Is x+y>0

So is x > -y


1)x(x+y)>0

x^2 + xy > 0
X^2 > -xy
x > -y

SUFF

2)y(x+y)>0

yx + y^2 >0
yx > -y^2
x > -y

SUFF

I choose D.

if u take x=-5 and y=-1 then it satisfies both eqns x+y<0
x=5 y=1 x+y>0

I think its E..
I dont have the OA myself..

Thanks Dmateer ,
But i see a flaw in this logic . You cannot cancel x from both sides without having an impact on sign of the equation. It would depend from equation to equation , but doing this
[ X^2 > -xy
x > -y
] and maintaining the sign i.e > may not be correct. Actually i feel the answer should be E .
1)x(x+y)>0
Either [ x is > 0 and (x+y) > 0] OR [x < 0 and (x +y) < 0].
if we consider from this the result could be anything.
Insufficient.
2. y(x+y) > 0
Either [ y is > 0 and (x+y) > 0] OR [y < 0 and (x +y) < 0].
Insufficient.

COMBINED
x(x+y) + y(x+y) > 0 i.e (x+y)(x+y) > 0 which again is inconclusive cause (x+y) can be positive or negative.

Any comments pls.

I choose E.

x^2 + xy > 0
X^2 > -xy
x > -y

Dmateer,

I dont think we can divide by x since we dont know the sign of x.If x is negative then the inequality reverses.


Stmt I

x(x+y)>0
x^2+xy>0

x=1 y=1
x+y>0 and x(x+y)>0
yes

x=-2 y=-1

x(x+y) > 0
x+y < 0

INSUFF

Stm II
Similar to above

INSUFF

Stmt I and II together

x(x+y)>0
y(x+y)>0

Add inequalities facing same direction:

x^2+xy+y^2+xy>0

(x+y)^2>0

|x+y|>0

x>-y or

-x-y>0

-x> y

Divide by -1 inequality reverse

x <-y

Dont know for sure anyhting about the signs of x and y still or their magnitudes(values)

Choose E)

Thanks for correcting me!

Thanks for correcting me!

No probs! Keep up the good work (I have seen some excellent solutions from u for various other problems)

Good luck!

NehaBhandari wrote:Is x+y>0

1)x(x+y)>0
2)y(x+y)>0

Let's use number properties.

When the product of two terms is positive, then both are positive or both are negative. Accordingly:

(1) either x and (x+y) are both positive or both negative... insufficient.

(2) either y and (x+y) are both positive or both negative.. insufficient.

Together: x and y could both be positive or both be negative.. insufficient.

Choose (E).

IMO E

dmateer25 wrote:Is x+y>0

So is x > -y


1)x(x+y)>0

x^2 + xy > 0
X^2 > -xy
x > -y

SUFF

2)y(x+y)>0

yx + y^2 >0
yx > -y^2
x > -y

SUFF

I choose D.

This is exactly how GMAT tricks us. By solving the equatins like this, we tend to forget that x^2 could also be -x * -x. Thanks for posting this dmateer