Problem 47 - Contemporary Precalculus

Question

Lillian borrows \(\$ 10,000 .\) She borrows some from her friend at \(8 \%\) annual interest, twice as much as that from her bank at \(9 \%,\) and the remainder from her insurance company at 5\%. She pays a total of \(\$ 830\) in interest for the first year. How much did she borrow from each source?

Step-by-Step Solution

Answer

Based on the given information and the solution above, Lillian borrowed $3,000 from her friend, $6,000 from the bank, and $1,000 from the insurance company.

Step 1: Define the Unknowns

Let x be the amount Lillian borrowed from her friend, y be the amount she borrowed from the bank, and z be the amount she borrowed from the insurance company.

Step 2: Create the Equations

We have the following information: 1. The total amount borrowed (\(\$10,000\)). 2. The amount borrowed from the bank is twice the amount borrowed from her friend. 3. The interest rates for each source and the total interest paid (\(\$830\)). Using this information, we can create three equations: 1. x + y + z = 10000 (Total amount borrowed) 2. y = 2x (Bank loan is twice the friend's loan) 3. 0.08x + 0.09y + 0.05z = 830 (Total interest paid)

Step 3: Solve the System of Equations

We'll first eliminate y from the first equation and third equation using the second equation. Substitute y = 2x in the first equation: x + 2x + z = 10000 3x + z = 10000 (Equation 1) Substitute y = 2x in the third equation: 0.08x + 0.09(2x) + 0.05z = 830 0.08x + 0.18x + 0.05z = 830 0.26x + 0.05z = 830 (Equation 2)

Step 4: Solve the Equations

Solve Equation 1 for z: z = 10000 - 3x Substitute the expression for z from Equation 1 into Equation 2: 0.26x + 0.05(10000 - 3x) = 830 Simplify the equation: 0.26x + 500 - 0.15x = 830 Combine like terms and solve for x: 0.11x = 330 x = 3000 Now that we have the value of x, we can find the values of y and z: y = 2x = 2(3000) = 6000 z = 10000 - 3x = 10000 - 3(3000) = 1000

Step 5: Final Answer

Lillian borrowed \(\$3,000\) from her friend, \(\$6,000\) from the bank, and \(\$1,000\) from the insurance company.

Key Concepts

Interest Calculation

Interest calculation is a fundamental aspect of managing loans and investments. When you borrow money, you often have to pay interest, a fee for using someone else's money. In the case of Lillian, she borrowed money from three different sources, each with a different interest rate. Understanding how to calculate the total interest paid is essential.
The formula to calculate simple interest is:

• Interest = Principal × Rate × Time

For Lillian:

• She pays an 8% annual interest rate on her friend's loan, a 9% rate on the bank loan, and a 5% rate on the insurance company's loan.

These rates mean that the interest charged is a percentage of the amounts she borrowed. To find the total interest Lillian pays, we multiply each borrowed amount by its respective interest rate and sum them up. This gives us the third equation in our system, which embodies how these interest calculations combine to represent the total interest paid during the first year.

Substitution Method

The substitution method is a common way to solve systems of linear equations. It involves solving one of the equations for one variable and substituting this expression into another equation. This method simplifies the process, reducing the number of variables.
In Lillian's case:

• We have three equations representing the total amount borrowed, the relationship between loans, and the interest paid.

To apply substitution here, we solve the second equation, which states that the bank loan is twice the friend's loan (y = 2x).
This relationship allows us to express y in terms of x, which we then substitute into the other equations. By substituting y = 2x into the first and third equations, we eliminate y, and this brings us to the simplified equations involving only x and z. This reduction makes solving easier, as you can handle fewer unknowns at once.

Algebraic Expressions

Algebraic expressions are crucial in formulating and solving mathematical problems. They contain variables, constants, and operations and are used to represent real-world situations in a mathematical form.
For Lillian's loan problem, algebraic expressions help in setting up the equations:

• The total money borrowed is expressed as x + y + z = 10,000.
• The relationship between the bank and friend's loans is expressed as y = 2x.
• The total interest equation is represented as 0.08x + 0.09y + 0.05z = 830.

These expressions are derived from the problem's conditions. Each uses variables to represent unknown amounts borrowed and constants for interest rates.
Understanding how to manipulate these expressions is key to finding solutions.
By substituting, simplifying, and solving these algebraic expressions, we determine the exact amounts Lillian borrowed from each source. Thus, algebraic expressions offer a powerful tool for translating word problems into manageable mathematical forms.