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One way to solve this quickly by testing values
A: x = 1.1 and y = 2 yes satisfies, hence Eliminate A
B: y = -0.9, x = -1, yes satisfies, Eliminate B
C: y = -0.9, RHS will be -2.7+2 i.e. -0.7 now x has to be less than this, but if you take x = -0.6 this again satisfies ---> Eliminate
D: Remember if |a| < = - a ---> this can only be true if a = 0 or a is negetive
Hence this implies that x - 2 < = 0 i.e x < = 2 so yes this works cause if x > 2 and then y is also positive then overall you wont get a negative value that the question requires. Correct
E: Square both sides you get y^2 < = y^2 + 2y + 1 hence you ll get y >= -1/2 but we saw that y = -0.9 satisfies hence Eliminate
Answer is D
A: x = 1.1 and y = 2 yes satisfies, hence Eliminate A
B: y = -0.9, x = -1, yes satisfies, Eliminate B
C: y = -0.9, RHS will be -2.7+2 i.e. -0.7 now x has to be less than this, but if you take x = -0.6 this again satisfies ---> Eliminate
D: Remember if |a| < = - a ---> this can only be true if a = 0 or a is negetive
Hence this implies that x - 2 < = 0 i.e x < = 2 so yes this works cause if x > 2 and then y is also positive then overall you wont get a negative value that the question requires. Correct
E: Square both sides you get y^2 < = y^2 + 2y + 1 hence you ll get y >= -1/2 but we saw that y = -0.9 satisfies hence Eliminate
Answer is D
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