Problem 75 - A Graphical Approach to College Algebra

Question

Investment problems such as those in Exercises \(75-80\) can be solved by using a method similar to the one explained in Example \(2,\) along with the simple- interest formula \(I=P R T\) where I is the interest earned, \(P\) is the initial amount of money deposited, \(R\) is the annual interest rate as a decimal, and \(T\) is the time the money is deposited in years. Solve each problem. Let \(T=1\) year for each exercise. Real-Estate Financing Cody Westmoreland wishes to sell a piece of property for \(\$ 240,000 .\) He wants the money to be paid off in two ways: a short-term note at \(6 \%\) interest and a long-term note at \(5 \% .\) Find the amount of each note if the total annual interest paid is \(\$ 13,000\).

Step-by-Step Solution

Answer

Short-term note: $100,000; Long-term note: $140,000.

Step 1: Define Variables

Let \( x \) be the amount for the short-term note at \( 6\% \) interest, and \( y \) be the amount for the long-term note at \( 5\% \) interest. According to the problem, \( x + y = 240,000 \).

Step 2: Set Up Interest Equations

Using the simple interest formula \( I = PRT \), the interest from the short-term note is \( 0.06x \) and from the long-term note is \( 0.05y \). The total interest paid, according to the problem, is \( 13,000 \), so we have the equation: \( 0.06x + 0.05y = 13,000 \).

Step 3: Solve the System of Equations

You have two equations: \( x + y = 240,000 \) and \( 0.06x + 0.05y = 13,000 \). Solve this system using substitution or elimination. First, express \( y \) in terms of \( x \): \( y = 240,000 - x \).

Step 4: Substitute and Solve

Substitute \( y = 240,000 - x \) into the interest equation: \( 0.06x + 0.05(240,000 - x) = 13,000 \). Simplify and solve for \( x \): 1. \( 0.06x + 12,000 - 0.05x = 13,000 \)2. \( 0.01x = 1,000 \)3. \( x = 100,000 \).

Step 5: Find the Value of y

Substitute \( x = 100,000 \) back into the equation \( y = 240,000 - x \):\( y = 240,000 - 100,000 = 140,000 \).

Step 6: Solution Verification

Verify that the solution satisfies both original equations:1. \( x + y = 100,000 + 140,000 = 240,000 \). 2. Interest: \( 0.06 \times 100,000 + 0.05 \times 140,000 = 6,000 + 7,000 = 13,000 \). Both equations are satisfied.

Key Concepts

Investment Problems

Investment problems are a common type of financial problem that involves determining the best way to allocate funds to achieve the desired financial outcome. Typically, these problems require the solver to decide how to invest a sum of money among different options that offer various interest rates and terms. For example, in the exercise, Cody wishes to divide the sale money of a property into two notes, each with different interest rates. This involves calculating the interest earned from these options to make sure the total interest aligns with the given target, ensuring that the capital is allocated optimally.
Investment problems usually utilize the simple-interest formula, which is represented as:

• \( I = PRT \)

Where:

• \( I \) is the interest earned.
• \( P \) is the principal or initial amount of money invested.
• \( R \) is the annual interest rate expressed as a decimal.
• \( T \) is the time period of the investment in years.

These types of problems require a clear understanding of establishing relationships using given data to ensure correct allocation.

System of Equations

A system of equations is a set of equations with multiple variables that are solved simultaneously to find the values of these variables. In investment problems, systems of equations are particularly useful for analyzing situations where an amount needs to be divided among different interest-bearing accounts.
In the given problem, the task was to find the amount invested in two notes, given their collective total and the total interest. With two unknowns, \( x \) for the short-term note and \( y \) for the long-term note, two equations can be formed:

• \( x + y = 240,000 \)
• \( 0.06x + 0.05y = 13,000 \)

These equations create a system that can be solved using either substitution or elimination methods. A system of equations is essential for determining the exact values when facing constraints from the problem, providing an organized way to reach a solution.

Algebraic Solution

An algebraic solution involves using algebraic methods to solve mathematical equations and find unknown quantities. It's a systematic and logical approach to deriving solutions. In the context of the exercise, after defining variables and setting up equations, the algebraic method comes into play to find the solution for the unknowns, \( x \) and \( y \).
Here's a step-by-step approach to reach an algebraic solution:

• Express one variable in terms of the other using one of the equations. In this scenario, we expressed \( y \) as \( y = 240,000 - x \).
• Substitute the expression for \( y \) into the second equation. This simplifies the system to one equation with one variable, \( x \).
  Using the interest equation:
  • \( 0.06x + 0.05(240,000 - x) = 13,000 \)
  • Simplify and solve for \( x \) to get \( x = 100,000 \).
• Once \( x \) is found, substitute back to find \( y \). Hence, \( y = 240,000 - 100,000 = 140,000 \).
• Verify the solution by plugging the values of \( x \) and \( y \) back into original equations to ensure the system holds true.

An algebraic solution is powerful as it provides a structured method to tackle complex problems, breaking them into manageable parts.