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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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23 Jul 2015, 00:19

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?

(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.

Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?

No.

1. You cannot write x = -b/a from ax + b = 0, because a can be 0 and division by 0 is not allowed. 2. Even if we had x = -b/a, minus sign does not indicate that x is negative. If b/a is negative itself then x = -b/a = -(negative) = positive.

If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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14 Aug 2015, 15:20

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

Thanks for your answer Bunuel, great as always. My approach was:

ax + b = 0 implies that either: ax = negative and b is positive {a = positive, x = negative or x = positive, a = negative} or ax = positive and b is negative {a = negative, x = negative or x = positive, a = positive} or ax = 0 and b = 0

1) a + b > 0 Insufficient because no conclusion can be reached on sign of a or b

2) a - b > 0 which means that a > b if a > b, then: b must be negative or zero. this means that we cannot reach conclusion on x. if b is negative, then ax = positive. This means that a = negative, x = negative or a = positive, x = positive if b is zero, then ax = 0, which means that x = 0 as a > b and cannot be 0. x still has wide range of possibilities.

1 + 2, a does not help in answering whether b must be negative or zero or tell us what sign a should be. Thus, question raised in 2 is still not solved. Therefore E

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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02 Sep 2015, 06:59

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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02 Sep 2015, 07:00

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

Show Tags

02 Sep 2015, 07:01

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

(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.

from 1+2, why can not we (a+b)-(a-b)>0 --> b>0 ? what is wrong in my thinking

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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23 Sep 2016, 10:27

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Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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23 Dec 2016, 03:11

Another way:

we know that x = - b/a

Is x>0?

for x to be positive we must know signs of b and a: if b/a > 0 (i.e. they have the same sign) then x<0 if b/a < 0 (i.e. one pos one neg) then x>0 In both cases we'd be able to answer.

1. a + b > 0. Clearly NS because they can be both positive, both negative or pos/neg 2. a -b > 0. Clearly NS because they can be both positive, both negative or pos/neg

1.+2. We only know that their sum is positive and one is bigger than the other: they can be both positive, both negative or pos/neg

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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23 Feb 2017, 05:59

If ax + b = 0, is x > 0

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

Prompt analysis ax+b=0 or x =-b/a

Super set The answer will be either yes or no.

Translation In order to find the exact answer, we need: 1# exact value of a and b. 2# any equation to find the a and b 3# any equation to find the exact or the range for -b/a

Statement analysis St 1: a +b >0 or -b/a<1 or x <1, therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT St 2: a -b >0 or b/a<1 or -x<1 or x>-1.therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT

St 1 & st2: -1 < x < 1. therefore x can be x can be greater than 0 or less than 0. INSUFFICIENT Option E

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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28 Mar 2017, 10:07

from the beginning we can make up an equation to x=-b/a 1)a+b>0 or a-ax>0. a(1-x)>0 not enough data 2) a-b>0 or a-ax>0. a(1+x)>0 not enough data 1+2) let's sum the equation, because they have the same signs. a>0, from this point we can try the other numbers: 1-x>0, x<1 1+x>0, x>-1 this gives us no answer if x is positive or negative

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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01 Apr 2017, 09:50

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

Bunuel 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

(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.

Bunuel 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

We have \(a(1-x)>0\). For a product of two numbers to be positive they must have the same sign. Two scenarios:

1. Both are positive: \(a>0\) and \(1-x>0\). \(1-x>0\) is the same as \(x<1\).

2. Both are negative: \(a<0\) and \(1-x<0\). \(1-x<0\) is the same as \(x>1\).
_________________

Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a - b > 0 [#permalink]

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17 Jun 2017, 03:03

Bunuel wrote:

uzzy12 wrote:

If ax + b = 0, is x > 0

(1) a + b > 0 (2) a - b > 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) --> \((a+b)+(a-b)>0\) --> \(a>0\), 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.

Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic

(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.

Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic

The highlighted part is not correct.

From \(b=-ax\) knowing that \(a>0\), does not necessarily mean that x is negative. Consider x = -1/2 and x = 1/2. Notice that we know nothing about b, we don't know whether it's positive or negative.
_________________

I just wanted to check if my approach is correct. Can you let me know? Below is how I solved it

Given:-

Statement 1:- a+b>0 a-ax>0 a(1-x)>0 i. Either a>0 , then x can be >0 or <0 ii. OR x<1 , x can be <0 or >0 iii. OR a>0 and x<1, we still can't say anything about about x

Insufficient

Statement 2:- a-b>0 a+ax>0 a(1+x)>0 i. Either a>0 , then x can be >0 or <0 ii. OR x>-1 , x can be <0 or >0 iii. OR a>0 and x>-1, we still can't say anything about about x

If I combine both statements, I can't conclude anything about x

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