SAT Mathematics Practice Test 5

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