Problem 71 - Precalculus: Mathematics for Calculus

Question

Investments \(A\) woman invests a total of \(\$ 20,000\) in two accounts, one paying \(5 \%\) and the other paying \(8 \%\) simple interest per year. Her annual interest is \(\$ 1180 .\) How much did she invest at each rate?

Step-by-Step Solution

Answer

Invested \$14,000 at 5\% and \$6,000 at 8\%.

Step 1: Define Variables

Let \( x \) be the amount invested at \( 5\% \) interest, and \( y \) be the amount invested at \( 8\% \) interest. We must find the values of \( x \) and \( y \).

Step 2: Set Up Equations

The total investment is \\(20,000: \( x + y = 20,000 \). The total interest from both investments is \\)1,180: \( 0.05x + 0.08y = 1,180 \). These give us two equations: \( x + y = 20,000 \) and \( 0.05x + 0.08y = 1,180 \).

Step 3: Solve the System of Equations

First solve the first equation for \( x \): \( x = 20,000 - y \). Substitute \( x \) in the second equation: \( 0.05(20,000 - y) + 0.08y = 1,180 \). Simplifying gives: \( 1,000 - 0.05y + 0.08y = 1,180 \) which simplifies to \( 0.03y = 180 \).

Step 4: Calculate \( y \)

Solve \( 0.03y = 180 \) for \( y \): \( y = \frac{180}{0.03} = 6,000 \). Thus, the amount invested at \( 8\% \) is \$6,000.

Step 5: Calculate \( x \)

Use \( x = 20,000 - y \): \( x = 20,000 - 6,000 = 14,000 \). The amount invested at \( 5\% \) is \$14,000.

Key Concepts

Simple Interest

Simple interest is a way to calculate the interest charge on a loan or investment based on the original amount. This type of interest does not take into account the effect of compounding, making it straightforward to calculate. The formula for simple interest is:
\[ I = P \times r \times t \]

• \(I\) represents the interest earned or paid.
• \(P\) is the principal amount, or the initial sum of money.
• \(r\) stands for the rate of interest per period.
• \(t\) is the time the money is invested or borrowed for, usually in years.

In a practical situation, like the one given in the exercise, simple interest helps determine how much extra money the investment generates over time. It is particularly useful for understanding the financial outcome in short or predictable periods.

Investment Problems

When dealing with investment problems, the aim is to find out how the total amount is distributed across different investment options. In many cases, different investments yield varying interest rates, which can complicate how the total future value is determined. For these problems:

• Identify the total amount to be invested.
• Recognize the different interest rates for each portion of the investment.
• Use equations to break down how the total is divided.

These steps allow you to solve how much is invested in each option to reach a specific financial return. This concept is crucial for organizing and managing personal or organizational finances, ensuring one's investments are optimized for desired returns.

Variable Substitution

Variable substitution is a mathematical method for solving equations by replacing a variable with another expression. This is especially useful in systems of equations where you have more than one equation and unknown. For example, in the provided exercise, we needed to solve the system of equations:

• \(x + y = 20,000\)
• \(0.05x + 0.08y = 1,180\)

To do this, we first solve one of the equations for one of the variables. We chose to express \(x\) in terms of \(y\):
\[ x = 20,000 - y \]Substitute this expression into the other equation to solve for \(y\). This makes the system simpler to handle, as it reduces one equation to a single variable, helping us find one of the unknowns.

Solving Equations

Solving equations involves finding the value of the unknowns that satisfy the equations. Especially in systems of linear equations, this requires manipulation to isolate and calculate each variable accurately. Once a single variable is isolated, solving it usually involves arithmetic operations.
For our exercise, we reached:
\[ 0.03y = 180 \]Solving for \(y\), we divided both sides by \(0.03\):
\[ y = \frac{180}{0.03} = 6,000 \]This gave us one part of the solution. Next was to find \(x\) using \(x = 20,000 - y\):
\[ x = 20,000 - 6,000 = 14,000 \]These solutions indicate the amounts invested in each interest rate, completing the solution to the investment problem. Each step depends on simplifying expressions and performing basic arithmetic to find answers systematically.