Problem 25 - Precalculus with Limits

Question

\(\begin{array}{l}{\text { Solving a System by Substitution In Exercises }} \\\ {25-28 \text { , you are given the total annual interest earned }} \\ {\text { from a total of } \$ 12,000 \text { invested in two funds paying the }} \\\ {\text { given rates of simple interest. Write and solve a system of }} \\\ {\text { equations to find the amount invested at each rate. }}\end{array}\) \(\begin{array}{l}{\text { Annual Interest }} \\ {\$ 500}\end{array}\) \( \begin{array}{ll}{\text { Rate } 1} & {\text { Rate } 2} \\ {3 \%} & {5 \%}\end{array} \)

Step-by-Step Solution

Answer

The solution to the problem is that $5000 was invested at an annual interest rate of 3%, and $7000 was invested at an annual interest rate of 5%.

Step 1: Set up the System of Equations

Let \( x \) represent the amount invested at an annual interest rate of 3%, and \( y \) will represent the amount invested at an annual interest rate of 5%. We get the following two equations based on the exercise: \n1) The equation to represent total amount invested: \( x+y=12000 \) dollars\n2) The equation to represent total interest earned: \( 0.03x+0.05y=500 \) dollars

Step 2: Using Substitution to Solve the System

The aim in this step is to express one variable in terms of the other variable and substitute that in the second equation. From the equation \( x+y=12000 \) can be rearranged to make \( x=12000-y \). We substitute \( x=12000-y \) into the second equation, thus obtaining \( 0.03(12000-y)+0.05y=500 \)

Step 3: Solve for y

Distribute 0.03 in the equation obtained above, we get \( 360-0.03y+0.05y=500 \). Arranging terms give us \( 0.02y=140 \). Then, solving for \( y \) by dividing by 0.02 gives us \( y=7000 \). This indicates that $7000 was invested at an annual interest rate of 5%.

Step 4: Substitute y=7000 into the first equation

We substitute \( y = 7000 \) into the first equation from the system of equations \( x+y = 12000 \) gives us \( x+7000 = 12000 \). Solving this equation for \( x \) we find that \( x = 5000 \). This indicates that $5000 was invested in the fund with an annual interest rate of 3%.

Key Concepts

Systems of Linear Equations

Systems of linear equations are integral to algebra and they consist of two or more linear equations with the same set of variables. In the context of financial mathematics, they are often used to model scenarios where multiple financial products are combined to achieve a certain goal, such as maximizing interest earnings from different investments.
When solving these systems, the goal is to find the values of the variables that satisfy all equations simultaneously. This concept is fundamental in understanding how different financial options can be evaluated together to determine the optimal investment strategy.

Simple Interest

Simple interest is a way to calculate the interest charge on a loan or investment based on the original principal. It is calculated by multiplying the daily interest rate by the principal by the number of days that elapse between payments. This kind of interest does not compound, meaning that it's calculated solely on the initial amount you invested or borrowed, not on any interest earned or accrued.
For example, if you invest \(1,000 at an annual simple interest rate of 5%, you would earn \)50 in interest per year. Understanding how simple interest works is crucial for making informed decisions about loans and investments.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, where one variable is expressed in terms of the other. Once a variable has been isolated, it can be substituted into the other equation, effectively reducing the number of variables and simplifying the problem.
This method is especially useful when the equations are not conveniently aligned for elimination. It is a step-by-step process: Solve one equation for one variable and substitute that expression into the other equation. It's a staple of algebraic problem solving, helping students and mathematicians alike to solve equations efficiently.

Algebraic Problem Solving

Algebraic problem solving is a systematic approach that involves exploring, conjecturing, and reasoning logically to solve problems formulated in algebraic language. It often requires the translation of word problems into algebraic expressions and equations.
Effective algebraic problem solving includes understanding the problem, devising a plan (often involving the creation of equations to represent the scenario), carrying out the plan (solving the equations), and then looking back to review the solution. This approach not only solves the given problem but also strengthens one's ability to tackle a wide range of problems using a similar methodology.