Problem 26 - Precalculus with Limits

Question

\(\begin{array}{l}{\text { Solving a System by Substitution In Exercises }} \\\ {25-28 \text { , you are given the total annual interest earned }} \\ {\text { from a total of } \$ 12,000 \text { invested in two funds paying the }} \\\ {\text { given rates of simple interest. Write and solve a system of }} \\\ {\text { equations to find the amount invested at each rate. }}\end{array}\) \(\begin{array}{l}{\text { Annual Interest }} \\ {\$ 630}\end{array}\) \( \begin{array}{ll}{\text { Rate } 1} & {\text { Rate } 2} \\ {4 \%} & {6 \%}\end{array} \)

Step-by-Step Solution

Answer

The amount invested at 4% is $4500 and the amount invested at 6% is $7500.

Step 1: Set Up The Equations

Using the information given, two equations can be set up. Let's denote the amount invested at 4% as \(x\) and the amount invested at 6% as \(y\). The first equation representing the total investment is: \(x + y = 12000\). The second equation representing the total interest earned is: \(0.04x + 0.06y = 630\)

Step 2: Solve Equation 1 for \(x\)

We can make the system of equations easier to solve by expressing one variable in terms of the other. We express \(x\) in terms of \(y\) from the first equation, which would give us \(x = 12000 - y\)

Step 3: Substitute \(x\) Into Equation 2

Now let us substitute the expression from step 2 into the second equation from step 1: \(0.04x + 0.06y = 630\) becomes \(0.04(12000 - y) + 0.06y = 630\)

Step 4: Solve For \(y\)

Now let's solve the equation formulated in step 3 for \(y\). Distribute the 0.04 to get \(480 - 0.04y + 0.06y = 630\). Combine like terms to get \(0.02y = 150\). Solve for \(y\) to get \(y = 7500\)

Step 5: Solve For \(x\)

Now let's substitute \(y = 7500\) into the first equation to solve for \(x\). \(x + 7500 = 12000\) gives \(x = 4500\)

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of linear equations. In this method, we solve at least one of the equations for one variable in terms of the others. Once we have a substitution equation, we replace this expression in the other equation. This method simply uses the concept of replacing variables to reduce the number of equations.

Here is how it generally works:

• Choose one of the equations and solve it for one of the variables.
• Substitute the expression from the first step into the other equation.
• Simplify and solve for the remaining variable.
• Substitute back to find the value of the other variable.

In our problem, we used the substitution method to find how much money was invested at two different rates. This method is particularly helpful because once you express a variable in terms of others, it simplifies the computation.

Simple Interest

Simple interest is a straightforward way to calculate the interest earned on a sum of money invested. The formula to compute simple interest is: \( I = P \times r \times t \) where:

• \( I \) is the interest earned,
• \( P \) is the principal amount (initial investment),
• \( r \) is the rate of interest per time period, expressed as a decimal, and
• \( t \) is the time the money is invested.

In this exercise, we compute the annual interest from different investments. Understanding simple interest is crucial as it allows us to determine how much each investment contributes to the total interest earned.

Linear Equations

Linear equations are mathematical expressions that form a line when graphed on a coordinate plane. In their simplest form, they are \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

In the context of our problem, two linear equations are used. These represent:

• The total amount invested, i.e., \(x + y = 12000\),
• The total interest earned, i.e., \(0.04x + 0.06y = 630\).

The solution of these equations gives the amounts invested at different interest rates. Understanding the limitations and properties of linear equations makes solving real-world investment problems easier.

Investment Problems

Investment problems often require you to figure out how money is distributed across different accounts or investments to achieve a certain financial outcome. These problems often involve systems of equations as they tie together several unknowns like amounts invested and interest rates.

For example, in our exercise we need to find the sums invested at different interest rates that yield a specific total interest. With

• multiple interest rates,
• a known total investment,
• and a given total interest,

we can set up equations to solve for unknowns. Mastering approaches like the substitution method and comprehending simple and linear equations is crucial for effectively tackling investment problems.