Problem 15 - Elementary and Intermediate Algebra

Question

Small-Business Loans. Chelsea took out three loans for a total of \(\$ 120,000\) to start an organic orchard. Her bank loan was at an interest rate of \(8 \%\) the small-business loan was at an interest rate of \(5 \%\) and the mortgage on her house was at an interest rate of \(4 \% .\) The total simple interest due on the loans in one year was \(\$ 5750 .\) The annual simple interest on the mortgage was \(\$ 1600\) more than the interest on the bank loan. How much did she borrow from each source?

Step-by-Step Solution

Answer

B = \$15,000, S = \$35,000, M = \$70,000.

Step 1: Define Variables

Let the amount borrowed from the bank be represented by \( B \), the small-business loan by \( S \), and the mortgage by \( M \).

Step 2: Set Up Equations for Loan Amounts

The total amount borrowed from all three sources is \( B + S + M = 120,000 \).

Step 3: Set Up Equations for Interest

The total simple interest in one year is given by \( 0.08B + 0.05S + 0.04M = 5750 \).

Step 4: Relate Mortgage and Bank Loan Interests

It is given that the annual interest on the mortgage was \ \$ 1600 more than the interest on the bank loan, which gives us \( 0.04M = 0.08B + 1600 \).

Step 5: Solve System of Equations

Now solve the system of equations: \( B + S + M = 120,000 \), \( 0.08B + 0.05S + 0.04M = 5750 \), and \( 0.04M = 0.08B + 1600 \). Start by isolating \( M \) in the third equation: \( M = 2B + 40,000 \).

Step 6: Substitute in Equations

Substitute \( M = 2B + 40,000 \) into the other two equations: \( B + S + (2B + 40,000) = 120,000 \) and \( 0.08B + 0.05S + 0.04(2B + 40,000) = 5750 \).

Step 7: Simplify Equations

Simplify the modified equations: \( 3B + S + 40,000 = 120,000 \) and \( 0.08B + 0.05S + 0.08B + 1600 = 5750 \). This simplifies to: \( 3B + S = 80,000 \) and \( 0.16B + 0.05S = 4150 \).

Step 8: Solve for S and B

Isolate \( S \) in the first simplified equation: \( S = 80,000 - 3B \). Substitute this into the second simplified equation: \( 0.16B + 0.05(80,000 - 3B) = 4150 \), resulting in \( 0.16B + 4,000 - 0.15B = 4150 \), which simplifies to \( 0.01B = 150 \), so \( B = 15,000 \).

Step 9: Find S

Substitute \( B = 15,000 \) back into \( S = 80,000 - 3B \): \( S = 80,000 - 45,000 = 35,000 \).

Step 10: Find M

Substitute \( B = 15,000 \) back into \( M = 2B + 40,000 \): \( M = 2(15,000) + 40,000 = 70,000 \).

Key Concepts

Simple Interest

Simple interest is a straightforward way to calculate the interest on an amount of money (loan) over a period of time. The formula to find simple interest is:
\text{Simple Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \( I = P \times R \times T \).
In this problem, we use simple interest for calculating how much Chelsea will pay on her loans. Each type of loan (bank, small-business, mortgage) has a different rate of interest. By knowing the total interest paid, we can determine how much she borrowed from each source.

Linear Equations

Linear equations are mathematical statements that showcase a relationship between two variables, generally in the form \( y = mx + c \).
In our problem, we use linear equations to set up relationships between the loan amounts and the interest paid. For example:
\[ B + S + M = 120,000 \]
This equation states that the total of the three loans equals $120,000. Similarly, we use another linear equation for the total interest:
\[ 0.08B + 0.05S + 0.04M = 5750 \].
Putting these equations together helps solve for the unknowns.

Loan Amounts

In this problem, Chelsea borrowed money from three different sources: a bank, a small-business loan, and a mortgage. Let’s call the amounts from these loans \( B \), \( S \), and \( M \) respectively.
Each loan contributes to the total amount borrowed:
\( B + S + M = 120,000 \).
This equation tells us how much money she altogether borrowed to start her orchard. The exercise also provides the interest rates for each loan type and the total interest paid, which helps us to further determine the specific amounts of each loan.

Substitution Method

The substitution method is a straightforward approach to solving systems of linear equations. It involves isolating one variable in an equation and then substituting that expression into another equation.
In this exercise, we isolate \( M \):
\( M = 2B + 40,000 \),
which is derived from the interest relationship: \( 0.04M = 0.08B + 1600 \).
Next, we substitute this value of \( M \) into the other equations to get simpler equations. This step-by-step substitution allows us to solve for the variables \( B \) and \( S \). Once \( B \) is determined, substituting back into the original equations lets us find the values of all other variables effortlessly.