Let's set up the system of equations based on the information provided.
Let \( x \) be the price per loaf of bread (in silver doubloons) and \( y \) be the price per bushel of apples (in silver doubloons).
From the first stand, we can formulate the equation based on the total cost:
- First stand:
\( 12x + 14y = 180 \) (Equation 1)
From the second stand, we can formulate another equation:
- Second stand:
\( 9x + 16y = 168 \) (Equation 2)
Now we have the system of equations:
\[ \begin{align*} 12x + 14y &= 180 \quad \text{(1)} \ 9x + 16y &= 168 \quad \text{(2)} \end{align*} \]
Solving the System of Equations
We can solve these equations using the substitution or elimination method. Here, we'll use elimination.
First, let's multiply both equations to align the coefficients of \( x \):
Multiply Equation 1 by 3:
\[ \begin{align*} 36x + 42y &= 540 \quad \text{(3)} \end{align*} \]
Multiply Equation 2 by 4:
\[ \begin{align*} 36x + 64y &= 672 \quad \text{(4)} \end{align*} \]
Now we have:
\[ \begin{align*} 36x + 42y &= 540 \quad \text{(3)} \ 36x + 64y &= 672 \quad \text{(4)} \end{align*} \]
Next, subtract Equation (3) from Equation (4):
\[ (36x + 64y) - (36x + 42y) = 672 - 540 \]
This simplifies to:
\[ 22y = 132 \]
Now, divide both sides by 22:
\[ y = 6 \]
Now that we have the value for \( y \), we can substitute \( y \) back into one of the original equations to find \( x \). We'll use Equation (1):
\[ 12x + 14(6) = 180 \]
This simplifies to:
\[ 12x + 84 = 180 \]
Now, subtract 84 from both sides:
\[ 12x = 96 \]
Finally, divide by 12:
\[ x = 8 \]
Solution
The price per loaf of bread \( x \) is 8 silver doubloons, and the price per bushel of apples \( y \) is 6 silver doubloons.
\[ \boxed{x = 8} \quad \text{and} \quad \boxed{y = 6} \]