Problem 17 - College Algebra with Corequisite Support

Question

For exercises 17 and 18, use this scenario: A retired woman has $$\$ 50,000$$ to invest but needs to make $$\$ 6,000$$ a year from the interest to meet certain living expenses. One bond investment pays \(15 \%\) annual interest. The rest of it she wants to put in a CD that pays \(7 \%\). If we let \(x\) be the amount the woman invests in the \(15 \%\) bond how much will she be able to invest in the CD?

Step-by-Step Solution

Answer

The woman will invest \$18,750 in the CD.

Step 1: Define the Variables

In this problem, let \( x \) be the amount the woman invests in the bond with \( 15\% \) interest. The remaining amount, \( \$ 50,000 - x \), will be invested in the CD with \( 7\% \) interest.

Step 2: Set Up the Equation

To meet her yearly expense goal of \( \$6,000 \), the total interest from both investments must equal this amount. The interest from the bond investment is \( 0.15x \) and from the CD investment is \( 0.07(50,000 - x) \). Set up the equation for the total interest: \[ 0.15x + 0.07(50,000 - x) = 6,000 \]

Step 3: Simplify the Equation

Distribute the \( 0.07 \) in the equation:\[ 0.15x + 3,500 - 0.07x = 6,000 \]Combine like terms to simplify:\[ 0.08x + 3,500 = 6,000 \]

Step 4: Solve for \( x \)

Subtract \( 3,500 \) from both sides to isolate terms involving \( x \):\[ 0.08x = 2,500 \]Now, solve for \( x \) by dividing both sides by \( 0.08 \):\[ x = \frac{2,500}{0.08} = 31,250 \]

Step 5: Calculate the CD Investment Amount

The amount invested in the CD is the remaining part of her \( \\( 50,000 \) after investing \( \\) 31,250 \) in the bond. Calculate this as:\[ \\( 50,000 - \\) 31,250 = \$ 18,750 \]

Key Concepts

Linear Equations

Linear equations play a crucial role when solving investment problems in algebra, such as determining how much money to invest in different accounts to reach a specific interest target. A linear equation involves variables where each term is either a constant or the product of a constant and a single variable.

In this exercise, the linear equation is derived from considering the total interest from two different investments. By letting one variable, say \(x\), represent the amount invested in a bond, and expressing the other amount in terms of \(x\), a straightforward equation can be created. This is because we want the sum of interests from both investments to meet a certain goal, in this case, \( \$6,000\) annually.

Thus, the linear equation used here is:

• \(0.15x + 0.07(50,000 - x) = 6,000\)

This equation represents the relationship between the different investment components and the goal, showing how algebra can be used to systematically find the unknown values by manipulating the equation.

Interest Rates

Understanding interest rates is vital for calculating returns on investments, which directly impacts financial decision-making. Interest rate is the percentage at which invested money grows over time.

In this context, the retired woman wants to balance her portfolio with two different interest rates: a bond at \(15\%\) and a CD at \(7\%\). The difference in interest rates indicates varying levels of risk and return. Higher interest rates typically yield higher returns, but they may also come with increased risk.

To calculate interest from each investment:

• Interest from the bond is \(0.15x\).
• Interest from the CD is \(0.07(50,000 - x)\).

These formulas show how each investment's rate contributes to meeting the desired income. By combining these, one can figure out the best combination of investments to achieve the desired interest return. Analyzing interest rates in this way provides insight into how various financial strategies can be optimized.

Algebraic Equations

Algebraic equations help in determining the desired amounts for each investment option in order to reach a specific financial goal, such as earning a certain amount of interest annually. An algebraic equation is a statement of equality that contains variables and constants connected by operations.

Solving an algebraic equation involves finding the value of the variable that makes the statement true. In this example, we start with the equation:

\[ 0.15x + 0.07(50,000 - x) = 6,000 \]
This equation combines the individual expressions for the interest from each investment.

To find the value of \(x\), the amount invested in the bond, we must manipulate the equation by distributing, combining like terms, and isolating \(x\):

• Distribute the \(0.07\) to get \(0.15x + 3,500 - 0.07x = 6,000\).
• Simplify: \(0.08x + 3,500 = 6,000\).
• Subtract \(3,500\) to isolate terms with \(x\): \(0.08x = 2,500\).
• Divide both sides by \(0.08\): \(x = 31,250\).

This value for \(x\) represents the amount of money to invest in the bond. By solving this equation, one can determine precisely how to distribute the initial \(\$50,000\) investment to achieve the financial income target efficiently.