1. **Stating the problem:** We are given a system of inequalities: $$\begin{cases} x + y \geq e \\ x - z \geq a \\ x + y + z \leq e \\ x, y, z \geq 0 \end{cases}$$ We want to understand the feasible region defined by these inequalities. 2. **Understanding the inequalities:** - The first inequality $x + y \geq e$ means the sum of $x$ and $y$ is at least $e$. - The second inequality $x - z \geq a$ means $x$ is at least $a$ more than $z$. - The third inequality $x + y + z \leq e$ means the total sum of $x$, $y$, and $z$ is at most $e$. - The last constraints $x, y, z \geq 0$ restrict variables to non-negative values. 3. **Analyzing the system:** - From $x + y \geq e$ and $x + y + z \leq e$, since $z \geq 0$, the only way for both to hold is if $z = 0$ and $x + y = e$. - From $x - z \geq a$ and $z = 0$, we get $x \geq a$. - Since $x, y \geq 0$ and $x + y = e$, $y = e - x$. 4. **Summary:** The feasible region reduces to points where: $$z = 0, \quad x \geq a, \quad y = e - x, \quad x, y \geq 0$$ 5. **Final feasible set:** $$\{(x, y, z) \mid z=0, a \leq x \leq e, y = e - x, x, y \geq 0\}$$ This is a line segment on the plane $z=0$ between points $(a, e - a, 0)$ and $(e, 0, 0)$, assuming $0 \leq a \leq e$.