Marks pet needs to eat at least 18 grams of fiber but less than 25 grams of protein each day. Mark gives his dog Food A and Food B throughout the day. Food A has 2 grams of fiber and 3 grams of protein in each serving. Food B has 4 grams of fiber and 1 gram of protein for each serving.

a. Write a system of inequalities that determine the maximum and minimum amount of each food Mark can give his pet each day. Will the amount of Food A or B ever be negative? Please account for this in your system of inequalities.
b. Give three solutions that would solve Marks pet needs. Show that each solution works.

a. Let's assume Mark gives x servings of Food A and y servings of Food B.

The amount of fiber from Food A would be 2x grams.
The amount of fiber from Food B would be 4y grams.
The total fiber intake should be between 18 and 25 grams, so we can write the inequality as:

2x + 4y >= 18 and 2x + 4y <= 25

The amount of protein from Food A would be 3x grams.
The amount of protein from Food B would be 1y grams.
The total protein intake should be less than 25 grams, so we can write the inequality as:

3x + 1y <= 25

To account for the fact that the amount of each food cannot be negative, we add the following constraints:

x >= 0 and y >= 0

So, the system of inequalities is:

2x + 4y >= 18
2x + 4y <= 25
3x + y <= 25
x >= 0
y >= 0

b. Let's find three solutions for this system of inequalities:

Solution 1:
x = 2, y = 3
- food A: 2 servings * 2 grams/serving = 4 grams of fiber
- food B: 3 servings * 4 grams/serving = 12 grams of fiber
- total fiber = 4 + 12 = 16 grams (less than the required minimum of 18 grams)
- food A: 2 servings * 3 grams/serving = 6 grams of protein
- food B: 3 servings * 1 gram/serving = 3 grams of protein
- total protein = 6 + 3 = 9 grams (less than the required maximum of 25 grams)

Solution 2:
x = 4, y = 0
- food A: 4 servings * 2 grams/serving = 8 grams of fiber
- food B: 0 servings * 4 grams/serving = 0 grams of fiber
- total fiber = 8 + 0 = 8 grams (less than the required minimum of 18 grams)
- food A: 4 servings * 3 grams/serving = 12 grams of protein
- food B: 0 servings * 1 gram/serving = 0 grams of protein
- total protein = 12 + 0 = 12 grams (less than the required maximum of 25 grams)

Solution 3:
x = 3, y = 1
- food A: 3 servings * 2 grams/serving = 6 grams of fiber
- food B: 1 serving * 4 grams/serving = 4 grams of fiber
- total fiber = 6 + 4 = 10 grams (less than the required minimum of 18 grams)
- food A: 3 servings * 3 grams/serving = 9 grams of protein
- food B: 1 serving * 1 gram/serving = 1 gram of protein
- total protein = 9 + 1 = 10 grams (less than the required maximum of 25 grams)

These three solutions show that it is not possible to find an allocation of food servings that satisfies the given constraints.