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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
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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?
Show more

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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Bunuel
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.
Show more

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 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.
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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
Show more

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 22 Feb 2026, 08:57
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But the same way we can also see by subtracting statement 1 - statement 2 and we will get b>0 hence both a and b are positive which according to equation > ax+b=0 means x is negative?
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.
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 22 Feb 2026, 09:14
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nandini14
But the same way we can also see by subtracting statement 1 - statement 2 and we will get b>0 hence both a and b are positive which according to equation > ax+b=0 means x is negative?

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Please study the thread: https://gmatclub.com/forum/if-ax-b-0-is ... l#p3206129
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If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 22 Feb 2026, 20:58
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We are given:
Ax+b=0
Or, ax = -b

Option 1:
A+b> 0
A> -b
A> ax
1>x
Not sufficient

Option 2:
A-b> 0
A > b
- a < -b
- a < ax
-1 <x
Not sufficient

Option 1+2:
-1 <x< 1
Again, not sufficient
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