An animal breeder can buy four types of food for Vietnamese pot-bellied pigs. Each case of Brand A contains 25 units of fiber, 40 units of protein, and 40 units of fat. Each case of Brand B contains 100 units of fiber, 130 units of protein, and 110 units of fat. Each case of Brand C contains 275 units of fiber, 350 units of protein, and 310 units of fat. Each case of Brand D contains 200 units of fiber, 260 units of protein, and 200 units of fat. How many cases of each brand should the breeder mix together to obtain a food that provides 3975 units of fiber, 5100 units of protein, and 4440 units of fat? cases of Brand D. There are four ways in which the breeder can mix brands to obtain a food that provides 3975 units of fiber, 5100 units of protein, and 4440 units of fat. If $$ w=0 $$, the solution is $$ (0,15,9,0) $$. If $$ w=1 $$, the solution is $$ (1,10,10,1) $$. If $$ w=2 $$, the solution is $$ (2,5,11,2) $$. If $$ w=3 $$, the solution is $$ (\square, \square, \square, 3) $$.

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Let x, y, z, w denote the number of cases of Brands A, B, C, and D, respectively. We are given:

Brand A: 25 fiber, 40 protein, 40 fat  
 Brand B: 100 fiber, 130 protein, 110 fat  
 Brand C: 275 fiber, 350 protein, 310 fat  
 Brand D: 200 fiber, 260 protein, 200 fat

The desired totals are 3975 fiber, 5100 protein, and 4440 fat. This gives us the system:

25x + 100y + 275z + 200w = 3975   (1)  
 40x + 130y + 350z + 260w = 5100   (2)  
 40x + 110y + 310z + 200w = 4440   (3)

We are told that one particular solution comes when w = 3, and our goal is to find the corresponding values of x, y, and z.

Step 1. Substitute w = 3 into the equations.

For fiber (1):  
 25x + 100y + 275z + 200(3) = 3975  
 25x + 100y + 275z + 600 = 3975  
 25x + 100y + 275z = 3975 − 600 = 3375  
Divide by 25:  
 x + 4y + 11z = 3375 ÷ 25 = 135   (4)

For protein (2):  
 40x + 130y + 350z + 260(3) = 5100  
 40x + 130y + 350z + 780 = 5100  
 40x + 130y + 350z = 5100 − 780 = 4320   (5)

For fat (3):  
 40x + 110y + 310z + 200(3) = 4440  
 40x + 110y + 310z + 600 = 4440  
 40x + 110y + 310z = 4440 − 600 = 3840   (6)

Step 2. Subtract (6) from (5) to eliminate x:  
 (40x + 130y + 350z) − (40x + 110y + 310z) = 4320 − 3840  
 (130y − 110y) + (350z − 310z) = 480  
 20y + 40z = 480  
Divide by 20:  
 y + 2z = 24   (7)  
So, y = 24 − 2z.

Step 3. Substitute y = 24 − 2z into (4):  
 x + 4(24 − 2z) + 11z = 135  
 x + 96 − 8z + 11z = 135  
 x + 96 + 3z = 135  
Solve for x:  
 x = 135 − 96 − 3z = 39 − 3z   (8)

Step 4. Use (6) to determine z. Substitute x and y from (8) and (7) into (6):  
 40(39 − 3z) + 110(24 − 2z) + 310z = 3840

Calculate term by term:  
 40 × 39 = 1560      40 × (−3z) = −120z  
 110 × 24 = 2640     110 × (−2z) = −220z

So, the equation becomes:  
 1560 − 120z + 2640 − 220z + 310z = 3840  
Combine like terms:  
 (1560 + 2640) + (−120z − 220z + 310z) = 4200 − 30z = 3840  
Now solve:  
 4200 − 30z = 3840  
 4200 − 3840 = 30z  
 360 = 30z  
 z = 360 ÷ 30 = 12

Step 5. Find y and x:  
 From (7): y = 24 − 2(12) = 24 − 24 = 0  
 From (8): x = 39 − 3(12) = 39 − 36 = 3

Thus, when w = 3, we have x = 3, y = 0, z = 12.

The solution is:  
 (Brand A, Brand B, Brand C, Brand D) = (3, 0, 12, 3).

So, the missing numbers are 3 cases of Brand A, 0 of Brand B, and 12 of Brand C when there are 3 cases of Brand D.
When $$ w = 3 $$, the breeder should mix 3 cases of Brand A, 0 cases of Brand B, 12 cases of Brand C, and 3 cases of Brand D to obtain the desired amounts of fiber, protein, and fat.

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