Solving Systems Of Equations A Coin Problem With Quarters And Dimes
This article delves into a classic mathematical problem involving a collection of coins, specifically quarters and dimes. We will explore how to set up and solve a system of equations to determine the quantity of each type of coin. This type of problem is not only a great exercise in algebra but also has practical applications in everyday scenarios involving money and finances. Let's dive into the fascinating world of coin combinations and mathematical problem-solving.
Setting Up the System of Equations
In this coin problem, understanding the fundamental concepts of system of equations is very important. The core of solving problems like this lies in translating the given information into mathematical equations. Kat has a total of 19 coins, which are a mix of quarters and dimes. We can represent the number of quarters with the variable q and the number of dimes with the variable d. The first piece of information tells us that the sum of the quarters and dimes is 19. This translates directly into our first equation:
q + d = 19
This equation represents the total number of coins. Now, we need to consider the value of the coins. Each quarter is worth $0.25, and each dime is worth $0.10. Kat's coins have a total value of $4. This information leads to our second equation:
0.25q + 0.10d = 4
This equation represents the total value of the coins. We now have a system of two equations with two variables:
q + d = 19
0. 25q + 0.10d = 4
This system of equations is the key to solving the problem. It encapsulates all the information provided and allows us to use algebraic techniques to find the values of q and d. The next step involves choosing a method to solve this system, such as substitution or elimination. Understanding how these equations represent the real-world scenario is crucial for setting up similar problems in the future. This initial setup is often the most challenging part, but with practice, it becomes more intuitive. Remember, the goal is to translate the words into mathematical expressions that accurately reflect the given information. The beauty of algebra lies in its ability to represent complex situations in a concise and manageable form.
Solving the System of Equations: Elimination Method
Applying the elimination method is a strategic approach to solve systems of equations, particularly effective when variables have coefficients that are easily manipulated. In our case, the system of equations is:
q + d = 19
0. 25q + 0.10d = 4
To use the elimination method, we aim to eliminate one variable by making its coefficients opposites in the two equations. Let's eliminate d. To do this, we can multiply the first equation by -0.10:
-0.10(q + d) = -0.10(19)
-0.10q - 0.10d = -1.9
Now we have the modified system:
-0.10q - 0.10d = -1.9
0. 25q + 0.10d = 4
Notice that the coefficients of d are now opposites (-0.10 and 0.10). We can add the two equations together:
(-0.10q - 0.10d) + (0.25q + 0.10d) = -1.9 + 4
Simplifying, we get:
0. 15q = 2.1
Now, we can solve for q by dividing both sides by 0.15:
q = 2.1 / 0.15
q = 14
So, Kat has 14 quarters. Now that we have the value of q, we can substitute it back into one of the original equations to find d. Let's use the first equation:
q + d = 19
14 + d = 19
Subtracting 14 from both sides, we get:
d = 19 - 14
d = 5
Therefore, Kat has 5 dimes. We have successfully used the elimination method to find the number of quarters and dimes. This method demonstrates the power of algebraic manipulation in solving real-world problems. By strategically eliminating one variable, we simplified the system and made it easy to find the solution. The elimination method is a valuable tool in any mathematician's or problem-solver's arsenal.
Solving the System of Equations: Substitution Method
Another powerful technique for solving systems of equations is the substitution method. This method involves solving one equation for one variable and then substituting that expression into the other equation. Let's revisit our system:
q + d = 19
0. 25q + 0.10d = 4
We can easily solve the first equation for one of the variables. Let's solve for q:
q = 19 - d
Now, we substitute this expression for q into the second equation:
0. 25(19 - d) + 0.10d = 4
Distribute the 0.25:
4. 75 - 0.25d + 0.10d = 4
Combine like terms:
-0. 15d = -0.75
Divide both sides by -0.15:
d = -0.75 / -0.15
d = 5
So, we find that Kat has 5 dimes. Now, we can substitute this value of d back into the equation we derived earlier:
q = 19 - d
q = 19 - 5
q = 14
Thus, Kat has 14 quarters. The substitution method provides an alternative approach to solving systems of equations. It is particularly useful when one equation can be easily solved for one variable. By substituting the expression into the other equation, we reduce the problem to a single equation with one variable, which is much easier to solve. This method highlights the flexibility of algebraic techniques and the importance of choosing the most efficient method for a given problem. Both the elimination and substitution methods are valuable tools, and understanding when to use each one can significantly improve your problem-solving skills. The substitution method, like the elimination method, is a cornerstone of algebraic problem-solving and essential for tackling various mathematical challenges.
Verification and Conclusion
Verifying the solution is a critical step in any mathematical problem-solving process. It ensures that the values we've found for the variables satisfy all the conditions of the original problem. In our case, we found that Kat has 14 quarters and 5 dimes. To verify this solution, we need to check if these values satisfy both equations in our system:
q + d = 19
0. 25q + 0.10d = 4
Let's substitute q = 14 and d = 5 into the first equation:
14 + 5 = 19
This equation holds true. Now, let's substitute the values into the second equation:
0. 25(14) + 0.10(5) = 4
3. 50 + 0.50 = 4
4. 00 = 4
This equation also holds true. Since our values for q and d satisfy both equations, we can confidently conclude that our solution is correct. Kat has 14 quarters and 5 dimes. This verification step is not just a formality; it's a crucial check against potential errors in our calculations or reasoning. It reinforces the accuracy of our solution and provides peace of mind. Moreover, the process of verification deepens our understanding of the problem and the relationships between the variables. It's a valuable habit to develop in any mathematical endeavor. In conclusion, we have successfully solved the coin problem using both the elimination and substitution methods, and we have verified our solution to ensure its accuracy. This exercise demonstrates the power of systems of equations in modeling and solving real-world problems. The ability to translate word problems into mathematical expressions and apply appropriate algebraic techniques is a fundamental skill in mathematics and has wide-ranging applications in various fields.
In this article, we explored a classic coin problem, demonstrating how to set up and solve a system of equations. We utilized both the elimination and substitution methods to find the solution and emphasized the importance of verifying the results. This type of problem-solving is a valuable skill in mathematics and beyond, offering a practical application of algebraic concepts.
Systems of equations, coin problems, elimination method, substitution method, algebra, mathematical problem-solving, quarters, dimes, verification, algebraic techniques.
