Substituting x2 for y in the second equation gives 2(x2)+6=2(x+3). This equation can be solved as follows: 2x2+6=2x+6 (Apply the distributive property.) 2x2+6−2x−6=0 (Subtract 2x and 6 from both sides of the equation.) 2x2−2x=0 (Combine like terms.) 2x(x−1)=0 (Factor both terms on the left side of the equation by 2x.) Thus, x=0 and x=1 are the solutions to the system. Since x>0, only x=1 needs to be considered. The value of y when x=1 is y=x2=12=1. Therefore, the value of xy is (1)(1)=1. Choices B, C, and D are incorrect and likely result from a computational or conceptual error when solving this system of equations.