Answer
$x = 7$
$y = 10$
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 $\overline{PR}$, $\overline{PT}$ is congruent to $\overline{TR}$. For $\overline{QS}$, $\overline{QT}$ is congruent to $\overline{TS}$. Let's set up the two equations reflecting this information:
$PT = TR$
$QT = TS$
Let's substitute in what we know:
$y = x + 3$
$2y = 3x - 1$
We can solve for both $x$ and $y$ by setting the two equations up as a system of equations:
$y = x + 3$
$2y = 3x - 1$
Let's get all the variables on one side and the constants on the other:
$-x + y = 3$
$-3x + 2y = -1$
Let's modify the first equation so that the $y$ variables are the same but differ only in sign. We can do this by multiplying the first equation by $-2$:
$2x - 2y = -6$
$-3x + 2y = -1$
Now, we add the two equations together:
$-x = -7$
Divide each side by $-1$ to solve for $x$:
$x = 7$
Now that we have the value for $x$, we can plug in this value for $x$ into one of the original equations to find $y$:
$y = 7 + 3$
Add to solve for $y$:
$y = 10$
