Problem 55 - 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 \(\$ 2,300\) into account \(1,\) and \(\$ 2,700\) into account 2 . If the total amount of interest after one year is \(\$ 254\) and account 2 has 1.5 times the interest rate of account \(1,\) what are the interest rates? Assume simple interest rates.

Step-by-Step Solution

Answer

The interest rates are 4% for account 1 and 6% for account 2.

Step 1: Define Variables

Let \( r_1 \) be the interest rate for account 1, and \( r_2 \) be the interest rate for account 2. We need to express everything in terms of these variables to set up our equations.

Step 2: Formulate the Interest Equation

For account 1, the interest earned is \( 2300 \times r_1 \) and for account 2, the interest is \( 2700 \times r_2 \). The total interest from both accounts is \( 254 \), so we have the equation \( 2300r_1 + 2700r_2 = 254 \).

Step 3: Use the Rate Relationship

From the problem statement, account 2 has 1.5 times the interest rate of account 1. This means \( r_2 = 1.5r_1 \).

Step 4: Substitute and Simplify

Substitute \( r_2 = 1.5r_1 \) into the total interest equation: \[ 2300r_1 + 2700(1.5r_1) = 254 \]. This simplifies to \[ 2300r_1 + 4050r_1 = 254 \].

Step 5: Solve for r1

Combine like terms: \( 6350r_1 = 254 \). Solve for \( r_1 \) by dividing both sides by 6350 to get \( r_1 = \frac{254}{6350} \approx 0.04 \).

Step 6: Solve for r2

Use \( r_2 = 1.5r_1 \): \( r_2 = 1.5 \times 0.04 = 0.06 \).

Step 7: Verify the Solution

Check if these values satisfy the original equation: \( 2300(0.04) + 2700(0.06) = 92 + 162 = 254 \). The values are correct.

Key Concepts

Understanding Simple Interest

Simple interest is a straightforward way to calculate the interest on an investment. Unlike compound interest, which involves interest on top of interest, simple interest is calculated solely on the initial principal. This makes it easy to understand and compute.

The formula for simple interest is:

• \( I = P \times r \times t \)

Where:

• \( I \) is the interest earned,
• \( P \) is the principal amount (the initial amount of money),
• \( r \) is the interest rate (expressed as a decimal),
• and \( t \) is the time the money is invested or borrowed for, typically in years.

In problems like these, you need to identify the given values and plug them into the formula to find the interest. The simplicity of this method allows one to focus on understanding the rates and time period involved. It's particularly useful for short-term loans or investments where the simplicity of calculation can be an advantage over more complex interest calculations.

Interest Rate Equations

Understanding the equations used to calculate interest rates is crucial for solving financial problems. With simple interest, each account's interest is calculated separately and then summed to find the total interest earned. This allows us to set up an equation to express relationships between different interest rates.

For instance, in our exercise, for two accounts—account 1 and account 2—we use:

• Account 1: Interest = Principal of account 1 \( \times \) Interest rate of account 1, which is \( 2300r_1 \)
• Account 2: Interest = Principal of account 2 \( \times \) Interest rate of account 2, which is \( 2700r_2 \)

These amounts must add up to the total interest, forming our primary interest equation: \( 2300r_1 + 2700r_2 = 254 \).Moreover, any relationships between interest rates, such as account 2 having 1.5 times the interest rate of account 1 (\( r_2 = 1.5r_1 \)), give further information to help solve for unknowns. By replacing one variable in terms of another using relationships, students create a solvable linear equation.

Solving Linear Systems with Augmented Matrices

Linear systems are a collection of linear equations involving the same set of variables. An efficient method for solving these systems is using augmented matrices, especially when dealing with two or more equations.

An augmented matrix consolidates the coefficients of the linear equations and their respective constants into a single matrix. For example, the system of equations \( 2300r_1 + 2700r_2 = 254 \) and \( r_2 = 1.5r_1 \) can be expressed in matrix form.The steps typically include:

• Setting up the matrix using the coefficients of the variables and distributing the constants as a separate column.
• Employing row operations to simplify the matrix—much like solving equations step-by-step until the matrix is in reduced row-echelon form.
• Interpreting the final matrix to find the values of the variables.

This method, though initially daunting, simplifies solving the system of equations, providing a visual and systematic way to find variable values. Practicing this approach will enhance your problem-solving toolkit, especially for more complex systems involving multiple equations.