To solve this, let’s analyze the given equation and how it corresponds to each graph.

Problem Setup and Equation:

The cost constraint equation given is: $11x + 6y = 200$ where:

• $x$ is the number of adult tickets, each costing $11.
• $y$ is the number of child tickets, each costing $6.
• The budget for the trip is $200.

Steps:

1. Find the Intercepts:

   • For $x$-intercept (where $y = 0$): $11x = 200 \Rightarrow x = \frac{200}{11} \approx 18.18$
   • For $y$-intercept (where $x = 0$): $6y = 200 \Rightarrow y = \frac{200}{6} \approx 33.33$

   These intercepts indicate that the boundary line should cross close to $x = 18$ and $y = 33$.

2. Analyze the Graphs:

   • The graph representing the equation should show a line passing near $x = 18$ on the x-axis and $y = 33$ on the y-axis.
   • The shaded region should represent the feasible region (all combinations of $x$ and $y$ that stay within the $200 budget).

Solution:

By observing the graphs:

• Graph D shows the line passing through approximately $(18, 0)$ and $(0, 33)$, with a shaded region below the line, indicating all combinations within the budget constraint of $200.

Thus, the correct choice is: Graph D.

Would you like more details on the steps or any further questions?

Follow-Up Questions:

1. How would the graph change if the budget increased?
2. What would the graph look like if the ticket prices were different?
3. How would the equation change if there was an additional fixed fee?
4. What if both ticket prices had discounts? How would that affect the solution?
5. How could we find the maximum number of adults allowed if we brought 10 children?

Tip: To determine feasible regions, always consider intercepts and understand the shading in inequality graphs.