Verified Answer :
- Define the variables \(h\) (hamburger) and \(d\) (hot dog) to create the system \(2h + 5d = 8.00\) and \(5h + 2d = 9.50\).
- Use the elimination method to isolate \(h\) by multiplying the equations to align the \(d\) coefficients: \(4h + 10d = 16.00\) and \(25h + 10d = 47.50\).
- Subtract the first from the second to find \(21h = 31.50\), which results in \(h = 1.50\).
- The final price of a hamburger is \(\boxed{\$1.50}\).
Explanation
- Analysis of the Problem
We are given two scenarios regarding the cost of hamburgers and hot dogs. Let \(h\) represent the price of a hamburger and \(d\) represent the price of a hot dog. Based on the information provided, we can establish the following system of linear equations:
1) \(2h + 5d = 8.00\)
2) \(5h + 2d = 9.50\)
Our goal is to find the value of \(h\), the price of a single hamburger.
- Eliminating the Variable d
To solve this system, we can use the elimination method to isolate one variable. We want to eliminate \(d\). We multiply equation (1) by \(2\) and equation (2) by \(5\):
Equation (1) multiplied by \(2\): \(4h + 10d = 16.00\)
Equation (2) multiplied by \(5\): \(25h + 10d = 47.50\)
Now, subtract the first modified equation from the second modified equation to eliminate \(d\):
\((25h + 10d) - (4h + 10d) = 47.50 - 16.00\)
\(21h = 31.50\)
- Solving for the Price of a Hamburger
Now, solve the resulting equation for \(h\):
\(h = \frac{31.50}{21} = 1.50\)
Thus, the price of a single hamburger is \(\$1.50\).
- Verification and Conclusion
To verify, we can find the price of a hot dog (\(d\)) using equation (1):
\(2(1.50) + 5d = 8.00\)
\(3.00 + 5d = 8.00\)
\(5d = 5.00\)
\(d = 1.00\)
Checking this in equation (2):
\(5(1.50) + 2(1.00) = 7.50 + 2.00 = 9.50\).
The values are consistent. The price of a hamburger is \(\$1.50\). The final answer is \(\boxed{\$1.50}\).
Examples
This system of equations represents a common real-life situation where you have limited information about individual costs but know the totals for different bundles. For example, if a store sells meal deals, you might know the price of two different combinations of burgers and fries. By setting up a system of linear equations like this, you can easily determine the individual price of each item, helping you make informed decisions when shopping or managing a budget. This method is fundamental for balancing costs in many practical scenarios, from personal finance to inventory management.