Step 1: Test the corner points of $3x+5y\le15$ in the first quadrant directly against $x+y\ge8$, instead of reasoning about the general region:
The region $3x+5y\le15,\,x\ge0,\,y\ge0$ is a triangle with corners $(0,0)$, $(5,0)$, $(0,3)$.
Step 2: Check $x+y$ at each corner of that triangle:
At $(0,0)$: $x+y=0$. At $(5,0)$: $x+y=5$. At $(0,3)$: $x+y=3$. The maximum of $x+y$ anywhere inside this triangle is at a corner, since $x+y$ is linear, so the largest value achievable in this whole triangle is $5$ (at $(5,0)$).
Step 3: Compare against the requirement $x+y\ge8$:
Since even the best point in the triangle only reaches $x+y=5$, and we need $x+y\ge8$, no point of the triangle can ever satisfy $x+y\ge8$.
Final Answer:
The two constraints never overlap, so the feasible region is empty and no minimum of $z$ exists.
\[ \boxed{\text{No feasible solution exists}} \]