Q.Find and shade the feasible region of the system $x+y\le 4,\ x\ge 0,\ y\ge 0$.

Shade each half-plane; the feasible region is their common overlap in the first quadrant.

Non-negativity. $x\ge0$ is the region right of the $y$-axis and $y\ge0$ is the region above the $x$-axis; together they restrict everything to the first quadrant (axes included).

The main constraint. The boundary of $x+y\le4$ is the line $x+y=4$, with intercepts $(4,0)$ and $(0,4)$; it is solid (weak $\le$). Test the origin: $0+0=0\le4$ true, so shade the side containing the origin.

Feasible region. The points satisfying all three inequations form the overlap: the triangle bounded by the two axes and the line $x+y=4$.

Corner points. Solve the boundaries in pairs: $x$-axis $\cap$ $y$-axis $=(0,0)$; line $\cap$ $x$-axis ($y=0\Rightarrow x=4$) $=(4,0)$; line $\cap$ $y$-axis ($x=0\Rightarrow y=4$) $=(0,4)$. So the corners are $(0,0),(4,0),(0,4)$. …

Ignoring $x\ge0,y\ge0$ and shading the whole strip $x+y\le4$ (which extends into other quadrants). The non-negativity c …