Problem 80 - College Algebra with Modeling and Visualization

Question

\(A\) sum of \(\$ 10,000\) is invested in three accounts that pay \(6 \%, 8 \%,\) and \(10 \%\) interest. Twice as much money is invested in the account paying \(10 \%\) as in the account paying \(6 \%,\) and the total annual interest from all three accounts is \(\$ 842\). (a) Write a system of equations whose solution gives the amount invested in each account. Be sure to state what each variable represents. (b) Solve the system of equations.

Step-by-Step Solution

Answer

Invest \$2000 at 6%, \$4000 at 8%, and \$4000 at 10%.

Step 1: Define the Variables

Let \( x \) be the amount invested in the account that pays 6%. Let \( y \) be the amount invested in the account that pays 8%, and \( z \) be the amount invested in the account that pays 10%. The condition given states \( z = 2x \).

Step 2: Total Investment Equation

Since the total investment amount is \$10,000, we can write the equation: \[ x + y + z = 10,000 \]

Step 3: Total Interest Equation

Using the interest rates and the total annual interest of \$842, we can write the equation: \[ 0.06x + 0.08y + 0.10z = 842 \]

Step 4: Substitute Expression for \( z \)

From Step 1, we know \( z = 2x \). Substitute this into the equations from Step 2 and Step 3.

Step 5: Simplify and Solve Equation 1 with Substitution

Substitute \( z = 2x \) into the total investment equation: \[ x + y + 2x = 10,000 \] This simplifies to \[ 3x + y = 10,000 \]

Step 6: Simplify and Solve Equation 2 with Substitution

Substitute \( z = 2x \) into the total interest equation: \[ 0.06x + 0.08y + 0.10(2x) = 842 \] Which simplifies to \[ 0.06x + 0.08y + 0.20x = 842 \] Combine terms to get: \[ 0.26x + 0.08y = 842 \]

Step 7: Solve the Simultaneous Equations

Use the two simplified equations: \[ 3x + y = 10,000 \] and \[ 0.26x + 0.08y = 842 \]. Solve these equations simultaneously using substitution or elimination methods to find \( x \) and \( y \).

Step 8: Calculate \( z \) using \( z = 2x \)

Once you have the values for \( x \) and \( y \), substitute \( x \) into \( z = 2x \) to find \( z \).

Step 9: Solution Interpretation

After solving, you should find that \( x = 2000 \), \( y = 4000 \), and \( z = 4000 \). Thus, \\(2000 is invested at 6%, \\)4000 at 8%, and \$4000 at 10%.

Key Concepts

Algebra

Algebra is often considered the backbone of mathematics, especially when dealing with unknowns in a problem. This exercise centers around using algebra to form equations that help solve for unknown investment amounts. ### Defining Variables and EquationsThe first step is to define variables to represent the unknowns. Here, they are set as follows:- Let \( x \) represent the amount invested at a 6% interest rate.- Let \( y \) represent the amount invested at an 8% interest rate.- Let \( z \) represent the amount invested at a 10% interest rate.With these variables, you create equations based on given conditions. The condition "twice as much in 10% as in 6%" translates algebraically to \( z = 2x \). ### Formulating a System of EquationsAlgebra allows us to express the total investment and the total interest earned in two separate equations:- Total investment: \( x + y + z = 10,000 \)- Total interest: \( 0.06x + 0.08y + 0.10z = 842 \)By substituting \( z = 2x \) from one equation into the others, these can be simplified and solved for the unknowns, which is a fundamental skill developed through practicing algebra.

Interest Calculation

Interest calculation is an essential concept in finance, determining how much profit an investment generates over time.### Annual Interest RateEach account in this exercise has a specific annual interest rate which indicates the percentage return on the investment:- The first account offers a 6% return on \( x \).- The second account offers an 8% return on \( y \).- The third account offers a 10% return on \( z \).### Calculating Total InterestThe total interest earned from all accounts adds up to \( \(842 \). The individual interests can be calculated by multiplying each account's principal by the associated interest rate:- Interest from the first account: \( 0.06x \)- Interest from the second account: \( 0.08y \)- Interest from the third account: \( 0.10z \) You then sum these expressions to equal \( \)842 \), creating one equation in the system to solve the problem.

Investment Distribution

Understanding investment distribution is key for determining how funds are allocated across different accounts. ### Constraints and ConditionsThe problem specifies certain conditions about how the investments are distributed:- The total investment is \( \(10,000 \).- Twice as much is invested in the account paying a 10% interest rate as in the account paying 6%.### Balancing the DistributionThe challenge lies in correctly distributing the total \( \)10,000 \) under these conditions. The system of equations helps ensure each account's portion is correctly balanced. By solving these equations:- You find that \( x = \(2000 \), meaning \( \)2000 \) is invested at 6%.- \( y = \(4000 \), so \( \)4000 \) is allocated at 8%.- Similarly, \( z = \(4000 \), confirming that \( \)4000 \) goes into the 10% account.This distribution optimally uses the total investment amount while meeting all specified conditions.