To solve this question, we need to find out how many pounds of soybean meal and commeal should be mixed to get a 320-pound mixture that is 9% protein.
### Step-by-Step Solution:
1. Identify the Variables:
- Let [tex]\( x \)[/tex] represent the pounds of soybean meal.
- Let [tex]\( y \)[/tex] represent the pounds of commeal.
2. Set Up the Equations:
- The total weight of the mixture is 320 pounds:
[tex]\[ x + y = 320 \][/tex]
- The total protein from the soybean meal and commeal must be equal to the protein content of a 320-pound mixture that is 9% protein:
[tex]\[ 0.16x + 0.08y = 0.09 \times 320 \][/tex]
3. Calculate the Total Protein Content Required:
- Since 9% of 320 pounds is protein:
[tex]\[ 0.09 \times 320 = 28.8 \text{ pounds of protein} \][/tex]
4. Form the Protein Content Equation:
- Now substitute the protein percentages:
[tex]\[ 0.16x + 0.08y = 28.8 \][/tex]
5. Solve the System of Equations:
- We have the equations:
1. [tex]\( x + y = 320 \)[/tex]
2. [tex]\( 0.16x + 0.08y = 28.8 \)[/tex]
6. Express One Variable in Terms of the Other:
- From the first equation, solve for [tex]\( y \)[/tex]:
[tex]\[ y = 320 - x \][/tex]
7. Substitute [tex]\( y \)[/tex] in the Protein Equation:
- Substitute [tex]\( y = 320 - x \)[/tex] into the second equation:
[tex]\[ 0.16x + 0.08(320 - x) = 28.8 \][/tex]
- Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 0.16x + 25.6 - 0.08x = 28.8 \][/tex]
[tex]\[ 0.08x = 3.2 \][/tex]
[tex]\[ x = 40 \][/tex]
8. Find [tex]\( y \)[/tex]:
- Substitute [tex]\( x = 40 \)[/tex] back into the equation [tex]\( y = 320 - x \)[/tex]:
[tex]\[ y = 320 - 40 = 280 \][/tex]
### Final Solution:
- Pounds of Soybean Meal: [tex]\( \boxed{40} \)[/tex] pounds
- Pounds of Commeal: [tex]\( \boxed{280} \)[/tex] pounds
So, to get a 320-pound mixture that is 9% protein, you need 40 pounds of soybean meal and 280 pounds of commeal.
