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(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.

x=-b/a; x>0 means a and b have to be different sign. I am usually not good at picking numbers, but in this case if you see a+b>0 and a-b>0 it is clear that 5+2>0 5-2>0 and 5+(-2)>0 5-(-2)>0 hence we cannot say that a and b will have the same sign. Hence E.
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(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.

you've made things easier bunuel. +1
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Consider giving me kudos if you find my explanations helpful so i can learn how to express ideas to people more understandable.

awesome explanations both bunuel and mainhoon --these two approaches would probably serve two different types of students really well. nice work!

there's also a potential blend of the two-- mainhoon's rephrase (are the signs of a and b different) and the stacking/adding of inequalities when evaluating choice c for that same rephrased question (if your head gets turned around when too many numbers are involved)
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(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.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

(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.

Great solution sir, but can you correct me?? I took option a) because to satisfy a+b>0. one of them ,a or b, must be negative. So, if a is positive b is negative. x should be positive. On the contrary, if b is positive, a must be negative. So the x should be positive to keep the negative sign of a alive. Thank you. Waiting for your assistance

Why does \(a+b>0\) imply that either a or b is negative? Why cannot both be positive?
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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 May 2014, 09:22

I don't understand fossey or bunuel's solution. Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < . Can Somebody please explain this

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

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13 May 2014, 01:16

nivi123 wrote:

I don't understand fossey or bunuel's solution. Bunuel's soln- I'm kinda confused on how you are marking as both either > and > or < and < . Can Somebody please explain this

Hi,

What is the product when 2 positive nos a,b are multiplied. The product is >0 When a positive number and a negative number is multipled the product is <0 When 2 negative nos are multiplied then the product is >0

Now in the given inequality arrived by simplifying the given expression we get that a*(1-x) >0 now this is possible only if either a>0 and (1-x)>0 or a <0 and (1-x) <0

Hope it helps
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“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

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30 Nov 2014, 21:24

we have a+b>0 a-b>0 if we add both the equations. a>0 if we subtract both the equations, b>0 then we know both have same signs.... hence can determine, sign of x=-b/a is >0 or not What is wrong in my explanation? some please help... Bunuel

we have a+b>0 a-b>0 if we add both the equations. a>0 if we subtract both the equations, b>0 then we know both have same signs.... hence can determine, sign of x=-b/a is >0 or not What is wrong in my explanation? some please help... Bunuel

For two inequalities, you can only apply subtraction when their signs are in the opposite directions: If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Choose two numbers: a=3, b=2. a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.

To get insufficiency you should get both a NO and an YES answers to the question. You have a NO answer, so to get that the statements are insufficient, you should choose numbers which give an YES answer to the question.
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The 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 < 0

Dividing 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) a-b>0

a>b---- for a>0 1<b/a Correct expression: 1 > b/a x<-1Correct 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<xCorrect 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.

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