- wholelottalove
- Newbie | Next Rank: 10 Posts
- Posts: 4
- Joined: Sat Jun 29, 2013 1:40 pm
I'm having difficulty understanding this problem. I understand why 1 and 2 are insufficient on their own. We know that x<0 and y<0. If x<0 and y<0 then I would say that |x|+|y| cannot equal zero. However, the correct answer states that I am to combine both statements and find the actual value of x and y (which both happen to be zero) to solve. Why can't I solve simply by looking at the signs of x and y and inferring that |x| + |y| cannot = 0? Thanks!!!
#1 x + 2|y| = 0
x + 2|y| = 0
2|y| = -x
|y| = -(x/2)
The only way for x to equal |y| is if x is negative. Therefore, we know the sign of x but nothing else.
INSUFFICIENT
#2 y + 2|x| = 0
The same steps taken with #1 can be applied here. To find the sign of y we can isolate |x|
y + 2|x| = 0
2|x| = -y
|x| = -(y/2)
For -(y/2) to be equal to an absolute value, y must be negative.
1+2)
if x<0 and y<0 then |x| + |y| ≠0 (This reasoning however is incorrect) The correct answer is as follows (why is the correct answer lol?)
1+2)
Knowing the signs of x and y, we can set x + 2|y| = 0 and y + 2|x| = 0 equal to one another and solve:
y + 2|x| = 0 = x + 2|y|
y - 2x = 0, x - 2y = 0
y = 2x
x - 2(2x) = 0
-3x = 0
3x=0
x=0
y-2x = 0
y - 2(0) = 0
y=0
so x=0, y=0
SUFFICIENT
(C)
#1 x + 2|y| = 0
x + 2|y| = 0
2|y| = -x
|y| = -(x/2)
The only way for x to equal |y| is if x is negative. Therefore, we know the sign of x but nothing else.
INSUFFICIENT
#2 y + 2|x| = 0
The same steps taken with #1 can be applied here. To find the sign of y we can isolate |x|
y + 2|x| = 0
2|x| = -y
|x| = -(y/2)
For -(y/2) to be equal to an absolute value, y must be negative.
1+2)
if x<0 and y<0 then |x| + |y| ≠0 (This reasoning however is incorrect) The correct answer is as follows (why is the correct answer lol?)
1+2)
Knowing the signs of x and y, we can set x + 2|y| = 0 and y + 2|x| = 0 equal to one another and solve:
y + 2|x| = 0 = x + 2|y|
y - 2x = 0, x - 2y = 0
y = 2x
x - 2(2x) = 0
-3x = 0
3x=0
x=0
y-2x = 0
y - 2(0) = 0
y=0
so x=0, y=0
SUFFICIENT
(C)













