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