Problem 62 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system for \(x, y,\) and \(z\). A recent college graduate took advantage of his business education and invested in three investments immediately after graduating. He invested $$\$ 80,500$$ into three accounts, one that paid \(4 \%\) simple interest, one that paid \(3 \frac{1}{8} \% \quad\) simple interest, and one that paid \(2 \frac{1}{2} \%\) simple interest. He earned $$\$ 2,670$$ interest at the end of one year. If the amount of the money invested in the second account was four times the amount invested in the third account, how much was invested in each account?

Step-by-Step Solution

Answer

$25,500 in the 4% account, $44,000 in the 3.125% account, $11,000 in the 2.5% account.

Step 1: Define Variables

Let \( x \) be the amount invested at \(4\%\), \( y \) be the amount invested at \(3\frac{1}{8}\% \), and \( z \) be the amount invested at \(2\frac{1}{2}\% \). Then, we need to find values for \( x, y, \) and \( z \).

Step 2: Set Up Equations

We have three main pieces of information. First, the total amount invested is \(80,500. Therefore,\[ x + y + z = 80,500 \]Second, the total interest earned is \)2,670 at the end of one year, so:\[ 0.04x + 0.03125y + 0.025z = 2,670 \]Lastly, the amount in the second account is four times that in the third account, so:\[ y = 4z \]

Step 3: Substitute and Simplify

Substitute \( y = 4z \) into the first and second equations:1. \( x + 4z + z = 80,500 \) simplifies to \( x + 5z = 80,500 \)2. \( 0.04x + 0.03125(4z) + 0.025z = 2,670 \) simplifies to \( 0.04x + 0.125z + 0.025z = 2,670 \), or \( 0.04x + 0.15z = 2,670 \)

Step 4: Solve the System

First, solve for \( x \) from \( x + 5z = 80,500 \):\[ x = 80,500 - 5z \]Substitute this into the interest equation:\[ 0.04(80,500 - 5z) + 0.15z = 2,670 \]Simplify and solve for \( z \):\[ 3,220 - 0.2z + 0.15z = 2,670 \]\[ -0.05z = -550 \]\[ z = 11,000 \]

Step 5: Find Remaining Variables

With \( z = 11,000 \), use \( y = 4z \) to find:\[ y = 4 \times 11,000 = 44,000 \]Then use \( x = 80,500 - 5z \):\[ x = 80,500 - 5 \times 11,000 = 25,500 \]

Step 6: Conclusion

The investment amounts are \( x = 25,500 \), \( y = 44,000 \), and \( z = 11,000 \). These satisfy all the given conditions about total investment, interest, and relationship between the second and third accounts.

Key Concepts

Simple Interest

Simple interest is a way to calculate the interest earned or paid on an amount of money over a specific period of time. It is straightforward because it uses a fixed percentage of the principal amount for each time period.
This means that the interest remains constant each year.

• The formula for simple interest is: \( I = P \times r \times t \)
• Where:
  • \( I \) is the interest earned,
  • \( P \) is the principal amount (initial investment),
  • \( r \) is the rate of interest per period as a decimal, and
  • \( t \) is the time period the money is invested or borrowed.

In the problem we considered, the simple interest for various accounts was calculated over one year. This involved applying the respective percentage rates directly to each principal amount. By working through these calculations, students can clearly see how interest values differ based on the principal and the rate used.

Investment Problems

Investment problems often involve deciding how to allocate money into different accounts or assets to achieve specific goals, such as maximizing returns or meeting certain financial constraints. The key to solving these problems is to understand the conditions and requirements connected to the investments.
In our scenario, the recent graduate invested in three accounts with different interest rates. Here are the details:

• The first account had a 4% rate.
• The second account had a 3\(\frac{1}{8}\)% rate.
• The third account had a 2\(\frac{1}{2}\)% rate.

Additionally, a piece of crucial information was that the second account's balance is four times the third account's. This allows us to create relationships and constraints necessary to solve the problem. By mastering setup and simplification of such systems of equations, students can efficiently solve similar investment-related issues.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves expressing one variable in terms of the others and substituting it into other equations to find the values of the remaining variables. Here's how it works in practice:
In our exercise, we had this condition:
\( y = 4z \)
We substituted this into the other two main equations to find values for all variables involved. This step-by-step process clarified dependencies and reduced the problem complexity:

• Replace \( y \) in equations with \( 4z \)
• Transform and solve the simplified equations
• Iteratively solve for one variable at each step

Through substitution, we isolate individual variables, simplifying the process of finding their specific values. Understanding this method is essential in algebra and is particularly useful when dealing with more complex systems that require variable reduction.

College Algebra

College Algebra encompasses a wide range of fundamental mathematical concepts, including equations, inequalities, functions, and graphing. In the context of system of equations, it ensures we understand how to apply algebraic skills to solve complex, real-world problems.
Key tools and concepts used in our exercise include:

• Setting up accurate equations based on problem conditions
• Simplifying and manipulating equations logically
• Strategically solving for variables using techniques like substitution

These tools are vital for calculating investment problem solutions and beyond, like optimizing budgets or distributions. By leveraging algebraic reasoning, students acquire the confidence to tackle intricate scenarios efficiently and accurately. Familiarity with these algebraic principles will greatly aid anyone preparing for a college-level algebra course or dealing with real-world financial scenarios.