Problem 69 - Algebra and Trigonometry

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

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

Step 1: Define Variables

Let \( x \) be the amount invested at the 5% interest rate, and let \( y \) be the amount invested at the 8% interest rate. We are tasked with finding the values of \( x \) and \( y \).

Step 2: Set Up Equations for Total Investment

According to the problem, the total investment is \$20,000. Therefore, we have the equation:\[ x + y = 20,000 \]

Step 3: Set Up Equations for Total Interest

The total interest earned is \$1180. The interest from the 5% investment is \( 0.05x \), and the interest from the 8% investment is \( 0.08y \). So, we have the equation:\[ 0.05x + 0.08y = 1180 \]

Step 4: Solve the System of Equations

We now have a system of two equations:1. \( x + y = 20,000 \)2. \( 0.05x + 0.08y = 1180 \)Substitute \( y = 20,000 - x \) from the first equation into the second equation:\[ 0.05x + 0.08(20,000 - x) = 1180 \]Simplify it:\[ 0.05x + 1600 - 0.08x = 1180 \]\[ -0.03x + 1600 = 1180 \]Rearrange and solve for \( x \):\[ -0.03x = 1180 - 1600 \]\[ -0.03x = -420 \]\[ x = \frac{-420}{-0.03} \]\[ x = 14,000 \]

Step 5: Calculate the Other Investment

Now we know \( x = 14,000 \), substitute it back into the equation \( x + y = 20,000 \) to find \( y \):\[ 14,000 + y = 20,000 \]\[ y = 20,000 - 14,000 \]\[ y = 6,000 \]

Step 6: Conclusion

The woman invested \\(14,000 in the account with a 5% interest rate and \\)6,000 in the account with an 8% interest rate.

Key Concepts

Simple Interest

Simple interest is a way to calculate the return on an investment based on the original amount or principal. It is not compounded, meaning the interest is earned solely on the initial amount invested, rather than on the principal plus any previously earned interest. This simplicity makes it an easy concept to grasp and apply.

In our investment problem, each account produces a different amount of interest because they have different rates. The interest for the account paying 5% is calculated as \(0.05 \times x\), where \(x\) is the amount invested in that account. Similarly, the interest for the 8% account is \(0.08 \times y\), where \(y\) is the amount invested at this higher rate.

Calculating simple interest is straightforward:

• Multiply the principal (amount invested) by the interest rate.
• The result gives the interest earned over the time period considered, usually a year.

This approach helps investors understand exactly how much return their investment generates without getting caught up in compounding complexities.

System of Equations

A system of equations consists of two or more equations with the same variables. Solving a system of equations means finding the values of the variables that satisfy all equations simultaneously. In investment problems, systems of equations are often used to model different conditions involving the same quantities.

In the exercise, the equations describe:

• The total amount invested: \(x + y = 20,000\)
• The total interest earned: \(0.05x + 0.08y = 1180\)

By solving these equations together, we can determine not only how much was invested in each account but also verify the consistency of both the total investment and the interest generated.

The power of this approach lies in its ability to handle multiple conditions simultaneously, ensuring solutions that satisfy all given constraints and requirements.

Algebraic Solutions

Algebraic solutions involve manipulating equations to isolate variables and solve for unknown values. This technique is essential in breaking down complex relationships into manageable parts.

In our example, substitution is an algebraic technique used to solve the system of equations. After expressing \(y\) in terms of \(x\) from the equation \(x + y = 20,000\), the expression \(y = 20,000 - x\) is substituted into the interest equation \(0.05x + 0.08y = 1180\).

This substitution makes solving the second equation easier by reducing it to one variable. Solving for \(x\) involves:

• Simplifying the equation.
• Using arithmetic operations to isolate \(x\).
• Dividing to find the value of \(x\).

Once \(x\) is known, it's straightforward to find \(y\) by substituting back into the initial equation. Algebraic methods offer a step-by-step path to uncover solutions, ensuring clarity and logical flow from problem to answer.