If ax + b = 0, is x > 0 ?

[geetesht]

If ax + b = 0, is x > 0 ?

(1) a + b > 0

(2) a - b > 0

The given information can be rephrased as ax = -b.

If i add statement 1 & 2, ( since two inequalities can be added) i get, a > 0 i.e a is positive . Hence i can say for the equation ax = -b to be true , x is negative. Since
positive (a) * negative(x) = negative(b)

My ans would be 'C'.

Could anyone please confirm i am thinking on the right track!

[RonPurewal]

If ax + b = 0, is x > 0 ?

(1) a + b > 0

(2) a - b > 0

The given information can be rephrased as ax = -b.

If i add statement 1 & 2, ( since two inequalities can be added) i get, a > 0 i.e a is positive . Hence i can say for the equation ax = -b to be true , x is negative. Since
positive (a) * negative(x) = negative(b)

My ans would be 'C'.

Could anyone please confirm i am thinking on the right track!

nope. incorrect.

the first thing we need to get on the table here is that "-B" is NOT necessarily a negative number. this is a very common misconception - probably the most common misconception in all of algebra, actually - but it's still a misconception.
* if "b" is positive, then "-b" is negative, as expected.
* if "b" is NEGATIVE, then "-b" is POSITIVE.

--

in this problem, we can deduce that "a" is positive (as you found), but we can't figure out anything about "b", which can be positive, negative, or zero. (for instance, if a = 3, then b can be anything between -3 and 3.)

if b is positive, then x must be negative.
if b is zero, then x must be zero.
if b is negative, then x must be positive.

so, insufficient; the answer is (e).

[geetesht]

Ron, Your the man ! The supreme GENIUS of GMAT :) .
Many thanks Coach! Appreciate it.

[Ben Ku]

Glad it helped!

[ishkaran88]

Thankyou.
It really helped me clear a serious flaw in my approach!

[RonPurewal]

Thankyou.
It really helped me clear a serious flaw in my approach!

glad it helped

[SagarK235]

If ax + b = 0, is x > 0 ?

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

- Rephrase the given condition in terms of the desired variable: x = -b/a
- Rephrase question: So x > 0 when a and b have opposite signs, do they have opposite signs?

1) a + b > 0
C1: a +ive, b +ive; x < 0 => no
C2: a +ive, b -ive (where a>b); x > 0 => yes
Insufficient

2) a - b > 0
C1: a +ive, b +ive (where a > b); x < 0 => no
C2 is the same as in 1)
Insufficient

1 +2 insufficient because same two cases were used to prove insufficiency individually
Note: the check for the sign of x was unnecessary in the cases, since we rephrased the question. But added it here anyway...

[Sage Pearce-Higgins]

Using that rephrase is a valid and interesting way to consider this problem. Thanks for your contribution to the discussion.