Problem 52 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Correspond to those in Exercises \(15-27\), Section \(5.1 .\) Use the results of your previous work to help you solve these problems. Michael Perez has a total of $$\$ 2000$$ on deposit with two savings institutions. One pays interest at the rate of $$6 \% /$$ year, whereas the other pays interest at the rate of $$8 \%$$ year. If Michael earned a total of $$\$ 144$$ in interest during a single year, how much does he have on deposit in each institution?

Step-by-Step Solution

Answer

Michael Perez has \(\$800\) deposited in the institution with a \(6\%\) interest rate and \(\$1200\) deposited in the institution with an \(8\%\) interest rate.

Step 1: Write down the system of equations based on the given information

The system of linear equations is: 1. \( x + y = 2000 \) 2. \( 0.06x + 0.08y = 144 \)

Step 2: Solve for one variable in terms of the other in equation 1

From the first equation, we have: \( x = 2000 - y \)

Step 3: Substitute this expression in equation 2

Substituting the expression for x in equation 2: \( 0.06(2000 - y) + 0.08y = 144 \)

Step 4: Solve for y

Now, we have an equation with one variable: \( 120 - 0.06y + 0.08y = 144 \) Combine the y terms: \( 0.02y = 24 \) Divide by 0.02: \( y = 1200 \)

Step 5: Use the value of y to find the value of x

Now that we have found the value of y, we can substitute it in the expression of x we found in Step 2: \( x = 2000 - 1200 \) So, \( x = 800 \)

Step 6: Interpret the results

Michael Perez has \(800 deposited in the institution with a \)6\%\( interest rate and \)1200 deposited in the institution with an \(8\%\) interest rate.

Key Concepts

Understanding System of Equations

When solving financial problems like Michael Perez's savings, we often come across a **system of equations**. This involves finding values that satisfy more than one equation at the same time. In this problem:

• The first equation \( x + y = 2000 \) combines two deposits into a total amount.
• The second \( 0.06x + 0.08y = 144 \) calculates total interest from both accounts.

These equations represent relationships between the variables, helping us figure out individual deposits that Michael made.

Simple Interest Problems

A common topic in applied mathematics is computing interest. **Simple interest problems** like this one involve calculating interest using a formula. Here, the interest earned is given as a known value, allowing us to reverse-engineer the problem:

• The interest formula used is \( \, \, \, \text{Interest} = \text{Principal} \times \text{Rate} \).
• Michael's interest from each deposit depends on its principal (deposit amount) and the rate of interest offered.
• By knowing the total interest Michael earned, we can work backward using our system of equations.

This method is helpful in breaking down interest-related problems into simpler terms.

Algebra Techniques

At the core of solving such problems is **algebra**. It involves manipulating equations to express variables in terms of one another. In our exercise:

• We first express one variable, \( x \), in terms of the other, \( y \), using one equation: \( x = 2000 - y \). This technique is called substitution.
• By substituting this into the other equation, it transforms it into a single-variable equation, which is easier to solve.
• Finally, we solve for \( y \) and backtrack to find \( x \), completing our solution.

Algebra helps us navigate through variables efficiently, reaching our answer systematically.

Working with Linear Equations

**Linear equations** are foundational in many algebra problems, like the one with Michael's deposits. They are equations where each term is either a constant or the product of a constant and a single variable:

• The equations \( x + y = 2000 \) and \( 0.06x + 0.08y = 144 \) are linear because they form straight-line graphs when plotted.
• Linear equations allow us to predict relationships between variables, assuming those relationships are proportional and additive.
• In this case, each deposit contributes linearly to the total deposit and to the earned interest.

This consistency makes linear equations easy to handle and very effective for modeling real-world financial scenarios.