Problem 72 - College Algebra with Corequisite Support

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. If an investor invests a total of $$\$ 25,000$$ into two bonds, one that pays \(3 \%\) simple interest, and the other that pays \(2 \frac{7}{8} \%\) interest, and the investor earns \(\$ 737.50\) annual interest, how much was invested in each account?

Step-by-Step Solution

Answer

Invest \$15,000 in the \(3\%\) bond and \$10,000 in the \(2\frac{7}{8}\%\) bond.

Step 1: Define Variables

Let's define the variables for the system of equations. Let \( x \) be the amount invested in the bond with \(3\%\) interest, and \( y \) be the amount invested in the bond with \(2 \frac{7}{8}\%\) interest.

Step 2: Create Equations

We have two pieces of information that will help us build our equations:1. The total investment is \\(25,000.2. The total annual interest earned is \\)737.50.Using these, we can form the following equations:Equation 1 (Total investment): \[ x + y = 25,000 \]Equation 2 (Total interest): \[ 0.03x + 0.02875y = 737.5 \]

Step 3: Solve for One Variable

Using Equation 1, we can express \( y \) in terms of \( x \):\[ y = 25,000 - x \]

Step 4: Substitute and Solve

Substitute \( y = 25,000 - x \) into Equation 2:\[ 0.03x + 0.02875(25,000 - x) = 737.5 \]Expand the equation:\[ 0.03x + 718.75 - 0.02875x = 737.5 \]Combine like terms:\[ 0.00125x + 718.75 = 737.5 \]Subtract 718.75 from both sides:\[ 0.00125x = 18.75 \]Divide by 0.00125 to solve for \( x \):\[ x = 15,000 \]

Step 5: Substitute Back to Find the Second Variable

Now that \( x = 15,000 \), substitute back to find \( y \):\[ y = 25,000 - x = 25,000 - 15,000 = 10,000 \]

Step 6: Verify the Solution

Check that these amounts satisfy both equations. The total interest from these investments should equal \$737.50:For \( x = 15,000 \) at \(3\%\) interest:\[ 0.03 \times 15,000 = 450 \]For \( y = 10,000 \) at \(2 \frac{7}{8}\%\) interest:\[ 0.02875 \times 10,000 = 287.5 \]Adding these interests together confirms:\[ 450 + 287.5 = 737.5 \]Both conditions are satisfied.

Key Concepts

Simple Interest

Simple interest is a way of calculating the interest charge on a loan or bond based only on the original amount, also known as the principal. In this particular exercise, we deal with investments that yield income through simple interest. It's different from compound interest where the interest amount increases because earned interest itself earns more interest over time.
With simple interest, the formula is straightforward:

• Interest = Principal × Rate × Time.

Here, the rate is expressed as a percentage or fraction, and the time is usually in years. This means when an investor places money in a bank or financial institution, the amount of interest they will earn each year is determined solely by the interest rate and the initial amount invested.
In our problem, the rates provided are 3% and \(2 \frac{7}{8}\%\), indicating that for every dollar invested, that percentage of a dollar will be earned over one year.

Investment Problems

Investment problems often involve finding how much money is allocated to different financial instruments or accounts to maximize returns, minimize risk, or, as in this exercise, determine the specific amounts invested from a given total.
Such problems require setting up equations based on the given returns or interests earned. These equations stem from the conditions of the problem—ensuring that both the total amount invested and the total interest income are accounted for.
Here, we must figure out how much was invested in two bonds with different interest rates, knowing the total principal and total interest.

• First, we recognize the total investment as a crucial equation \( x + y = 25,000 \) \, where \( x \) and \( y \) represent amounts invested in two different bonds, respectively.
• The second equation arises from the total interest earned \( 0.03x + 0.02875y = 737.5 \).

By understanding these equations, we lay the groundwork for solving the problem using algebraic methods.

Two Variables

An equation with two variables is commonplace in investment problems. In mathematics, when two unknowns are to be calculated, a system of linear equations is used. Each variable represents an unknown quantity; in our exercise, these quantities are the amounts invested in each bond.
Analyzing each equation:

• The first equation \( x + y = 25,000 \) represents the total amount, with \( x \) and \( y \) being the variables for amounts invested in each bond.


• The second equation \( 0.03x + 0.02875y = 737.5 \) relates to the annual interest income.

These variables allow us to form equations necessary to solve for the unknowns, ultimately revealing how much was individually invested in each bond. Using algebraic methods such as substitution or elimination, the unknown values can be derived.

Algebraic Equations

Algebraic equations are powerful tools for solving problems involving unknown variables. In a system of equations, each equation must be satisfied for the values of the variables. This approach was used in solving the investment problem.
Here’s how it works:

• First, express one variable in terms of the other. We started with \( y = 25,000 - x \) using the first equation \( x + y = 25,000 \).


• Next, substitute this expression into the second equation \( 0.03x + 0.02875y = 737.5 \).


• This leaves a single variable equation that can be solved to find one of the unknowns.

These steps involved require simplifying and rearranging terms, followed by solving the equation, allowing us to find precise values for both investments.