To represent the scenario with a system of linear equations, we can analyze the information given for both platters of sandwiches.
For the first platter:
- The cost is $3838.
- It contains 22 turkey sandwiches and 33 roast beef sandwiches.
This gives us the equation: \[ 22x + 33y = 3838 \]
For the second platter:
- The cost is $3737.
- It contains 33 turkey sandwiches and 22 roast beef sandwiches.
This gives us the equation: \[ 33x + 22y = 3737 \]
Combining these, we get the correct system of linear equations:
- \( 22x + 33y = 3838 \)
- \( 33x + 22y = 3737 \)
Thus, the correct answer is:
D. \( 22x + 33y = 3838 \)
\( 33x + 22y = 3737 \)
Part 2: Solving the System
Now, let's solve the system of equations to find the values of \( x \) (cost of each turkey sandwich) and \( y \) (cost of each roast beef sandwich).
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The first equation is: \[ 22x + 33y = 3838 \]
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The second equation is: \[ 33x + 22y = 3737 \]
To solve this system, we can use the method of substitution or elimination. Here, we'll use elimination.
To eliminate \( y \), we can multiply the first equation by 2 and the second by 3:
- \( 2(22x + 33y) = 2(3838) \) → \( 44x + 66y = 7676 \)
- \( 3(33x + 22y) = 3(3737) \) → \( 99x + 66y = 11211 \)
Now, we can subtract the first modified equation from the second: \[ (99x + 66y) - (44x + 66y) = 11211 - 7676 \] This simplifies to: \[ 55x = 3535 \] So: \[ x = \frac{3535}{55} = 64.27 \]
Now that we have \( x \), we can substitute it back into either of the original equations to find \( y \). Let's use the first equation: \[ 22(64.27) + 33y = 3838 \] Calculating \( 22 \times 64.27 \): \[ 1414.94 + 33y = 3838 \] Subtracting \( 1414.94 \) from both sides: \[ 33y = 3838 - 1414.94 = 2423.06 \] Now, divide by 33: \[ y = \frac{2423.06}{33} = 73.36 \]
Summary of Results:
- Cost of each turkey sandwich \( x \): approximately $64.27
- Cost of each roast beef sandwich \( y \): approximately $73.36