#### Answer

$y = 11$
$x = 3$

#### Work Step by Step

Theorem 6-8 states that in a quadrilateral that is a parallelogram, its opposite sides are congruent.
We can now deduce that $\overline{AB}$ is congruent to $\overline{DC}$, and $\overline{AD}$ is congruent to $\overline{BC}$. Let's set up the two equations reflecting this information:
$AB = DC$
$AD = BC$
Let's substitute in what we know:
$2y + 2 = 3y - 9$
$3x + 6 = y + 4$
Let's turn our attention to the first equation because we can solve for $y$ in this equation:
$2y + 2 = 3y - 9$
Subtract $2$ from each side of the equation to isolate constants on one side of the equation:
$2y = 3y - 11$
Subtract $3y$ from each side of the equation to isolate the variable on the left side of the equation:
$-y = -11$
Divide each side of the equation by $-1$ to solve for $x$:
$y = 11$
Now that we have the value for $y$, we can plug in this value for $y$ into our second equation to find $x$:
$3x + 6 = 11 + 4$
Subtract $6$ from each side of the equation to isolate constants on the right side of the equation:
$3x = 11 + 4 - 6$
Simplify the right side of the equation by adding or subtracting from left to right:
$3x = 9$
$x = 3$