Q.Graph the system of linear inequalities $x + y \le 4$, $x \ge 0$, $y \ge 0$, and identify the feasible region.

Combine the half-plane of $x+y\le4$ with the non-negative quadrant to get the feasible region.

The line $x+y=4$ meets the axes at $(4,0)$ and $(0,4)$; testing the origin gives $0\le4$, true. Combined with the first-quadrant restriction, the feasible region is the triangle with vertices $O(0,0)$, $(4,0)$, $(0,4)$.

The line $x+y=4$ has x-intercept $(4,0)$ and y-intercept $(0,4)$; testing the origin in $x+y\le4$ gives $0\le4$, true, so its side is shaded. Restricting to $x\ge0$ and $y\ge0$ keeps only the first-quadrant part of that shaded half-plane. The resulting feasible region is the closed triangle bounded by the y-axis, the x-axis, and the line $x+y=4$, with corner points $O(0,0)$, $(4,0)$, $(0,4)$.

As with any system, shading the union rather than the intersection of the three half-planes is the most common error; a second is forgetting to check that all three corner points genuinely satisfy every inequality in the system, not just the one whose boundary they lie on.