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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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10 Apr 2015, 04:19



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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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12 Apr 2015, 06:15
I am getting 2 different equations here.. Please help where I am going wrong..
ax + b = 0
So X=b/a
1) a+b>0
So, a>b for a>0
1>b/a 1>x or a>b for a<0 1<x x<1
2) ab>0
a>b for a>0 1<b/a x<1
a>b for a<0 1<b/a 1<x



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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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27 Apr 2015, 23:38
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Dear sheolokeshThe mistake you made was at the part highlighted in red: sheolokesh wrote: 1) a+b>0
So, a>b for a>0
1>b/a 1>x
or a>b for a<0 1<x x<1
Correct processing of the case a < 0 would be as follow: From Statement 1, a + b > 0 > a >  b . . . (1) Case: a < 0Dividing both sides of an inequality with a negative number changes the sign of inequality.So, dividing both sides of Inequality 1 with a, we get: 1 < \(\frac{b}{a}\) Substituting b/a = x, we get: 1 < x That is, x > 1 So, from Statement 1, we see that If a > 0, then x < 1 And, if a < 0, then x > 1 So, we cannot say for sure if x is positive or not. Similarly, in your analysis of Statement 2, you got confused between the impact of a being positive or negative on the sign of inequality: sheolokesh wrote: 2) ab>0
a>b for a>0 1<b/a Correct expression: 1 > b/a x<1 Correct expression: 1 > x Correct expression: 1 < x, that is, x > 1
a>b for a<0 Correct expression: a > b 1<b/a Correct expression: 1 < b/a (multiplying both sides by negative number changes sign) 1<x Correct expression: 1 < x Correct expression: 1 > x, that is, x < 1 (multiplying both sides by negative number changes sign)
It is okay to make mistakes as long as we learn from them. And, the important takeaway from our discussion of this mistake is: Don't skip steps when multiplying or dividing terms on both sides of an inequality. Because, this step is particularly prone to errors.
Hope our discussion was helpful! Best Regards Japinder
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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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?



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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, 02:34
sinhap07 wrote: 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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. Hope it's clear.
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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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



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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 E (this was my first post, hello GMATClub )



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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 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



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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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28 Mar 2017, 10:07
from the beginning we can make up an equation to x=b/a 1)a+b>0 or aax>0. a(1x)>0 not enough data 2) ab>0 or aax>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: 1x>0, x<1 1+x>0, x>1 this gives us no answer if x is positive or negative
Answer is E



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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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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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 1x<0



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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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01 Apr 2017, 10:05
jamescath wrote: 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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 1x<0 We have \(a(1x)>0\). For a product of two numbers to be positive they must have the same sign. Two scenarios: 1. Both are positive: \(a>0\) and \(1x>0\). \(1x>0\) is the same as \(x<1\). 2. Both are negative: \(a<0\) and \(1x<0\). \(1x<0\) is the same as \(x>1\).
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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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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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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



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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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17 Jun 2017, 04:04
KARISHMA315 wrote: 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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.
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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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31 Aug 2017, 23:34
Hey BunuelI 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 aax>0 a(1x)>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: ab>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 Answer E
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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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25 Oct 2017, 10:11
Hi Bunuel Not very sure how you got x>1 region on the numberline. From stmt 1, I got aax>0 or a(1x)>0 or a(x1)<0 and on the number line ()1(+) ie x<1. Pls clarify. 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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.



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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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25 Oct 2017, 11:01
sinhap07 wrote: Hi Bunuel Not very sure how you got x>1 region on the numberline. From stmt 1, I got aax>0 or a(1x)>0 or a(x1)<0 and on the number line ()1(+) ie x<1. Pls clarify. 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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. \(aax>0\) > \(a(1x)>0\) So, a and 1x have the same sign: Both are positive: \(a>0\) and \(1x>0\), so \(x<1\) Both are negative: \(a<0\) and \(1x<0\), so \(x>1\).
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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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26 Oct 2017, 05:11
uzzy12 wrote: If ax + b = 0, is x > 0
(1) a + b > 0 (2) a  b > 0 (1) a + b > 0Let a = 2 & b = 1.........2+1 > 0...........2x+1=0....x =1/2 < 0............Answer is NO Let a = 2 & b =1.........2 1 > 0...........2x  1=0....x =1/2 > 0............Answer is Yes Insufficient (2) a  b > 0 Let a = 2 & b = 1.........2  1 > 0...........2x+1=0....x =1/2 < 0............Answer is NO Let a = 2 & b =1.........2 + 1 > 0...........2x  1=0....x =1/2 > 0............Answer is Yes Insufficient Combine 1 &2 Use same points above........No clear answer Insufficient Answer: E



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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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15 Feb 2018, 09:17
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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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. Hi Bunuel Why are we doing OR in ur statements? b=ax. Hence, aax>0 then a(1x)>0. therefore, a>0 or 1x >0 so 1>x. Thats all..How can we say x<1?



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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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15 Feb 2018, 09:21
zanaik89 wrote: 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) \(a+b>0\) > \(aax>0\) > \(a(1x)>0\) > either \(a>0\) and \(1x>0\), so \(x<1\) OR \(a<0\) and \(1x<0\), so \(x>1\). Not sufficient. (2) \(ab>0\) > \(a+ax>0\) > \(a(1+x)>0\) > 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) > \((a+b)+(ab)>0\) > \(a>0\), 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. Hi Bunuel Why are we doing OR in ur statements? b=ax. Hence, aax>0 then a(1x)>0. therefore, a>0 or 1x >0 so 1>x. Thats all..How can we say x<1? We have that \(a(1x)>0\), so the product of two multiples a and 1  x is positive. This to be true either both must be positive OR both must be negative.
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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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21 May 2018, 00:59
x = b/a
x will be greater than 0 only if a and b are of opposite signs.
(1) a + b > 0 Both a and b can be positive, can be negative, or one positive other negative.
(2) a > b Again, both a and b can be positive, can be negative, or one positive other negative.
Combining,
Adding the two inequalities, we get a > 0
But if a > 0, then b can still be positive or negative. For example, (a,b) can be (2,1) or (2,1). So, still not sufficient. Hence, E.




Re: If ax + b = 0, is x > 0 (1) a + b > 0 (2) a  b > 0
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