#### Answer

$x = 6$
$y = 8$

#### Work Step by Step

Theorem 6-6 states that in a quadrilateral that is a parallelogram, its diagonals bisect one another.
We can now deduce that for bisector $\overline{ab}$, $\overline{ab}$ is congruent to $\overline{ab}$. For $\overline{ab}$, $\overline{ab}$ is congruent to $\overline{ab}$. Let's set up the two equations reflecting this information:
$PT = TR$
$QT = TS$
Let's substitute in what we know:
$2x = y + 4$
$x + 2 = y$
We can solve for both $x$ and $y$ by setting the two equations up as a system of equations:
$2x = y + 4$
$x + 2 = y$
Let's get all the variables on one side and the constants on the other:
$2x - y = 4$
$x - y = -2$
Let's modify the second equation so that the $y$ variables are the same but differ only in sign. We can do this by multiplying the second equation by $-1$:
$2x - y = 4$
$-x + y = 2$
Now, we add the two equations together:
$x = 6$
Now that we have the value for $x$, we can plug in this value for $x$ into one of the original equations to find $y$:
$6 - y = -2$
Subtract $6$ from each side of the equation to isolate constants on the right side of the equation:
$-y = -8$
Divide each side by $-1$ to solve for $y$:
$y = 8$