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Stat1 It's the equation of a line. Apart the line x=0 and y=0, all lines are passing through 2 cadran (a cadran changes the sign of another cadran on either x, y or x and y).
Line : y = x - 1/2
Stat2 x/y > 1
Here too, it's under the line y=x when x is potive and y positive, and it's above the line y=x when x is negative and y is negative. Thus, x and y could represent a point in the cadran I partially, in the cadran III partially.
Both 1 and 2 The line y = x - 1/2 cut since x=1/2 and x increasing represents the solution of the 2 statments.
C (1) 2x - 2y = 1 ---> 2(x-y) = 1 ---> x-y = 1/2 x and y can be +ves or -ves and still sum to .5. Ins
(2) x and y can be neg or pos and be greater than 1 -4/-2 = 2 or 4/2 = 2. Ins
1 and 2 In order for x/y to be greater than 1 amd x - y to equal 1/2 x and y must both be positive.
Here's where I made the mistake. I rearranged statement (2) as x > y.
Now, if you take x = 0 and y = -1/2, the equation from (1) is still satisfied but x and y are not positive. So I chose E. Obviously this is a trap and I fell for it by rearranging (2). When should you rearrange the terms in an inequality and when shouldn't you ('cause sometimes rearranging terms in a inequality does help)? Btw, OA is C.