Problem 53 - College Algebra

Question

For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. You invest \(\$ 10,000\) into two accounts, which receive 8\(\%\) interest and 5\(\%\) interest. At the end of a year, you had \(\$ 10,710\) in your combined accounts. How much was invested in each account?

Step-by-Step Solution

Answer

Invest $7,000 at 8% and $3,000 at 5%.

Step 1: Define the Variables

Let's define the variables first. Let \( x \) represent the amount invested at 8\% interest, and \( y \) represent the amount invested at 5\% interest.

Step 2: Set Up the Equations

We have two conditions. First, the total amount invested is \\(10,000: \[ x + y = 10,000 \]. Second, the total interest earned after a year is \\)710, so \[ 0.08x + 0.05y = 710 \].

Step 3: Convert the System to Standard Form

Write the system of equations in standard form: 1. \( x + y = 10,000 \)2. \( 0.08x + 0.05y = 710 \)

Step 4: Set Up Matrices for Cramer's Rule

Define the coefficient matrix \( A \), constant matrix \( C \), and replace matrices for Cramer's Rule:\[ A = \begin{bmatrix} 1 & 1 \ 0.08 & 0.05 \end{bmatrix}, C = \begin{bmatrix} 10000 \ 710 \end{bmatrix} \]

Step 5: Calculate Determinant of Matrix A

Compute \( \text{det}(A) \):\[ \text{det}(A) = (1)(0.05) - (1)(0.08) = 0.05 - 0.08 = -0.03 \]

Step 6: Find Determinants for Numerator Matrices

Find determinant for matrix \( A_x \) (replace 1st column of \( A \) with \( C \)):\[ A_x = \begin{bmatrix} 10000 & 1 \ 710 & 0.05 \end{bmatrix} \]\[ \text{det}(A_x) = (10000)(0.05) - (1)(710) = 500 - 710 = -210 \]Find determinant for matrix \( A_y \) (replace 2nd column of \( A \) with \( C \)):\[ A_y = \begin{bmatrix} 1 & 10000 \ 0.08 & 710 \end{bmatrix} \]\[ \text{det}(A_y) = (1)(710) - (10000)(0.08) = 710 - 800 = -90 \]

Step 7: Calculate the Solution Using Cramer's Rule

Apply Cramer's Rule:\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-210}{-0.03} = 7000 \]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-90}{-0.03} = 3000 \]

Step 8: Verify the Solution

Check the solution by substituting back into original equations:- \( x + y = 10,000 \) becomes \( 7000 + 3000 = 10000 \), which is correct.- \( 0.08x + 0.05y = 710 \) becomes \( 0.08(7000) + 0.05(3000) = 560 + 150 = 710 \), which is also correct.

Key Concepts

System of Linear Equations

Understanding a system of linear equations is essential for tackling problems like determining investment allocations. A system of linear equations involves multiple linear equations working together. Each equation consists of one or more variables. The goal is to find values for these variables that satisfy all equations simultaneously.

In simple terms:

• A solution to a system of equations is a set of values for the variables that makes all equations true.
• For two variables, you typically have two equations forming a system.

In this exercise, two equations describe the investment behavior:

• The first equation, \( x + y = 10,000 \), signifies that the total investment is \(10,000.
• The second equation, \( 0.08x + 0.05y = 710 \), indicates the total interest earned is \)710.

The power of systems of equations lies in their ability to model real-world scenarios, allowing us to solve practical problems with mathematical techniques.

Interest Calculation

Interest calculation is crucial in finance and, in this scenario, helps determine how much money was allocated to each investment account. Here, simple interest is calculated, which is a straightforward method of applying interest based on the initial principal amount.

Simple interest is given by the formula:

• \[I = P imes r imes t\]

where:

• \( I \): Interest earned
• \( P \): Principal amount (initial investment)
• \( r \): Rate of interest (expressed as a decimal)
• \( t \): Time period

In our example:

• For the investment at 8% (0.08), the expected interest for \( x \) dollars after one year is \( 0.08x \).
• For the investment at 5% (0.05), the interest for \( y \) dollars is \( 0.05y \).

The combination of these individual interests gives a total interest of $710, forming our interest equation. This method helps in systematically understanding and calculating the rewards from investments.

Matrix Determinants

Matrix determinants are mathematical values calculated from a square matrix, crucial for solving systems of equations using Cramer's Rule. A determinant lets us determine if a system of equations has a unique solution.

To compute a determinant for a 2x2 matrix:

• For a matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]

The determinant \( ext{det}(A) \) is calculated as:

• \[a \times d - b \times c\]

In the exercise, the coefficient matrix \( A \) was:

• \[\begin{bmatrix} 1 & 1 \ 0.08 & 0.05 \end{bmatrix}\]

Calculating its determinant:

• \[1 \times 0.05 - 1 \times 0.08 = 0.05 - 0.08 = -0.03\]

Determinants are used to find variable values by substituting columns of \( A \) with the constant matrix and calculating new determinants:

• \( A_x \) and \( A_y \) are determinant matrices used to solve for \( x \) and \( y \).

This methodology, dubbed Cramer's Rule, is incredibly powerful for checking consistent systems of equations, guiding us to definitive solutions efficiently.