The local florists never had more than a combined total of 40 daisy bouquets and carnation bouquets, and she never has more than 12 carnation bouquets. Write a system of inequalities that represents this situation.

Let's define two variables to represent the number of bouquets:

- \( x \) = number of daisy bouquets
- \( y \) = number of carnation bouquets

From the problem, we are given the following information:

1. The combined total of daisy and carnation bouquets never exceeds 40.
   \[
   x + y \leq 40
   \]
   
2. The number of carnation bouquets is never more than 12.
   \[
   y \leq 12
   \]
   
Additionally, since the number of bouquets cannot be negative, we include the non-negativity constraints:
   \[
   x \geq 0 \quad \text{and} \quad y \geq 0
   \]

Thus, the system of inequalities is:

\[
\begin{aligned}
x + y &\leq 40 \\
y &\leq 12 \\
x &\geq 0 \\
y &\geq 0
\end{aligned}
\]

Would you like further details or have any questions?

Here are five related questions:

1. How can you graph the system of inequalities?
2. What would happen if the maximum number of carnation bouquets increased to 15?
3. If the florist had 25 daisy bouquets, how many carnation bouquets could she have?
4. Can this system be solved using a different method like substitution or elimination?
5. What happens if the total number of bouquets cannot exceed 30?

**Tip:** When graphing inequalities, always pay attention to boundary lines and shading to represent feasible regions!

Learn how to write a system of inequalities for a florist's inventory problem involving daisy and carnation bouquets. Understand key algebraic concepts such as inequalities, linear systems, and non-negativity constraints. Suitable for students in Grades 7-10.

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