Question 1209886
Let's analyze the inequalities to determine the region.

1.  **$y \ge |x|$**

    * This represents the region above the V-shaped graph of $y = |x|$.
    * $y = x$ for $x \ge 0$
    * $y = -x$ for $x < 0$

2.  **$y \le -|2x + 1| + 6$**

    * This represents the region below the inverted V-shaped graph of $y = -|2x + 1| + 6$.
    * To find the vertex of $y = -|2x + 1| + 6$, we set $2x + 1 = 0$, which gives $x = -\frac{1}{2}$.
    * When $x = -\frac{1}{2}$, $y = -|0| + 6 = 6$. So, the vertex is $(-\frac{1}{2}, 6)$.
    * For $2x + 1 \ge 0$, i.e., $x \ge -\frac{1}{2}$, $y = -(2x + 1) + 6 = -2x + 5$.
    * For $2x + 1 < 0$, i.e., $x < -\frac{1}{2}$, $y = -(-2x - 1) + 6 = 2x + 7$.

Now, we need to find the intersection points of the graphs.

**Intersection of $y = |x|$ and $y = -|2x + 1| + 6$:**

* **Case 1: $x \ge 0$ and $x \ge -\frac{1}{2}$ (i.e., $x \ge 0$)**
    * $x = -2x + 5$
    * $3x = 5$
    * $x = \frac{5}{3}$
    * $y = \frac{5}{3}$
    * Intersection point: $(\frac{5}{3}, \frac{5}{3})$

* **Case 2: $x < 0$ and $x \ge -\frac{1}{2}$ (i.e., $-\frac{1}{2} \le x < 0$)**
    * $-x = -2x + 5$
    * $x = 5$ (This is not in the interval, so no intersection)

* **Case 3: $x \ge 0$ and $x < -\frac{1}{2}$ (Impossible)**

* **Case 4: $x < 0$ and $x < -\frac{1}{2}$ (i.e., $x < -\frac{1}{2}$) **
    * $-x = 2x + 7$
    * $-3x = 7$
    * $x = -\frac{7}{3}$
    * $y = \frac{7}{3}$
    * Intersection point: $(-\frac{7}{3}, \frac{7}{3})$

The intersection points are $(\frac{5}{3}, \frac{5}{3})$ and $(-\frac{7}{3}, \frac{7}{3})$.

Now, we need to find the area of the region.

We can split the area into two triangles.

* Triangle 1: Vertices $(-\frac{7}{3}, \frac{7}{3})$, $(-\frac{1}{2}, 6)$, and $(0, 0)$.
* Triangle 2: Vertices $(0, 0)$, $(-\frac{1}{2}, 6)$, and $(\frac{5}{3}, \frac{5}{3})$.

Area of Triangle 1:
Using the determinant formula:
$$ \frac{1}{2} \left| (-\frac{7}{3})(6 - 0) + (-\frac{1}{2})(0 - \frac{7}{3}) + 0(\frac{7}{3} - 6) \right| $$
$$ \frac{1}{2} \left| -14 + \frac{7}{6} \right| = \frac{1}{2} \left| \frac{-84 + 7}{6} \right| = \frac{1}{2} \left| \frac{-77}{6} \right| = \frac{77}{12} $$

Area of Triangle 2:
$$ \frac{1}{2} \left| 0(6 - \frac{5}{3}) + (-\frac{1}{2})(\frac{5}{3} - 0) + (\frac{5}{3})(0 - 6) \right| $$
$$ \frac{1}{2} \left| -\frac{5}{6} - 10 \right| = \frac{1}{2} \left| \frac{-5 - 60}{6} \right| = \frac{1}{2} \left| \frac{-65}{6} \right| = \frac{65}{12} $$

Total area:
$$ \frac{77}{12} + \frac{65}{12} = \frac{142}{12} = \frac{71}{6} $$

Final Answer: The final answer is $\boxed{\frac{71}{6}}$