Q.Solve graphically: Maximize $Z = 2x + 3y$ subject to $x + 2y \le 10,\ 2x + y \le 10,\ x + y \le 6,\ x, y \ge 0$.

Constraints: $x+2y\le10,\ 2x+y\le10,\ x+y\le6,\ x,y\ge0$.

At $y=0$: the three lines give $x\le10,\ x\le5,\ x\le6$; binding is $x\le5$, vertex $(5,0)$.

At $x=0$: the three lines give $y\le5,\ y\le10,\ y\le6$; binding is $y\le5$, vertex $(0,5)$.

$(x+2y=10)\cap(2x+y=10)$: solving gives $\left(\dfrac{10}{3},\dfrac{10}{3}\right)$; checking $x+y\le6$: $\dfrac{20}{3}\approx6.67>6$ — infeasible, discarded.

$(x+2y=10)\cap(x+y=6)$: subtracting, $y=4,\ x=2$, giving $(2,4)$; checking $2x+y\le10$: $4+4=8\le10$ ✓ feasible. …

With three non-trivial constraints, compute all three pairwise line intersections, keep only those that also satisfy the remaining constraint, order …

Keeping the $(x+2y=10)\cap(2x+y=10)$ intersection as a vertex without checking it against $x+y\le6$, which would wrongly add an infeasi …