Problem 55 - College Algebra

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: Interpret the Problem

We need to find the interest rates for two accounts. The first account has an investment of $2,300, and the second has an investment of $2,700. After one year, the total interest earned is $254. Additionally, account 2's interest rate is 1.5 times that of account 1. We will set up equations to describe this situation.

Step 2: Define the Variables

Let the interest rate of account 1 be \( r_1 \) and the interest rate of account 2 be \( r_2 \). We can then express the conditions that account 2 has 1.5 times the interest rate of account 1 as \( r_2 = 1.5 \, r_1 \).

Step 3: Set Up Equations

Using the formula for simple interest, \( \text{Interest} = \text{Principal} \times \text{Rate} \), we can create equations for each account: 1. Account 1: \( 2300 \, r_1 \) 2. Account 2: \( 2700 \, r_2 = 2700 \, (1.5 \, r_1) = 4050 \, r_1 \)The total interest is \$254, so we have: \[ 2300 \, r_1 + 4050 \, r_1 = 254 \]

Step 4: Combine and Solve the Equation

Combine the terms to form a single equation:\[ 2300 \, r_1 + 4050 \, r_1 = 6350 \, r_1 = 254 \]To solve for \( r_1 \), divide both sides by 6350:\[ r_1 = \frac{254}{6350} \approx 0.04 \]

Step 5: Find the Second Interest Rate

Since \( r_2 = 1.5 \, r_1 \), substitute the value of \( r_1 \) to find \( r_2 \):\[ r_2 = 1.5 \, (0.04) = 0.06 \]

Step 6: Summarize the Solution

The interest rate for account 1, \( r_1 \), is 4%, and for account 2, \( r_2 \), it is 6%.

Key Concepts

Augmented Matrix

An augmented matrix is a very useful tool when solving systems of linear equations. It simplifies the process by combining the coefficients and constants of the equations into a single compact matrix form. Imagine you have a set of linear equations that look something like this:

• Equation 1: \( ax + by = c \)
• Equation 2: \( dx + ey = f \)

The augmented matrix takes these equations and represents them as \[\begin{bmatrix}a & b & | & c \d & e & | & f \\end{bmatrix}\]In this matrix, the vertical line acts as a visual separator between the coefficients of the variables and the constants from the right side of the equations. Augmented matrices are crucial in settings where multiple linear equations need to be solved systematically and efficiently. They allow for straightforward application of techniques like Gaussian elimination to find the solutions. Such matrices can be instrumental, especially in complex problems involving several equations.

Simple Interest

Simple interest is a fundamental concept in finance that calculates the interest earned or paid on a principal sum of money. Unlike compound interest, simple interest doesn’t add interest on the earned interest. The formula to calculate simple interest is \[\text{Interest} = ext{Principal} \times ext{Rate} \times ext{Time}\]- **Principal** is the initial amount of money.- **Rate** is the annual interest rate.- **Time** usually expressed in years.Simple interest calculations are straightforward and rely only on the original principal. For example, if you invest \\(2,000 at a rate of 5% per year, after 1 year, the interest will be:\[\text{Interest} = 2000 \times 0.05 \times 1 = 100\]This means you earn \\)100 after one year. Simple interest is beneficial in scenarios where the additional interest earned over the period isn't compounded, making it easier to track and calculate.

Solving Linear Equations

Solving linear equations is a basic yet powerful technique in algebra. It involves finding the values of variables that satisfy the given equations, often forming the core of many mathematical and real-world problems. A linear equation in two variables, for example, can typically be written as:\( ax + by = c \)The most common strategies to solve linear equations include:

• Substitution: Solve for one variable in terms of another and substitute it back into the other equation.
• Elimination: Combine equations to eliminate one of the variables, making it simpler to solve.
• Graphical method: Plot both equations on a graph and find the point where they intersect.

In the context of our example, after setting up two equations from the financial problem, we applied elimination by combining both interest equations. The goal was to find the rates that satisfied the total interest condition using very basic algebraic manipulations. Successfully solving linear equations helps determine not just interest rates, but also countless other relationships in mathematics and science.