Is X + Y 0

Question

Is X + Y > 0

1.  3^{(x+y)^2} >9^{xy} 
2. y=3

Salsanousi · Answer

Statement 1)

(x+y)^2 > 2xy

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

x^2 + y^2 > 0 x and y could be positive or negative. So x+y > 0 or x+y < 0

Insufficient

Try 2)

Even if I know y = 3 I have no information on x so insufficient.

Combine x^2 + 9 > 0

x^2 > -9 x could be -4 or 4

-4 + 3 < 0
4 + 3 > 0

Insufficient

Answer choice E.

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alldaytestprep · Answer

Start with statement 2

2) Know nothing about x -----> INSUFFICIENT
- eliminate B & D

1) 3^(x+y)^2 > 9^xy
- break bases up into their prime factors: 3^(x+y)^2 > (3^2)^xy ----> 3^(x+y)^2 > 3^2xy
- since bases are the same all we need to look at are the exponents: (x + y)^2 > 2xy
- distribute the left side: x^2 + 2xy + y^2 > 2xy
- x^2 + y^2 > 0 -----> since any number squared will always be positive or 0, all this tells us is that x & y each aren't equal to zero.

Since they can be any combo of positive and negative numbers this is insufficient ----> eliminate A

1 & 2) Since x can still be anything but zero this does not help us much.

Answer: E

SaquibHGMATWhiz · Answer

Solution:      
  Pre Analysis:  
  We are asked if  x+y>0  or not

Statement 1:    3^{(x+y)^2} >9^{xy} 
 ⇒3^{(x+y)^2}>(3^2)^{xy} 
 ⇒3^{(x+y)^2}>3^{2xy} 
  Since the base is the same and not equal to -1, 0 or 1, we can equate the powers 
  (x+y)^2>2xy 
 ⇒x^2+y^2+2xy>2xy 
 ⇒x^2+y^2>0  
 From  x^2+y^2>0 , we cannot be sure if  x+y>0  or not 
 Thus,  statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2:    y=3 
  No information about x 
 Thus,  statement 2 alone is also not sufficient

Combining:   
  From statement 1, we get  x^2+y^2>0  
 From statement 2, we get  y=3  
 Upo combining, we get  x^2+3^2>0  or  x^2>-9  which means  x can be any real number  and  y = 3  
 If  x=-5  and  y=3 , then  x+y=-5+3=-2<0  
 If  x=-1  and  y=3 , then  x+y=-1+3=2>0

Hence the right answer is    Option E

Is X + Y > 0

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Is X + Y > 0

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