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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Hi Bunuel

Not very sure how you got x>1 region on the numberline.

From stmt 1, I got a-ax>0 or a(1-x)>0 or a(x-1)<0 and on the number line ---(-)----1-----(+)-------- ie x<1.

Pls clarify.

\(a-ax>0\) --> \(a(1-x)>0\)

So, a and 1-x have the same sign:

Both are positive: \(a>0\) and \(1-x>0\), so \(x<1\)

Both are negative: \(a<0\) and \(1-x<0\), so \(x>1\).
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Hi Bunuel

Why are we doing OR in ur statements?

b=-ax. Hence, a-ax>0 then a(1-x)>0. therefore, a>0 or 1-x >0 so 1>x. Thats all..How can we say x<1?

We have that \(a(1-x)>0\), 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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Given ax + b = 0, hence x = -b/a, & is x > 0?

Statement 1:

a + b > 0

a = 1, b = 0, x = 0
a = -1, b = 2, x = 2 > 0

Statement 1 alone is Not Sufficient.

Statement 2:

a - b > 0

a = 1, b = 0, x = 0
a =1, b = -1, x = 1 > 0

Statement 2 alone is Not Sufficient.

Combining both statements,
a = 1, b = 0, x = 0
a = 2, b = -1, x = 1/2

Combing both statements is Not Sufficient.

Answer E.


Thanks,
GyM
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A video explanation can be found here:
https://www.youtube.com/watch?v=cYPgNTx-XK8&t=5s

For DS questions, always ask "What would I need to know in order to answer this question?" In this case, isolate x:

ax + b = 0

ax = -b

x = -b/a

Whether x is positive depends upon the signs of a and b -- The real question is, "What are the signs of a and b?"

(1) either both numbers are positive, or one is positive and one is negative. Insifficient.

(2) same problem. Insufficient.

Together, if you add the two inequalities you'll have 2a > 0, which means a > 0, but that's only half of what we need to know...

Answer E.
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I love this question. Probably, because one can solve it under 1 minute if you are clear on the rules of positive & negative (in inequality).

The question is saying the following: ax + b =0 => b=-ax or => -b/a = x

So the rephrased question is not to check x > 0 but is -b/a > 0. Now this has two implications as follows:-

1. Case I: If a > 0, b < 0 [Note: ',' indicates and]
2. Case II: If a <0, b > 0

So another deep down will give us the following phrasing:- Is Case I or Case II true (If either one of them is true, then you can see we will get our answer to the question)

Now, Statement A: a + b > 0 i.e. we don't the 'sign' of either a or b: both could be positive; one could be large positive, one could be small negative. So clearly, Not Sufficient

Moving on to statement B: a - b > 0 i.e. same as the above statement only with a minor tweak, hence not sufficient

Combining both the statement: Now inequalities can be added in the same direction; therefore, we get the following:-
2a > 0 => a > 0
Do you think this information is enough to answer the question?. The answer is a resounding 'No'. Why is that?
Well, the statement only gives us the sign of a; however, we don't know whether b is either > 0 or less < 0 or even equal to 0.

Kudos, if you liked the solution

Regards,
Abheek

Bunuel
On your solution of (1)+(2) in which you sum (1) and (2) as they have same sign>> (a+b)+(a−b)>0 and get a>0.
Question is whenever I encountered same sign, can I also subtract LHS terms >> (a+b)-(a−b)>0 and get b>0.
I was mistakenly did so with this Q and from (1)+(3); I got both a>0 (combine LHS) and b>0 (Subtract LHS), which results in X<0, if ax + b = 0, corresponding to ans. C

Please correct me on this point so that next time I can apply only the combination of LHS and not subtraction in context of equation with same sign.

Thank you!

Check : Manipulating Inequalities (adding, subtracting, squaring etc.).

For more check Ultimate GMAT Quantitative Megathread

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Target question: Is x > 0

Given: ax + b = 0

Statement 1: a + b > 0
At this point, we have 1 equation, 1 inequality, and THREE variables.
Even if we had 2 EQUATIONS and 3 variables, we probably wouldn't be able to make any conclusions about whether x is positive or negative.
Given this, let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: a - b > 0
This is the same scenario as statement 1, so let's TEST some values
There are several values of a, b and x that satisfy statement 1 (and the given equation ax + b = 0). Here are two:
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: a = 2, b = -1 and x = 0.5. In this case, the answer to the target question is YES, x is positive
Case b: a = 2, b = 1 and x = -0.5. In this case, the answer to the target question is NO, x is NOT positive

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Solution:

The easiest method to solve a DS question depends on the question. For certain questions (especially for statements which are not sufficient to provide an answer), the question can be easily solved by plugging in numbers. Other questions require you to use algebraic manipulations to arrive at the correct answer. In either case, you can use a combination of both methods; i.e. use algebraic manipulations to get some information about the statements and plug in numbers only after that so that you don’t waste time trying unnecessary cases.

Let me show what I mean on one of the examples you provided.

In the first example, assuming that a is non-zero, we can write x = -b/a. This expression tells us that x is -1 times the ratio of b to a; i.e. if b and a have the same sign then x is negative and if b and a have opposite signs, then x is positive. When plugging in numbers for a and b, this information tells me to only look for examples where a and b have the same and opposite signs, so that if I tried a = 1 and b = -1, I won’t waste time trying a = -1 and b = 1 since I know both will produce a positive x.

Statement One Alone:

For an example where a and b have the same sign, we can simply take a = b = 1. In this case, x + 1 = 0 and hence, x = -1 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 2 and b = -1. In this case, 2x - 1 = 0 and hence, x = 1/2 i.e. x > 0 is true.

Statement one alone is not sufficient.

Statement Two Alone:

For an example where a and b have the same sign, we can take a = 2 and b = 1. In this case, 2x + 1 = 0 and hence, x = -1/2 i.e. x > 0 is not true.

For an example where a and b have opposite signs, we can take a = 1 and b = -1. In this case, x - 1 = 0 and hence, x = 1 i.e. x > 0 is true.

Statement two alone is not sufficient.

Statements One and Two Together:

Notice that if a + b > 0 and a - b > 0 are true at the same time, then adding the inequalities side by side, we determine that 2a > 0; i.e. a > 0.

If a = 2 and b = 1, then 2x + 1 = 0 and hence, x = -1/2. Notice that a = 2 and b = 1 satisfy both a + b > 0 and a - b > 0. In this case, x < 0.

If a = 2 and b = -1, then 2x - 1 = 0 and hence, x = 1/2. Notice that a = 2 and b = - 1 satisfy both a + b > 0 and a - b > 0. In this case, x > 0.

Statements one and two together is not sufficient.

Answer: E

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ax + b = 0
ax = -b
x = -b/a

x > 0?
=> -b/a > 0?
=> Do a & b have opposite signs?

1+2)

Add both inequalities
1) a + b > 0
2) a - b > 0 => a > b

2a > 0
a > 0 (3rd inequality)

b could be + or - since a > b.
or
add 3rd and 1st inequality
2a + b > 0 => 2a > - b => 2 > -b/a => Not Sufficient.


ANSWER: E

But , together if you substract the inequalities — you will easily get b>0 ,

a + b -a+b >0 —> 2b > 0 , so b > 0

So itn’t C the Answer ?

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