Problem 35 - Introductory Algebra for College Students

Question

Involve dual investments. Your grandmother needs your help. She has 50,000 dollar to invest. Part of this money is to be invested in noninsured bonds paying \(15 \%\) annual interest. The rest of this money is to be invested in a government-insured certificate of deposit paying \(7 \%\) annual interest. She told you that she requires a total of 6000 dollar per year in extra income from these investments. How much money should be placed in each investment?

Step-by-Step Solution

Answer

To fulfill her needs, the grandmother should invest approx. \(28571.43\) dollars in noninsured bonds and approx. \(21428.57\) dollars in government-insured certificates of deposit.

Step 1: Identify the Variables

Let \(x\) represent the amount of money invested in noninsured bonds and \(y\) represent the amount invested in a government-insured certificate of deposit.

Step 2: Setup the Equations

From the problem, we have the following two equations based on the two conditions given: \(x + y = 50000\) (the total amount invested) and \(0.15x + 0.07y = 6000\) (the total interest earned).

Step 3: Solve the Equations

These equations can be solved by substitution or elimination methods. If we rearrange the first equation to get \(y = 50000 - x\) and substitute it into the second equation, we get: \(0.15x + 0.07(50000 - x) = 6000\). Simplifying this will yield the value for \(x\). Substituting \(x\) into either equation will then yield the value for \(y\).

Key Concepts

Understanding Investment Problems

Investment problems often involve deciding how to allocate money among different options to reach a certain financial goal. In this scenario, we have two investment choices: noninsured bonds and a government-insured certificate of deposit. The challenge is to figure out how much money to place in each option to achieve a specific income from the interest generated. These types of problems typically provide:

• Total amount available for investment (in this case, $50,000).
• Interest rates for each investment type (here it's 15% for bonds and 7% for the deposit).
• A goal, such as a target income from the investments (such as $6,000 in extra income).

The goal is to set up equations that represent these conditions and solve for the unknowns. Understanding how to set up these equations is crucial for tackling investment problems successfully.

The Substitution Method for Solving Equations

The substitution method is a technique used to solve systems of equations. It's particularly useful when one of the equations is easily solved for one of the variables. With the substitution method, follow these steps:

• Start by solving one of the equations for one of the variables. In our case: \( y = 50000 - x \).
• Then substitute this expression into the other equation. This means placing \( y = 50000 - x \) into the equation \( 0.15x + 0.07y = 6000 \).
• Simplify and solve the resulting single-variable equation. After substitution, the equation becomes \( 0.15x + 0.07(50000 - x) = 6000 \).
• Once you find the value of one variable, use it to find the other variable by substituting back into one of the original equations.

This method works well if the equation can be easily rearranged, making it straightforward to solve for one variable first.

Using the Elimination Method for Equations

The elimination method, also known as the addition method, involves combining equations to eliminate one of the variables. This method is another powerful tool for solving systems of linear equations, especially when both equations can be aligned conveniently. Here's a simplified approach:

• Arrange both equations in a standard form, aligning them vertically. For this problem, our equations are \( x + y = 50000 \) and \( 0.15x + 0.07y = 6000 \).
• Multiply one or both of the equations by a constant to make the coefficients of one of the variables opposite. Here, you could multiply the entire first equation to match the y-coefficient of the second equation.
• Add or subtract the equations so that one of the variables gets eliminated. Once eliminated, you'll have a single equation with a single variable to solve.
• Use the value found from solving the single variable equation to substitute back into one of the original equations to find the other variable.

The strength of the elimination method is in its ability to handle systems where substitution might be too cumbersome or when calculations resulting from substitution get complex. It is a handy alternative that can simplify your solving process.