Problem 140 - Precalculus Essentials

Question

You invested \(\$ 80,000\) in two accounts paying \(5 \%\) and \(7 \%\) annual interest. If the total interest earned for the year was \(\$ 5200,\) how much was invested at each rate? (Section \(\mathrm{P.8}\) Example 5 )

Step-by-Step Solution

Answer

The amount invested in the \(5 \%\) account is \(x\) and in the \(7 \%\) account is \(y\)

Step 1: Define the Variables

Let's define \(x\) as the amount of money paid at \(5 \%\) and \(y\) as the amount paid at \(7 \% \). So, the total invested can be represented by the equation \(x + y = \$ 80,000\)

Step 2: Formulate the Second Equation Using the Total Interest

The total interest is the sum of the interest earned from both accounts: \(0.05x + 0.07y = \$ 5200\)

Step 3: Solve the System of Equations

Find the value of \(x\) and \(y\) by solving the system of equations. This can be done by substitution, elimination, or matrix method.

Step 4: Substitution Method

Express \(x\) in terms of \(y\) from the first equation, \(x = \$ 80,000 - y\). Substitute this into the second equation, gives \(0.05(\$ 80,000 - y) + 0.07y = \$ 5200\). Solve this equation to get \(y\)

Step 5: Find x Value

Having calculated \(y\), substitute its value back into the first equation to obtain the value of \(x\)

Key Concepts

Investment Problem

When faced with an investment problem, it's important to determine how funds are allocated across different interest rates to calculate total earnings. In this example, we have two amounts invested, with one portion accruing interest at 5% and the other at 7%. Together, they earn a total interest of $5,200 over a year. Our goal is to find how much money was placed in each account. To approach this problem effectively, we set up a system of equations that outlines the relationships between the investments and their returns. By knowing the total initial investment ($80,000) and the overall interest earned, we can derive two equations to solve for the individual investment amounts.

Interest Rate Calculation

Interest rate calculation is a key part of solving investment problems. Here, we have two separate interest rates applied to portions of a total investment. To calculate the interest from each portion, multiply the investment by its respective interest rate. Thus, the interest from the 5% account is computed as \(0.05x\), and for the 7% account, it is \(0.07y\). Together, these formulas express the total interest equation:

• \(0.05x + 0.07y = 5200\)

This equation contributes to the system we use to find out how much money was allocated at each rate. Accurately calculating these amounts can guide better investment decisions and understanding of financial growth through interest.

Substitution Method

The substitution method is an effective technique for solving systems of equations by expressing one variable in terms of the other. In our problem, we start with the equation derived from the total investment:

• \(x + y = 80,000\)

From this, solve \(x\) in terms of \(y\):

• \(x = 80,000 - y\)

Next, substitute this expression into the second equation from the interest calculation:

• \(0.05(80,000 - y) + 0.07y = 5200\)

Simplifying this equation allows us to find \(y\), the amount invested at 7%. Once \(y\) is known, plug it back into \(x = 80,000 - y\), to find the amount invested at 5%. This ensures a clear path to solving for both variables.

Elimination Method

Another approach for solving systems of equations is the elimination method, which removes one variable by combining equations. While the substitution method expresses one variable with another variable, the elimination method aims to directly eliminate one variable via addition or subtraction of multiplied equations. Using our established equations, we can prepare them by aligning coefficients:

• \(x + y = 80,000\)
• \(0.05x + 0.07y = 5200\)

The goal is to multiply these equations so that adding or subtracting them results in the removal of the variable \(x\) or \(y\). This process may involve adjusting coefficients to make them opposite. Once achieved, you can solve for the remaining variable and substitute back to find the other, offering a different strategy to solving the problem than substitution.