2. Suppose that $x$ and $y$ are real numbers that satisfy all the following three conditions: $3 x-2 y=4-p ; 4 x-3 y=2+p ; x>y$. What are the possible values of $p$ ?
(A) $p>-1$
(B) $p1$
(E) $\quad p$ can be any real number
2. Answer: (D)
Solving the simultaneous equations $3 x-2 y=4-p$ and $4 x-3 y=2+p$, we obtain $x=8-5 p, y=10-7 p$. Since $x>y, 8-5 p>10-7 p$, so $p>1$.