- Define variables as \(d\) for dimes and \(q\) for quarters to form the system: \(10d + 25q = 6000\) and \(25d + 10q = 8700\).
- Simplify the system to \(2d + 5q = 1200\) and \(5d + 2q = 1740\).
- Use elimination to solve for \(d\) and \(q\), resulting in \(d = 300\) and \(q = 120\).
- Verify that the values satisfy the original conditions to conclude the result \(\boxed{d=300, q=120}\).
Explanation
- Analysis of the Problem
We are given that Travis has a total of \(\$60.00\) in dimes and quarters. Let \(d\) represent the number of dimes and \(q\) represent the number of quarters. Since a dime is worth \(\$0.10\) and a quarter is worth \(\$0.25\), the total value is represented by the equation \(0.10d + 0.25q = 60.00\). We are also told that if the number of dimes and quarters were switched, the total value would be \(\$87.00\), which gives us the second equation: \(0.10q + 0.25d = 87.00\). Our goal is to solve this system of linear equations to find the values of \(d\) and \(q\).
- Simplifying the System of Equations
To make the equations easier to work with, we first eliminate the decimals by multiplying each equation by 100:
1) \(10d + 25q = 6000\)
2) \(25d + 10q = 8700\)
Next, we can simplify these equations by dividing by their greatest common divisors:
1) Dividing by 5: \(2d + 5q = 1200\)
2) Dividing by 5: \(5d + 2q = 1740\)
- Solving the System
We now have the system:
i) \(2d + 5q = 1200\)
ii) \(5d + 2q = 1740\)
To solve this using elimination, we can make the coefficients of \(q\) the same in both equations. Multiply equation (i) by 2 and equation (ii) by 5:
i') \(4d + 10q = 2400\)
ii') \(25d + 10q = 8700\)
Subtract equation (i') from equation (ii'):
\((25d + 10q) - (4d + 10q) = 8700 - 2400\)
\(21d = 6300\)
\(d = 300\)
Now, substitute \(d = 300\) into equation (i):
\(2(300) + 5q = 1200\)
\(600 + 5q = 1200\)
\(5q = 600\)
\(q = 120\)
- Conclusion and Verification
We found that \(d = 300\) and \(q = 120\).
Checking with the original values:
Initial value: \(0.10(300) + 0.25(120) = 30 + 30 = 60\). (Matches \(\$60\))
Switched value: \(0.10(120) + 0.25(300) = 12 + 75 = 87\). (Matches \(\$87\))
Travis has 300 dimes and 120 quarters. The final answer is \(\boxed{d=300, q=120}\).
Examples
This type of algebraic system is frequently used in commerce and inventory management, such as balancing cash registers or reconciling accounts. For example, if a cashier knows the total value of the money in the register and the total count of certain coin types, they can use systems of linear equations to determine the exact number of each coin denomination present, ensuring that transactions have been processed correctly. Learning to solve these systems provides a fundamental tool for managing financial accuracy in various professional settings.