Problem 53 - College Algebra with Corequisite Support

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

$7,000 in 8% and $3,000 in 5% accounts.

Step 1: Understanding the Problem

We have a total of $10,000 invested, split between two accounts. The two accounts earn 8% and 5% interest, respectively. At the end of the year, the combined value of both accounts is $10,710.

Step 2: Setting Up the Variables

Let \( x \) be the amount invested at 8% interest, and \( y \) be the amount invested at 5% interest. We have two variables representing our unknowns.

Step 3: Formulating System of Equations

We need to translate the information given into equations. The equation for the total investment is: \( x + y = 10,000 \). The equation for the total final amount after receiving interest is: \( 0.08x + 0.05y = 710 \) (since the interest earned is $710).

Step 4: Writing Augmented Matrix

From the equations \( x + y = 10,000 \) and \( 0.08x + 0.05y = 710 \), the augmented matrix is: \[\begin{bmatrix} 1 & 1 & | & 10,000 \0.08 & 0.05 & | & 710 \end{bmatrix}.\]

Step 5: Calculating Determinants

First, find the determinant of the coefficient matrix: \( D = \begin{vmatrix} 1 & 1 \ 0.08 & 0.05 \end{vmatrix} = (1)(0.05) - (1)(0.08) = 0.05 - 0.08 = -0.03 \). Next, find \( D_x \) and \( D_y \):Determine \( D_x \): \[D_x = \begin{vmatrix} 10,000 & 1 \ 710 & 0.05 \end{vmatrix} = (10,000)(0.05) - (710)(1) = 500 - 710 = -210.\]Determine \( D_y \): \[D_y = \begin{vmatrix} 1 & 10,000 \ 0.08 & 710 \end{vmatrix} = (1)(710) - (10,000)(0.08) = 710 - 800 = -90.\]

Step 6: Applying Cramer's Rule

Using Cramer's Rule, solve for \( x \) and \( y \):\[ x = \frac{D_x}{D} = \frac{-210}{-0.03} = 7,000, \]\[ y = \frac{D_y}{D} = \frac{-90}{-0.03} = 3,000. \]

Step 7: Conclusion

You invested \( 7,000 \) dollars in the account with 8% interest and \( 3,000 \) dollars in the account with 5% interest.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. In our case, we are dealing with a financial investment situation, where we need to determine how much was invested in two different accounts with different interest rates. The equations describe relationships between these monetary amounts.
For example, if you have two variables representing amounts, such as \( x \) for the amount invested at 8% interest and \( y \) for the amount invested at 5% interest, the system of equations helps us express the conditions given in the problem, such as the total amount initially invested and the interest accumulated over time.
Typically, a system of linear equations can be displayed as follows:

• Equation 1: \( x + y = 10,000 \)
• Equation 2: \( 0.08x + 0.05y = 710 \)

These equations provide a mathematical representation of the problem, which will be solved to find the values of \( x \) and \( y \).

Determinants

A determinant is a special number that is calculated from a square matrix. In linear algebra, determinants are primarily used to determine the solvability of a system of equations and are essential for applying Cramer's Rule.
In our example, the determinant of the coefficient matrix \( D \) is calculated using the numbers from the system of linear equations. The coefficient matrix from our problem is \( \begin{vmatrix} 1 & 1 \ 0.08 & 0.05 \end{vmatrix} \), and its determinant is calculated as follows: \[ D = (1)(0.05) - (1)(0.08) = -0.03 \] This determinant tells us that the system of equations does not have an infinite number of solutions and can be solved using Cramer's Rule. By using determinants of modified matrices that replace the original matrix columns successively, we can solve for each variable individually.

Linear Algebra

Linear algebra involves the study of vectors, vector spaces, and linear equations. It is a crucial mathematical tool in understanding complex systems and problems, such as financial investment scenarios.
In the context of our problem, linear algebra allows us to use matrices and determinants to efficiently solve systems of linear equations. With techniques like Cramer's Rule, which is deeply rooted in linear algebra, we can solve for unknown variables in an organized fashion.
Using these concepts, we created a matrix from our system of equations and calculated determinants that assisted in determining how much was invested in each account. The skills arising from linear algebra are broadly applied from theoretical mathematics to practical applications like financial analysis.

Financial Investment Problem

Financial investment problems involve optimizing the allocation of assets in different investment opportunities to achieve certain financial goals. In our example, the main goal was to determine how much was invested at two different interest rates.
By setting up a system of linear equations reflecting the total available funds and the interest accumulated, we were able to administer linear algebra techniques to find exact investment amounts.
Key points to consider in such problems include:

• Understanding the terms like interest rates and total final amounts
• Translating word problems into mathematically represented equations
• Solving these equations to make informed financial decisions

This type of problem-solving is not only relevant in academics but also equips individuals with skills necessary to manage personal finances and professional financial consultations effectively.