To model the problem using a system of equations, we have two types of items (burgers and hot dogs) and two constraints given by the problem (the total number of items and the total cost).
Let:
- \( x \) be the number of burgers
- \( y \) be the number of hot dogs
From the problem, we can form the following equations:
-
The total number of items equation: \[ x + y = 7 \]
-
The total cost equation: \[ 5x + 3y = 27 \]
Now, we have the system of equations: \[ \begin{align*}
- & \quad x + y = 7 \quad \text{(1)} \
- & \quad 5x + 3y = 27 \quad \text{(2)} \end{align*} \]
Solving the System of Equations
We can solve this system using substitution or elimination. Let's use substitution by solving equation (1) for \( y \):
\[ y = 7 - x \]
Now, we can substitute \( y \) in equation (2):
\[ 5x + 3(7 - x) = 27 \]
Expanding and simplifying:
\[ 5x + 21 - 3x = 27 \]
\[ 2x + 21 = 27 \]
Subtracting 21 from both sides:
\[ 2x = 6 \]
Dividing by 2:
\[ x = 3 \]
Now plug \( x \) back into equation (1) to find \( y \):
\[ 3 + y = 7 \]
Subtracting 3 from both sides:
\[ y = 4 \]
Conclusion
The customer purchased:
- 3 burgers
- 4 hot dogs
Verification of the Solution
To verify, we can check both equations with \( x = 3 \) and \( y = 4 \):
-
\( x + y = 7 \): \[ 3 + 4 = 7 \quad \text{(True)} \]
-
\( 5x + 3y = 27 \): \[ 5(3) + 3(4) = 15 + 12 = 27 \quad \text{(True)} \]
Both equations are satisfied, confirming that the solutions are correct.