Solving a System of Equations for Chicken and Fish Prices
To solve this question, we will set up a system of equations based on the information provided. The photo indicates the stall purchases a combination of chicken and fish across two different days. We'll let \( C \) represent the price per pound of chicken and \( F \) represent the price per pound of fish.
From the photo, on Monday the stall purchases 8 pounds of chicken and 3 pounds of fish for a total of $52. On Tuesday, they purchase 5 pounds of chicken and 5 pounds of fish for a total of $60. Using this information, we can create the following equations:
1. For Monday's purchase:
\( 8C + 3F = 52 \)
2. For Tuesday's purchase:
\( 5C + 5F = 60 \)
Now we can solve this system of equations to find the values of \( C \) and \( F \). One way to solve the system is by using the method of substitution or elimination. I'll use the elimination method here by multiplying the second equation by \( 3/5 \), in order to get the same coefficient for F in both equations, and then subtract the first equation from the new second equation:
First, modify the second equation:
\( (3/5) \times (5C + 5F) = (3/5) \times 60 \)
\( 3C + 3F = 36 \)
Now subtract the first equation from this new equation:
\( (3C + 3F) - (8C + 3F) = 36 - 52 \)
\( 3C - 8C + 3F - 3F = -16 \)
\( -5C = -16 \)
Divide by -5 to find \( C \):
\( C = 16/5 \)
\( C = 3.20 \)
Now that we know the price per pound of chicken, we can substitute this value into one of the original equations to find the price per pound of fish. Let's use the first equation:
\( 8 \times 3.20 + 3F = 52 \)
\( 25.60 + 3F = 52 \)
Subtract 25.60 from both sides to solve for \( F \):
\( 3F = 52 - 25.60 \)
\( 3F = 26.40 \)
Divide by 3 to find \( F \):
\( F = 26.40/3 \)
\( F = 8.80 \)
Therefore, the price per pound for chicken is $3.20 and the price per pound for fish is $8.80.
