Solving Systems Of Equations Percy's Part-Time Job Problem

In the realm of mathematics, we often encounter real-world scenarios that can be elegantly modeled and solved using systems of equations. Let's delve into a practical example involving Percy, a diligent student working two part-time jobs to finance his college education. This article will show you the background, math and solutions for the problem.

Percy's Part-Time Job Dilemma: A Mathematical Modeling Approach

Percy's situation presents a classic problem that can be solved using a system of linear equations. Our main goal here is to determine how much Percy earns per hour at each of his jobs: the library and the coffee cart. This is a common scenario for students and individuals balancing multiple jobs, making it a relatable and valuable problem to solve.

To tackle this, we'll first define our variables. Let 'x' represent Percy's hourly wage at the library and 'y' represent his hourly wage at the coffee cart. These variables are the unknowns we aim to find. Next, we translate the given information into mathematical equations. On Monday, Percy works 3 hours at the library and 2 hours at the coffee cart, earning $36.50. This translates to the equation 3x + 2y = 36.50. Similarly, on Tuesday, he works 2 hours at the library and 5 hours at the coffee cart, earning $50, which gives us the equation 2x + 5y = 50. Now, we have a system of two linear equations with two variables:

3x + 2y = 36.50
2x + 5y = 50

This system of equations represents Percy's earnings and forms the foundation for solving for his hourly wages. Understanding how to set up these equations is crucial for applying this method to other similar problems. The ability to translate real-world scenarios into mathematical models is a key skill in algebra and problem-solving. In the following sections, we will explore different methods to solve this system of equations and find the values of 'x' and 'y', revealing Percy's hourly earnings at each job.

Unveiling Percy's Hourly Wages: Methods to Solve Systems of Equations

Now that we've established the system of equations, we need to find the values of 'x' and 'y'. There are several methods to solve systems of linear equations, each with its own advantages. We'll focus on two common methods: the substitution method and the elimination method. Understanding these methods not only helps in solving this specific problem but also provides valuable tools for tackling a wide range of mathematical challenges. Mastery of these techniques is crucial for success in algebra and beyond. Let's dive into each method and see how they apply to Percy's situation.

Method 1: The Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can be easily solved. Let's apply this to Percy's problem. We can choose either of our equations to start with. Let's take the first equation, 3x + 2y = 36.50, and solve for x. To isolate x, we first subtract 2y from both sides, giving us 3x = 36.50 - 2y. Then, we divide both sides by 3 to get x = (36.50 - 2y) / 3. Now we have an expression for x in terms of y.

Next, we substitute this expression for x into the second equation, 2x + 5y = 50. Replacing x with (36.50 - 2y) / 3, we get 2 * [(36.50 - 2y) / 3] + 5y = 50. This equation now contains only one variable, y. To solve for y, we first simplify the equation. Multiply both sides of the equation by 3 to eliminate the fraction: 2 * (36.50 - 2y) + 15y = 150. Distribute the 2: 73 - 4y + 15y = 150. Combine like terms: 73 + 11y = 150. Subtract 73 from both sides: 11y = 77. Finally, divide by 11 to find y: y = 7. So, Percy earns $7 per hour at the coffee cart. Now that we know y, we can substitute it back into our expression for x: x = (36.50 - 2 * 7) / 3 = (36.50 - 14) / 3 = 22.50 / 3 = 7.50. Therefore, Percy earns $7.50 per hour at the library. The substitution method provides a systematic way to solve for the variables by expressing one in terms of the other.

Method 2: The Elimination Method

The elimination method involves manipulating the equations so that when they are added together, one of the variables is eliminated. This is achieved by multiplying one or both equations by a constant so that the coefficients of one of the variables are opposites. Let's apply this method to Percy's equations:

3x + 2y = 36.50
2x + 5y = 50

To eliminate x, we can multiply the first equation by 2 and the second equation by -3. This will make the coefficients of x opposites: (2 * (3x + 2y)) = (2 * 36.50) gives us 6x + 4y = 73, and (-3 * (2x + 5y)) = (-3 * 50) gives us -6x - 15y = -150. Now we have the system:

6x + 4y = 73
-6x - 15y = -150

Adding these two equations eliminates x: (6x - 6x) + (4y - 15y) = 73 - 150, which simplifies to -11y = -77. Divide both sides by -11 to solve for y: y = 7. Again, we find that Percy earns $7 per hour at the coffee cart. Now, substitute y = 7 into either of the original equations to solve for x. Let's use the first equation: 3x + 2 * 7 = 36.50. Simplify: 3x + 14 = 36.50. Subtract 14 from both sides: 3x = 22.50. Divide by 3: x = 7.50. So, Percy earns $7.50 per hour at the library. The elimination method is particularly useful when the coefficients of one of the variables are easily made opposites, streamlining the solution process.

Verifying the Solution and Real-World Implications

After solving for the variables, it's crucial to verify our solution to ensure accuracy. This step helps prevent errors and reinforces our understanding of the problem. We can verify our solution by plugging the values of x and y back into the original equations. If the equations hold true, then our solution is correct. For Percy's situation, we found x = 7.50 (hourly wage at the library) and y = 7 (hourly wage at the coffee cart).

Let's plug these values into the first equation: 3x + 2y = 36.50. Substituting, we get 3 * 7.50 + 2 * 7 = 22.50 + 14 = 36.50, which is correct. Now let's check the second equation: 2x + 5y = 50. Substituting, we get 2 * 7.50 + 5 * 7 = 15 + 35 = 50, which is also correct. Since both equations hold true, we can confidently say that our solution is accurate. This verification step is a fundamental practice in mathematics and problem-solving, ensuring that the answers we obtain are reliable.

Beyond just finding the numerical solution, it's important to consider the real-world implications of our findings. In Percy's case, understanding his hourly wages allows him to manage his finances effectively. He can now calculate his weekly income, plan his expenses, and ensure he has enough money to cover his college classes. This problem-solving exercise highlights the practical application of systems of equations in everyday life. Whether it's budgeting, calculating costs, or making financial decisions, the ability to solve systems of equations can be a valuable asset.

Conclusion: Mastering Systems of Equations for Real-World Problem Solving

In conclusion, we've successfully tackled a real-world problem using the power of systems of equations. By translating Percy's work schedule and earnings into mathematical equations, we were able to determine his hourly wages at both the library and the coffee cart. We explored two methods for solving systems of equations: the substitution method and the elimination method. Both methods yielded the same result, demonstrating the flexibility and robustness of these techniques. This process reinforces the idea that mathematics is not just an abstract concept but a practical tool for solving everyday problems. Whether it's managing personal finances, optimizing business operations, or making informed decisions, the ability to model and solve systems of equations is a valuable skill.

Furthermore, we emphasized the importance of verifying our solutions to ensure accuracy. This step is crucial in any mathematical problem-solving process, as it helps prevent errors and builds confidence in our results. By plugging our solutions back into the original equations, we confirmed that Percy earns $7.50 per hour at the library and $7 per hour at the coffee cart. This exercise demonstrates the interconnectedness of mathematics and real-world scenarios, highlighting the practical applications of algebraic concepts. As we continue to explore mathematical concepts, it's essential to remember their relevance and utility in solving the challenges we face in our daily lives.