Because PQRS is a parallelogram, you know that \(2x-1=y-1\) and \(3y-11=x+4\). (This is because PT and TR are congruent, and the same goes for ST and TQ.)
Writing the equations in standard form, you have
\(\begin{cases}2x-y=0\\x-3y=-15\end{cases}\)
Solving this system gives \(x=3\) and \(y=6\). So,
\(PT=TR=5\)
\(ST=TQ=7\)
