If x!=-y, is [x-y]/[x+y] > 1
[1] x > 0
[2] y < 0
Inequalities
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Let's say [x-y]/[x+y] = IKItuz wrote:Can you help me solve this problem?
If x!=-y, is [x-y]/[x+y] > 1
[1] x > 0
[2] y < 0
S1: pick numbers
x = 5, y=-1 (I>1) , y=1 (I<1) ==> insufficient
S2: y = -5, x = -1 (I<1), x=1 (I<1)
y=-1, x=5 (I>1)
S1& S2
x=5 & y = -2 ==> yes
x=5 & y=(-1/2) ==> NO
IMO E
Another approach to combine S1 & S2:
x>0; -y>0 ==> add ==> (x-y)>0
But we cannot say anything about (x+y) ==> IMO E
Note: is it x!= -y? i am afraid, i consider it as x = -y.KItuz wrote:If x!=-y, is [x-y]/[x+y] > 1
[1] x > 0
[2] y < 0
From stat1: x > 0;
As x + y = 0, then y must be < 0. so, say x = 0.4, y = -0.4 then [ x - y ] = 0. and [ x + y] = 0
therefore, 0/0 which is an indeterminate form!
say if x = 2 and y = -2, [x - y ] = 4 and [x + y] = 0 => 4/0 = infinity which is also an indeterminate form!
Hence sufficient.
Similarly, stat2 also sufficient.
Answer D!