C

X-Y > -2 => x>-2 +Y

X-2y<-6 => x<-6+2Y => -x> 6-2Y

Add them 0 >4-Y => y>4


x>-2 +Y = > X must be positive since Y >4
so C

Tryingmybest wrote:X-Y > -2 => x>-2 +Y

X-2y<-6 => x<-6+2Y => -x> 6-2Y

Add them 0 >4-Y => y>4

This means X> 4 so C

X> 4 , how did you find this ?

Corrected now. But still answer is C .

Tryingmybest wrote:Corrected now. But still answer is C .

Big Brother is watching you.

C


explanation:

for xy>0 you need both x,y with same sign (i.e. both positive or both negative).

(1) x-y > -2 ==> x+2>y
this doesn't lead to x,y with same sign, so A and D are out.

(2) x-2y<-6 ==> x+6<2y
again, doesn't lead to x,y with same sign, so B is out.

need to eliminate C, so let's check both statements together.

start with (2):
x-2y<-6
x+6<2y
x+2+4<2y

use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<2y
2<y

so you have:
y>2

and
x+2>y
x+2>y>2
x+2>2
x>0

so both x, y are positive, thus xy>0 .

dilbert wrote:C

so you have:
y>2


x>0

so both x, y are positive, thus xy>0 .

Dilbert, check your calculations. Be careful with inserting one ineq into another. According your solution

y>2 and x>0 BUT y=3 and x=1 WILL NOT satisfy both of equations.

In case you want to see the problem on X-Y plane

logitech -
thanks for pointing this out!
indeed, I had a mistake in my calculations. here is the corrected one:

use y<x+2 (from 1), and you have:
x+2+4<2y
y+4<x+2+4<2y
y+4<2y
4<y <== this is corrected line, earlier I had mistake here


so you have:
y>4

and
x+2>y
x+2>y>4
x+2>4
x>2


so the correct solution for the two ineq is
x>2
y>4

and answer is C .

we see from the graph logitech attached that the intersection is (2,4), so x>2;y>4 is the area that solves both ineq.

again - thank you logitech for paying attention to my fault.
this time I was lucky to get C with a mistake, but I won't be that lucky in the real test...

dilbert wrote:logitech -
thanks for pointing this out! .

You are welcome