Problem 62 - College Algebra

Question

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 4\(\%\) simple interest, one that paid 3\(\frac{1}{8} \%\) 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

Invested: $25,500 at 4%, $44,000 at 3⅛%, $11,000 at 2½%.

Step 1: Define Variables

Let's define variables for the amounts invested in each account. Let \( x \) represent the amount invested at 4\(\%\), \( y \) the amount invested at 3\(\frac{1}{8}\%\), and \( z \) the amount invested at 2\(\frac{1}{2}\%\).

Step 2: Set Up Equations

We know the total investment is \(\\(80,500\). Therefore, the equation is:\[ x + y + z = 80,500 \] Also, the total interest earned is \(\\)2,670\). The equation for interest is:\[ 0.04x + 0.03125y + 0.025z = 2,670 \] The condition given is that the second account's investment is four times the third account. Hence:\[ y = 4z \]

Step 3: Substitute and Simplify

Substitute \( y = 4z \) into the first equation:\[ x + 4z + z = 80,500 \rightarrow x + 5z = 80,500 \] Substitute \( y = 4z \) into the interest equation:\[ 0.04x + 0.03125(4z) + 0.025z = 2,670 \]This simplifies to:\[ 0.04x + 0.125z + 0.025z = 2,670 \rightarrow 0.04x + 0.15z = 2,670 \]

Step 4: Solve for x and z

We have two equations:1. \( x + 5z = 80,500 \)2. \( 0.04x + 0.15z = 2,670 \)From the second equation, multiply through by 100 to clear decimals:\[ 4x + 15z = 267,000 \]Rearrange the first equation for \( x \):\[ x = 80,500 - 5z \]Substitute \( x = 80,500 - 5z \) into \( 4x + 15z = 267,000 \):\[ 4(80,500 - 5z) + 15z = 267,000 \]\[ 322,000 - 20z + 15z = 267,000 \]\[ -5z = -55,000 \]\[ z = 11,000 \]

Step 5: Find x and y

Substituting \( z = 11,000 \) into \( x = 80,500 - 5z \):\[ x = 80,500 - 5(11,000) \]\[ x = 80,500 - 55,000 = 25,500 \]Using \( y = 4z \):\[ y = 4(11,000) = 44,000 \]

Step 6: Conclusion

The amount invested in the first account (4\(\%\)) is \(\\(25,500\), the second account (3\(\frac{1}{8}\%\)) is \(\\)44,000\), and the third account (2\(\frac{1}{2}\%\)) is \(\$11,000\).

Key Concepts

Investment Equations

Investment equations play a crucial role in financial mathematics. They help us understand how money grows over time through interest earned.
In the context of simple interest, the formula \[I = P imes r imes t\]shows how much interest (\(I\)) is earned where \(P\) is the principal amount, \(r\) is the interest rate, and \(t\) is the time period for which the money is invested. In our problem, investment equations are used to balance the total amount invested and the total interest earned. Here are the equations from the problem:

• Total Investment: \( x + y + z = 80,500 \)
• Total Interest: \( 0.04x + 0.03125y + 0.025z = 2,670 \)
• Dependency: \( y = 4z \)

These equations structure the problem into a solvable format, allowing us to determine the amounts in each account. By carefully setting up these equations, we can find solutions to complex investment situations.

Algebraic Substitution

Algebraic substitution is a powerful technique to simplify and solve equations. It's particularly useful when dealing with systems of equations as it reduces complexity.
In this problem, we use substitution to replace one variable with something equivalent in terms of another. Specifically, we replace \(y\) using the relationship \(y = 4z\). Here's how it works in the equation:

• Substituting \(y = 4z\) in \( x + y + z = 80,500 \) results in a simplified equation: \( x + 5z = 80,500 \)
• Substituting \(y = 4z\) in the interest equation: \( 0.04x + 0.03125y + 0.025z = 2,670 \) simplifies to \( 0.04x + 0.15z = 2,670 \)

By substitution, we reduce the number of variables in consideration, making it straightforward to solve the system by other methods, such as solving linear equations.

Solving Linear Equations

Solving linear equations involves finding the values of variables that satisfy all given equations simultaneously. This problem demonstrates solving linear equations with both algebraic substitutions and rearranging terms.
First, we substitute to eliminate one variable, reducing our system to two equations:

• \( x + 5z = 80,500 \)
• \( 0.04x + 0.15z = 2,670 \)

These equations are then solved step-by-step. Starting with clearing the decimals in the interest equation by multiplying everything by 100:\[4x + 15z = 267,000\]Next, using substitution for \(x\), we replace it from the first equation to further progress:\[x = 80,500 - 5z\]This is then substituted back into the updated interest equation:\[4(80,500 - 5z) + 15z = 267,000\]After simplifying, solve for \(z\):\[z = 11,000\]Finally, use \(z\)'s value to solve for \(x\) and \(y\), completing the solutions for all variables. This systematic approach ensures all equations work together to give the correct investment values.