Problem 54 - College Algebra with Corequisite Support

Question

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. You invested \(\$ 10,000\) into two accounts: one that has simple \(3 \%\) interest, the other with \(2.5 \%\) interest. If your total interest payment after one year was \(\$ 283.50,\) how much was in each account after the year passed?

Step-by-Step Solution

Answer

$6,700 was in the 3% account, and $3,300 was in the 2.5% account.

Step 1: Define Variables

Let \( x \) be the amount of money invested at 3% interest, and \( y \) be the amount invested at 2.5% interest.

Step 2: Set Up Equation for Total Investment

The total investment is \$10,000, so the equation is: \( x + y = 10000 \).

Step 3: Set Up Equation for Total Interest

The total interest after one year is \$283.50, so the equation based on interest is: \( 0.03x + 0.025y = 283.50 \).

Step 4: Form the Augmented Matrix

Write the system of equations as an augmented matrix: \[ \begin{bmatrix} 1 & 1 & | & 10000 \ 0.03 & 0.025 & | & 283.50 \end{bmatrix} \]

Step 5: Use Row Operations to Solve

To solve this, we will perform row operations. First, we can multiply the second row by 100 to eliminate decimals: \[ \begin{bmatrix} 1 & 1 & | & 10000 \ 3 & 2.5 & | & 28350 \end{bmatrix} \]Next, we need to eliminate \( y \) from the first row. Multiply the first row by 2.5 and subtract from the second row (3 times the row), then: 1. Multiply first row by 2.5: \( 2.5 \times [1, 1, |, 10000] = [2.5, 2.5, |, 25000] \)2. Subtract this from the second row result: \[3x + 2.5y - 2.5x - 2.5y = 28350 - 25000 \0.5x = 3350 \]Solve for \( x \): \[ x = \frac{3350}{0.5} = 6700 \]

Step 6: Solve for y

Substitute \( x \) back into the first equation: \[ 6700 + y = 10000 \]Solve for \( y \): \[ y = 10000 - 6700 = 3300 \]

Step 7: Verify the Solution

Verify the calculated values by checking the total interest: \( 0.03(6700) + 0.025(3300) = 201 + 82.5 = 283.50 \)As the calculations meet the original conditions, the solution is verified.

Key Concepts

Understanding Simple Interest

Simple interest is a straightforward method for calculating the interest charged or earned on a principal amount over time. In this exercise, simple interest allows us to determine how much money was earned in interest from two different investments. Here’s how it works.

Simple interest is calculated using the formula:

• \( I = P imes r imes t \)

where:

• \( I \) is the interest amount.
• \( P \) is the principal amount or initial investment.
• \( r \) is the rate of interest as a decimal.
• \( t \) is the time in years.

In our scenario, we had two investments totaling \(10,000, with one at a 3% interest rate and another at 2.5%. After a year, the total interest earned was \)283.50. This information aids in setting up equations that help us figure out the investment breakdown.

Solving Systems of Equations

A system of equations is a set of equations with multiple variables that need to be solved simultaneously. In this situation, we address the problem of finding out how much was invested in each account using two equations.

The two crucial equations are:

• \( x + y = 10000 \)
• \( 0.03x + 0.025y = 283.50 \)

The first equation represents the total investment amount, while the second one represents the total interest from both accounts.
To solve this system, we must find values for \( x \) and \( y \) that satisfy both equations. Each variable represents the amount of money deposited in each account, providing a clear picture of money distribution between the two interest plans.

Understanding Row Operations in Augmented Matrices

An augmented matrix is a tool used to solve systems of linear equations. It conveniently organizes coefficients and constants from the equations into a matrix format, allowing us to perform row operations.

Here's the augmented matrix set up from our equations:

• \(\begin{bmatrix} 1 & 1 & | & 10000 \ 0.03 & 0.025 & | & 283.50 \end{bmatrix}\)

We perform row operations to simplify the matrix:

• Multiply row 2 by 100 to remove decimals, resulting in \( \begin{bmatrix} 1 & 1 & | & 10000 \ 3 & 2.5 & | & 28350 \end{bmatrix}\)
• Then, use row operations to eliminate a variable, in this case, eliminating \( y \).
• This yields the solution for \( x \).

Through these transformations, we can easily solve for both variables, helping to see how funds were allocated in each account.