
uzzy12
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0

Hi Bunuel, why can't we add both equations and subtract both equations to get a>0 and b>0 simultaneously, and hence we will get x<0. Can you explain the error in my explanation?

You can add the inequalities to get a > 0 because the sign are in the same direction. However, you cannot subtract. You can subtract one inequality from another, when their signs are in the opposite directions.

Check Manipulating Inequalities for more.

Hope it helps.

Thanks, learnt it for the first time
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 07 Aug 2023, 11:42
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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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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 16 Jan 2024, 02:27
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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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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 16 Jan 2024, 14:51
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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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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 17 Jan 2025, 19:34
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