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? Let x represent the number of cases of Brand A, y represent the number of cases of Brand B, z represent the number of cases of Brand C, and w represent be the number of 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 \( (\square, \square, \square, 0) \).

To solve the problem, we need to set up a system of equations based on the information provided about the food brands and the nutritional requirements. Let: - \( x \) = number of cases of Brand A - \( y \) = number of cases of Brand B - \( z \) = number of cases of Brand C - \( w \) = number of cases of Brand D The nutritional content for each brand is as follows: - Brand A: 25 units of fiber, 40 units of protein, 40 units of fat - Brand B: 100 units of fiber, 130 units of protein, 110 units of fat - Brand C: 275 units of fiber, 350 units of protein, 310 units of fat - Brand D: 200 units of fiber, 260 units of protein, 200 units of fat The total nutritional requirements are: - Fiber: 3975 units - Protein: 5100 units - Fat: 4440 units We can set up the following equations based on the nutritional content: 1. For fiber: \[ 25x + 100y + 275z + 200w = 3975 \] 2. For protein: \[ 40x + 130y + 350z + 260w = 5100 \] 3. For fat: \[ 40x + 110y + 310z + 200w = 4440 \] Since we are given that \( w = 0 \), we can substitute \( w \) into the equations: 1. For fiber: \[ 25x + 100y + 275z = 3975 \] 2. For protein: \[ 40x + 130y + 350z = 5100 \] 3. For fat: \[ 40x + 110y + 310z = 4440 \] Now we will solve this system of equations for \( x \), \( y \), and \( z \). Let's solve these equations together. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}25x+100y+275z=3975\\40x+130y+350z=5100\\40x+110y+310z=4440\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=159-4y-11z\\40x+130y+350z=5100\\40x+110y+310z=4440\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}40\left(159-4y-11z\right)+130y+350z=5100\\40\left(159-4y-11z\right)+110y+310z=4440\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}6360-30y-90z=5100\\6360-50y-130z=4440\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=42-3z\\6360-50y-130z=4440\end{array}\right.\) - step5: Substitute the value of \(y:\) \(6360-50\left(42-3z\right)-130z=4440\) - step6: Simplify: \(4260+20z=4440\) - step7: Move the constant to the right side: \(20z=4440-4260\) - step8: Subtract the numbers: \(20z=180\) - step9: Divide both sides: \(\frac{20z}{20}=\frac{180}{20}\) - step10: Divide the numbers: \(z=9\) - step11: Substitute the value of \(z:\) \(y=42-3\times 9\) - step12: Calculate: \(y=15\) - step13: Substitute the values of \(y,z:\) \(x=159-4\times 15-11\times 9\) - step14: Calculate: \(x=0\) - step15: Calculate: \(\left\{ \begin{array}{l}x=0\\y=15\\z=9\end{array}\right.\) - step16: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=15\\z=9\end{array}\right.\) - step17: Rewrite: \(\left(x,y,z\right) = \left(0,15,9\right)\) The solution to the system of equations, given that \( w = 0 \), is: \[ (x, y, z) = (0, 15, 9) \] This means the breeder should mix: - 0 cases of Brand A - 15 cases of Brand B - 9 cases of Brand C - 0 cases of Brand D Thus, the final answer is \( (0, 15, 9, 0) \).