Problem 61 - Contemporary Precalculus

Question

An investor has part of her money in an account that pays \(-9 \%\) annual interest and the rest in an account that pays \(11 \%\) annual interest. If she has \(\$ 8000\) less in the higher-paying account than in the lower-paying one and her total annual interest income is \(\$ 2010\), how much does she have invested in each account?

Step-by-Step Solution

Answer

Answer: The investor has $144,500 invested in the -9% annual interest account and $136,500 invested in the 11% annual interest account.

Step 1: Define the variables

Let \(x\) represent the amount invested in the \(-9\%\) annual interest account, and \(y\) represent the amount invested in the \(11\%\) annual interest account.

Step 2: Create the equations from given information

Given that she has $8000 less in the higher-paying account than in the lower-paying one, \(y = x - 8000\) And, her total annual interest income is \(\$ 2010\), \(-0.09x + 0.11y = 2010\)

Step 3: Solve the system of equations using substitution or elimination

We can use the substitution method to solve these equations. Substitute the first equation into the second equation: \(-0.09x + 0.11(x - 8000) = 2010\)

Step 4: Solve for x

Distribute and combine like terms to find the value of x: \(-0.09x + 0.11x - 880 = 2010\) \(0.02x = 2890\) Divide both sides by \(0.02\): \(x = \dfrac{2890}{0.02}\) \(x = 144500\)

Step 5: Solve for y

Now, substitute the value of x we found into the first equation to find the value of y: \(y = 144500 - 8000\) \(y = 136500\)

Step 6: Interpret the solution

The investor has \(\$144,500\) invested in the \(-9\%\) annual interest account and \(\$136,500\) invested in the \(11\%\) annual interest account.

Key Concepts

Linear Equations

Linear equations are fundamental in mathematics, often used to describe relationships involving constant rates of change. In our exercise, we utilize linear equations to express the relationships between different investment amounts and interest rates. These equations take the form of expressions like \(y = x - 8000\), where one variable depends linearly on another.
Linear equations can be graphed as straight lines, and their solutions involve finding values where these lines cross or meet. In the context of our problem, the linear equations help us determine the amounts invested in each account by expressing the relationship between them mathematically.

Interest Income

Interest income is the money one earns from an investment based on a particular interest rate, usually expressed as a percentage. In our exercise, two accounts pay different interest rates: \(-9\%\) and \(11\%\). This means that for every dollar invested, the investor either earns or loses a proportion according to these rates over a given time period, generally a year.
The total interest income in the problem, \(\$2010\), is the sum of interest earned or lost across both accounts. It acts as a pivotal reference to form the second equation: \(-0.09x + 0.11y = 2010\). This equation ties together the amounts invested in both accounts and their respective interest rates to equate to the investor's total interest income.

Substitution Method

The substitution method is a fundamental technique in solving systems of equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation. In our investment problem, we solve the first equation \(y = x - 8000\) for \(y\) and substitute it into the second equation: \(-0.09x + 0.11(x - 8000) = 2010\).

• This substitution simplifies the second equation into one that involves only one variable, \(x\).
• Once we solve for \(x\), we can easily find \(y\) using the relationship already established.

Employing this method simplifies the process of finding the investment amounts in each account. It eliminates one variable, allowing step-by-step progression toward the solution.

Investment Problems

Investment problems often involve determining how to allocate funds between different investment vehicles to achieve desired outcomes, like a specific interest income. In the given problem, the challenge is distributing the funds between accounts with negative and positive interest rates.

• Two key relationships to solve include the difference in amounts invested due to \(\\(8000\) less in one account and the total interest income amounting to \(\\)2010\).
• Analyzing such problems helps in understanding how money can grow (or diminish) based on varying interest rates and initial capital distribution.

Investment problems combine concepts from algebra and finance, showcasing real-world applications of mathematical solutions. They build skills in making informed financial decisions, understanding risks, and maximizing returns based on given constraints.