We can use the triangle mid-segment theorem to understand this problem better.
The midsegment of a triangle is parallel to the third side of the triangle and it’s always equal to 1/2 of the length of the third side.
From the image, we can say:
[tex]QS=\frac{1}{2}TU[/tex]Given the information for QS and TU, we can find the value of z:
[tex]\begin{gathered} QS=\frac{1}{2}TU \\ z=\frac{1}{2}(z+2) \\ z=\frac{1}{2}z+\frac{1}{2}(2) \\ z=\frac{1}{2}z+1 \\ z-\frac{1}{2}z=1 \\ \frac{1}{2}z=1 \\ z=\frac{1}{\frac{1}{2}} \\ z=2 \end{gathered}[/tex]Thus, the value of z is
z = 2