The problem asks to find the area of the region enclosed by the line $4x + 3y = 12$, the x-axis, and the y-axis.

Steps to solve:

1. Find the intercepts of the line:

   • The x-intercept occurs when $y = 0$. Substituting $y = 0$ into the equation $4x + 3(0) = 12$, we get: $4x = 12 \quad \Rightarrow \quad x = 3$ So, the x-intercept is at $(3, 0)$.
   • The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the equation $4(0) + 3y = 12$, we get: $3y = 12 \quad \Rightarrow \quad y = 4$ So, the y-intercept is at $(0, 4)$.

2. Plot the line and identify the enclosed region: The region is a right triangle with vertices at $(0, 0)$, $(3, 0)$, and $(0, 4)$, where the x-intercept and y-intercept form the base and height of the triangle, respectively.

3. Calculate the area of the triangle: The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ Here, the base is 3 (the x-intercept) and the height is 4 (the y-intercept), so: $\text{Area} = \frac{1}{2} \times 3 \times 4 = 6 \, \text{square units}$

Thus, the area of the region enclosed by the line, the x-axis, and the y-axis is 6 square units.

Let me know if you'd like further clarification or details!

Here are 5 questions that expand on this concept:

1. How would the area change if the equation of the line was $2x + 3y = 6$?
2. Can you calculate the slope of the line $4x + 3y = 12$?
3. What would be the area of the enclosed region if the line passed through $(0, 6)$ instead of $(0, 4)$?
4. How can the distance between the x-intercept and y-intercept be calculated?
5. What other methods can be used to find the area of a triangle besides the base-height formula?

Tip: Always remember that the x- and y-intercepts are key in identifying the area of triangles formed with linear equations.