Step 1: Setting up in matrix/substitution form:
At the binding intersection of both constraints (both active simultaneously), solve the 2x2 system \(x+3y=3\) and \(x+y=2\) using elimination: subtracting gives \(2y=1\), so \(y=1/2\), then \(x=2-1/2=3/2\).
Step 2: Testing feasibility of all axis-intercepts first:
Before assuming the two other corners, verify: \((0,1)\) fails \(x+y\ge2\); \((2,0)\) fails \(x+3y\ge3\) — both excluded, leaving \((0,2)\) and \((3,0)\) as the genuine boundary corners along with the intersection point.
Step 3: Comparing Z values to find the minimum:
\(Z(0,2)=10,\ Z(3/2,1/2)=7,\ Z(3,0)=9\); the smallest of these is \(7\), and since the feasible region only extends outward (unbounded) from these corners with Z's coefficients positive, no interior/outer point can give a smaller Z.
Final Answer:
\[ \boxed{Z_{min}=7 \text{ at } (3/2,\,1/2)} \]