Problem 266 - Elementary Algebra

Question

Arnold invested \(\$ 64,000\), some at \(5.5 \%\) interest and the rest at \(9 \% .\) How much did he invest at each rate if he received \(\$ 4,500\) in interest in one year?

Step-by-Step Solution

Answer

He invested \(36000 at 5.5% and \)28000 at 9%.

Step 1 - Define Variables

Let the amount invested at 5.5% be denoted as \( x \), and the amount invested at 9% be denoted as \( y \). Thus, the equations based on the problem statement will be: 1. \( x + y = 64000 \) 2. \( 0.055x + 0.09y = 4500 \)

Step 2 - Express One Variable in Terms of the Other

From the first equation, solve for \( y \) in terms of \( x \): \( y = 64000 - x \)

Step 3 - Substitute the Expression

Substitute \( y = 64000 - x \) into the second equation: \( 0.055x + 0.09(64000 - x) = 4500 \)

Step 4 - Simplify the Equation

Distribute and combine like terms in the substituted equation: \( 0.055x + 5760 - 0.09x = 4500 \) Simplifying further, this gives: \( -0.035x + 5760 = 4500 \)

Step 5 - Solve for \( x \)

Isolate \( x \): \( -0.035x = 4500 - 5760 \) \( -0.035x = -1260 \) Divide both sides by \( -0.035 \): \( x = \frac{-1260}{-0.035} \) \( x = 36000 \)

Step 6 - Find \( y \)

Use the value of \( x \) to find \( y \): \( y = 64000 - x \) \( y = 64000 - 36000 \) \( y = 28000 \)

Step 7 - Verify the Solution

Check the values by substituting back into the interest equation: \( 0.055(36000) + 0.09(28000) = 4500 \) \( 1980 + 2520 = 4500 \). The solution satisfies the equation.

Key Concepts

Linear Equations

Linear equations form the backbone of many mathematical problems. In this exercise, we have two linear equations that represent the amount of investment and the total interest earnings.

• Equation 1: The total investment amount, which is split between two different interest rates.
• Equation 2: The total interest earned from both investments after one year.

The key concept here is finding the values of two variables that satisfy both equations. By expressing one variable in terms of the other and substituting this into the second equation, we can solve linear equations effectively. This method leverages substitution and simplification principles that make linear equations a fundamental part of algebra.

Investment Problems

Investment problems often require understanding how to distribute a total amount across different rates to reach a given financial goal. In this scenario, Arnold's total investment is split between two interest rates (5.5% and 9%).

To solve these problems:

• Define variables for the amounts invested at different rates.
• Set up equations that represent the total investment and the total interest earned.
• Solve the equations by substitution or elimination methods.

This type of problem also helps strengthen the practical application of algebra in real-world financial contexts. By understanding how different rates affect overall returns, one can make more informed investment decisions.

Elementary Algebra

Elementary algebra involves the fundamental principles needed to manipulate and solve equations. In the given exercise, understanding algebraic operations like substitution, isolation of variables, and simplification are critical.

Important steps in solving algebraic problems include:

• Defining the variables appropriately.
• Setting up equations that correctly represent the problem.
• Using algebraic methods like substitution and combination to simplify and solve these equations.

Applying these algebraic skills to investment and interest rate problems showcases how foundational concepts in algebra are used to address practical, everyday issues. By mastering elementary algebra, students gain the tools needed to tackle more complex algebraic challenges in their academic and professional lives.