Problem 27 - Precalculus with Limits

Question

\(\begin{array}{l}{\text { Solving a System by Substitution In Exercises }} \\\ {25-28 \text { , you are given the total annual interest earned }} \\ {\text { from a total of } \$ 12,000 \text { invested in two funds paying the }} \\\ {\text { given rates of simple interest. Write and solve a system of }} \\\ {\text { equations to find the amount invested at each rate. }}\end{array}\) \(\begin{array}{l}{\text { Annual Interest }} \\ {\$ 396}\end{array}\) \( \begin{array}{ll}{\text { Rate } 1} & {\text { Rate } 2} \\ {2.8 \%} & {3.8 \%}\end{array} \)

Step-by-Step Solution

Answer

The amount invested at a 2.8% interest rate, which is \( x \), is $6000. The amount invested at a 3.8% interest rate, standing for \( y \), is also $6000.

Step 1: Establish the Equations

Let \( x \) represent the amount invested at 2.8%, and \( y \) represent the amount at 3.8%. Since the total amount invested is $12,000, the first equation in the system will be: \( x + y = 12000 \). To get the second equation, use the data about the interest earned: 2.8% of \( x \) plus 3.8% of \( y \) equals $396, can be represented as: \( 0.028x + 0.038y = 396 \).

Step 2: Solve the System by Substitution

Make \( x \) the subject of the first equation: \( x = 12000 - y \). Substitute this into the second equation: \( 0.028(12000 - y) + 0.038y = 396 \).

Step 3: Simplify and solve for \( y \)

Expand and resolve: \( 336 - 0.028y + 0.038y = 396 \). This simplifies to: \( 0.01y = 60 \). Divide both sides by 0.01: \( y = 6000 \).

Step 4: Determine \( x \)

Substitute \( y = 6000 \) into \( x + y = 12000 \) to obtain: \( x = 12000 - 6000 \).

Key Concepts

Simple Interest Calculations

Understanding simple interest is crucial for managing personal finances and solving various mathematical problems, including systems of equations related to investments. Simple interest can be calculated using the formula:

\( I = P \times r \times t \),
where \( I \) is the interest earned or paid, \( P \) is the principal amount (the initial sum of money), \( r \) is the annual interest rate (in decimal form), and \( t \) is the time in years the money is invested or borrowed.

For example, if you invest \$1,000 at a simple interest rate of 5% per year for 3 years, the interest you would earn is calculated as \( 1000 \times 0.05 \times 3 = \$150 \).
Applying this to the textbook exercise, investing a certain amount at different simple interest rates and determining the total interest earned involves creating and solving a system of equations that reflects these relationships.

Linear Equations

Linear equations form the backbone of algebra and are the simplest type of equations to handle. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can be written in the form:

\( ax + b = 0 \),
where \( a \) and \( b \) are constants, and \( x \) is the variable. If \( a \) is not equal to zero, the equation can be rearranged to solve for \( x \).

Applying Linear Equations

Solving linear equations is part of everyday problem-solving. In our textbook example, the equation \( x + y = 12000 \) is linear. It expresses the total amount invested as \$12,000, which is the sum of two parts invested at different interest rates, represented by \( x \) and \( y \). The other linear equation, \( 0.028x + 0.038y = 396 \), represents the total annual interest earned from these investments.
Linear equations are essential tools in creating models for real-world situations and translating problems into a mathematical context that can be analyzed and solved.

Systems of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The goal is to find the numerical values of these variables that satisfy all equations in the system simultaneously.

In the context of our example, the system of linear equations is:
\( x + y = 12000 \) (Equation 1)
\( 0.028x + 0.038y = 396 \) (Equation 2)

To solve this system, we can use several methods, including substitution, elimination, and using matrices or graphing.

Substitution Method

This method involves solving one of the equations for one variable and then substituting the result into the other equation. In our scenario, solving Equation 1 for \( x \) and substituting into Equation 2 helps us find \( y \). Once \( y \) is found, we substitute it back into Equation 1 to find \( x \).
Systems of linear equations can represent countless real-world situations where multiple conditions must be met simultaneously, making them invaluable in not only mathematics but in fields such as economics, engineering, and science.