Problem 61 - Elementary Algebra

Question

Set up an algebraic equation and then solve. Jane has her \(\$ 5,400\) savings invested in two accounts. She has part of it in a CD at \(3 \%\) annual interest and the rest in a savings account that earns \(2 \%\) annual interest. If the simple interest earned from both accounts is \(\$ 140\) for the year, then how much does she have in each account?

Step-by-Step Solution

Answer

Jane has $3,200 in the CD account and $2,200 in the savings account.

Step 1: Define the Variables

Let \( x \) be the amount Jane invested in the CD account at \( 3\% \), and \( y \) be the amount invested in the savings account at \( 2\% \). We know the total investment is \( x + y = 5400 \).

Step 2: Set Up Interest Equation

The interest from the CD account is \( 0.03x \) and from the savings account is \( 0.02y \). The total interest earned is \( 0.03x + 0.02y = 140 \).

Step 3: Solve the System of Equations

We have two equations: \( x + y = 5400 \) and \( 0.03x + 0.02y = 140 \). Solve the first equation for \( y \): \( y = 5400 - x \). Substitute \( y \) in the second equation: \( 0.03x + 0.02(5400 - x) = 140 \).

Step 4: Simplify and Solve for x

Substitute and simplify: \( 0.03x + 108 - 0.02x = 140 \). Combine like terms: \( 0.01x = 32 \). Solve for \( x \): \( x = \frac{32}{0.01} = 3200 \).

Step 5: Find the Value of y

Substitute \( x = 3200 \) back into the equation \( y = 5400 - x \): \( y = 5400 - 3200 = 2200 \).

Key Concepts

Understanding Simple Interest

Simple interest is a basic yet essential concept in finance that helps us understand how money grows over time when it is invested or borrowed. It is calculated using the formula:

• Simple Interest = Principal × Rate × Time

Here, the "Principal" is the initial amount of money invested or borrowed, the "Rate" is the percentage of interest per annum, and "Time" is the duration for which the money is invested or borrowed, expressed in years.
In the problem involving Jane, her interest earned is a result of her investments in two accounts with different interest rates. By using simple interest, we can calculate how much she earns annually from each of her accounts based on the amount invested.
Recognizing the importance of simple interest calculations in real-life financial decisions can help individuals make better investment choices. For instance, knowing how much you will earn or owe at the end of a period is crucial for financial planning.

Solving with System of Equations

A system of equations is a set of equations with multiple variables. Solving these equations simultaneously allows us to find unknown values given particular conditions.
In Jane's scenario, we have two equations:

• Equation 1: \( x + y = 5400 \) (total investment)
• Equation 2: \( 0.03x + 0.02y = 140 \) (total interest)

These represent Jane's total investment amount and her total interest from both accounts, respectively. Solving a system of equations involves finding the numerical values of all unknown variables (in this case, \( x \) and \( y \)).
The common method used here is substitution. We solved Equation 1 for \( y \) and substituted it in Equation 2. This process allowed us to express everything in terms of \( x \) and solve for its value.
Using systems of equations is a powerful technique in tackling all kinds of problems. It's especially useful in finance, engineering, and mathematics where relationships between different quantities need to be understood and managed.

Effective Mathematical Problem-Solving

Mathematical problem-solving revolves around understanding the problem, designing a strategy, and executing the solution effectively. In the context of the exercise, this involves multiple steps:

• Comprehending the problem statement: identifying Jane's total investment and the respective interest rates of her accounts.
• Developing a plan: establishing variables, setting up equations based on given data, and deciding on which method to use to solve them.
• Executing the solution: applying mathematical concepts like substitution to solve the equations for desired values.
• Checking solutions: substituting back to ensure the results meet all original conditions.
  This systematic approach ensures accuracy and encourages logic-based solutions to complex problems. Ultimately, mathematical problem-solving is a skill that improves with practice and can enhance one's analytical capabilities in everyday life.