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Is xy > 0 ? (1)x y > -2 (2)x 2y < -6

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Is xy > 0 ? (1)x y > -2 (2)x 2y < -6 [#permalink] New post 23 Feb 2007, 10:05
1. Is xy > 0 ?
(1)x – y > -2
(2)x – 2y < -6
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 [#permalink] New post 23 Feb 2007, 10:42
(C) for me :)

I prefer an approach with XY plan :)

x*y>0

From1
x-y > -2
<=> y < x + 2

This means that the points (x,y), represented by this inequation, are situated "bellow" the line y = x+2 (Fig 1).

We can be:
> partially in the cadran 1 (x>0, y>0)
> smally in the cadran 2 (x<0, y>0)
> partially in the cadran 3 (x<0, y<0)
> wholly in caran 4 (x>0, y<0)

INSUFF.

From2
x – 2y < -6
<=> y > x/2 + 3

This means that the points (x,y), represented by this inequation, are situated "above" the line y = x/2+3 (Fig 2).

We can be:
> partially in the cadran 1 (x>0, y>0)
> partially in the cadran 2 (x<0, y>0)
> partially in the cadran 3 (x<0, y<0)

INSUFF.

Both (1) and (2)
o y > x/2 + 3
o y < x + 2

We are only in cadran 1. (Fig 3)

SUFF.
Attachments

Fig1_y bellow x plus 2.gif
Fig1_y bellow x plus 2.gif [ 2.88 KiB | Viewed 307 times ]

Fig2_y above x on 2 plus 3.gif
Fig2_y above x on 2 plus 3.gif [ 2.63 KiB | Viewed 307 times ]

Fig3_both.gif
Fig3_both.gif [ 2.8 KiB | Viewed 307 times ]

  [#permalink] 23 Feb 2007, 10:42
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Is xy > 0 ? (1)x y > -2 (2)x 2y < -6

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