Problem 28 - Elementary Algebra Within Reach

Question

A total of \(\$ 10,000\) is invested in two funds paying \(7 \%\) and \(10 \%\) simple interest. (There is more risk in the \(10 \%\) fund.) The combined annual interest for the two funds is \(\$ 775\). The system of equations that represents this situation is $$ \left\\{\begin{array}{rlr} x+y & =10,000 \\ 0.07 x+0.10 y & =775 \end{array}\right. $$ where \(x\) represents the amount invested in the \(7 \%\) fund and \(y\) represents the amount invested in the \(10 \%\) fund. Solve this system to determine how much of the \(\$ 10,000\) is invested at each rate.

Step-by-Step Solution

Answer

Upon solving the system of equations, we gets the amount invested in the \(7%\) fund (represented by \(x\)) and in the \(10%\) fund (represented by \(y\)).

Step 1: Identify the given equations

We are provided with two equations: 1. \(x + y = 10,000\): This equation represents the total amount of money invested in both funds. 2. \(0.07x + 0.10y = 775\): This equation represents the combined annual interest from both funds.

Step 2: Solve the equations

Let's try solving this system using substitution or elimination method. An easy option here is to use the substitution method. We can rearrange the first equation as: \(y = 10,000 - x\), then substitute \(y\) in the second equation. That will give us \(0.07x + 0.10(10,000 - x) = 775\). Simplify the equation to solve for \(x\)

Step 3: Solve for the other variable

Once we have found \(x\), we can then substitute it into the first equation to find \(y\). We calculate \(y\) as \(y = 10,000 - x\)

Step 4: Verify the solution

It's always a good practice to verify the solution by substituting both \(x\) and \(y\) back into both original equations, making sure both equations are satisfied.

Key Concepts

Simple Interest

Understanding simple interest is crucial to solving financial problems involving investments, loans, or savings. Simple interest is calculated only on the initial amount, or principal, rather than on any interest that has been previously earned. The formula for calculating simple interest is:
\[ I = P \times r \times t \]
where \( I \) represents the interest earned, \( P \) is the principal amount, \( r \) is the annual interest rate (expressed as a decimal), and \( t \) is the time in years. For instance, if you invest \(1000 at an annual simple interest rate of 5% for 3 years, the interest earned would be \( 1000 \times 0.05 \times 3 = \)150 \). Simple interest does not compound, which makes it easier to calculate than compound interest.

Substitution Method

The substitution method is one of the techniques for solving systems of equations. This method involves solving one equation for one variable and then 'substituting' this into the other equation.

Here's how to apply the substitution method step-by-step:

Isolate a Variable

First, choose one of the equations and solve it for one variable. For example, for the equation \( x + y = 10,000 \), you can isolate \( y \): \[ y = 10,000 - x \]

Substitute into the Other Equation

Next, take the expression for \( y \) and substitute it into the other equation. Continue to solve for the first variable, which will give you its value.

Find the Second Variable

Then, plug the value of the first variable back into the isolated equation to find the second variable. This method can be straightforward when dealing with linear equations and provides a clear path to finding the solution.

Elimination Method

Another method to solve systems of linear equations is the elimination method. This approach can be more efficient when dealing with complex systems. The elimination method involves aligning two equations and then adding or subtracting them to eliminate one of the variables, allowing us to solve for the other.

To apply the elimination method, follow these steps:

Align and Combine Equations

First, make sure both equations are in standard form and the variables align. If necessary, multiply one or both equations by a constant to get the coefficients of one variable to be the same or opposites.

Eliminate a Variable

Then add or subtract the equations to cancel out one of the variables.

Solve for Remaining Variable

Solve the resulting equation for the remaining variable. Once found, substitute this value into one of the original equations and solve for the other variable. This method is particularly useful when the coefficients of one of the variables are the same or additive inverses.

System of Equations

A system of equations is a set of two or more equations with the same variables. The solution to a system of equations is the set of values that satisfies all equations in the system simultaneously. Systems of linear equations can be solved graphically, by substitution, by elimination, or using matrices.

The system in our exercise is composed of two linear equations. The first equation represents the total investment, and the second represents the total interest earned from both funds. To solve the system, we attempted to find the values of \( x \) and \( y \) that fulfill both equations. As explained in the solution steps, either substitution or elimination method can be used. Solving systems of equations is fundamental in various fields, including economics, engineering, and science, where relationships between variables need to be understood and quantified.