Problem 71 - A Graphical Approach to College Algebra

Question

Solve each application. Financing Expansion To get funds necessary for a planned expansion, a small company took out three loans totaling \(\$ 12,500 .\) The company was able to borrow some of the money at \(2 \% .\) It borrowed \(\$ 1000\) more than \(\frac{1}{2}\) the amount of the \(2 \%\) loan at \(3 \%\) and the rest at \(2.5 \% .\) The total annual interest was \(\$ 305 .\) How much did the company borrow at each rate?

Step-by-Step Solution

Answer

The company borrowed $5,000 at 2%, $3,500 at 3%, and $4,000 at 2.5%.

Step 1: Define Variables

Let's define the variables for the amounts borrowed at each interest rate. Let \( x \) be the amount borrowed at \(2\%\), \( y \) be the amount borrowed at \(3\%\), and \( z \) be the amount borrowed at \(2.5\%\).

Step 2: Set Up Equations from the Problem Statement

From the problem, we have three key pieces of information: 1. The total amount borrowed: \( x + y + z = \\( 12,500 \)2. \( y = \frac{1}{2}x + 1000 \) as \(1000\) more than half of \(x \).3. The total interest paid was \\) 305. So, the interest equation is:\( 0.02x + 0.03y + 0.025z = \$ 305 \).

Step 3: Substitute the Second Equation into Others

Substitute the expression for \( y \) from the second equation into both the first and third equations.- First equation becomes: \( x + (\frac{1}{2}x + 1000) + z = \\( 12,500 \)- Interest equation becomes: \[ 0.02x + 0.03(\frac{1}{2}x + 1000) + 0.025z = \\) 305 \]

Step 4: Simplify Substituted Equations

Simplify the equations:- Total amount equation: \( 1.5x + 1000 + z = \\( 12,500 \) simplifies to \( 1.5x + z = \\) 11,500 \)- Interest equation:\[ 0.02x + 0.015x + 30 + 0.025z = 305 \]\( 0.035x + 0.025z = 275 \)

Step 5: Solve the System of Equations

Use the equations \( 1.5x + z = 11,500 \) and \( 0.035x + 0.025z = 275 \) to solve for \( x \) and \( z \):- From the first equation, express \( z \) as \( z = 11,500 - 1.5x \).- Substitute \( z \) in the interest equation;\[ 0.035x + 0.025(11,500 - 1.5x) = 275 \]Simplify to find \( x \):\[ 0.035x + 287.5 - 0.0375x = 275 \]\[ -0.0025x = -12.5 \]\( x = 5,000 \)

Step 6: Calculate Remaining Loan Amounts

Substitute \( x = 5,000 \) to find \( y \) and \( z \):- \( y = \frac{1}{2} \times 5000 + 1000 = 3500 \)- Use \( z = 11,500 - 1.5 \times 5000 \)\( z = 11,500 - 7,500 = 4,000 \)

Key Concepts

Systems of Equations

A system of equations is a set of equations with multiple variables that you solve together. In our exercise, we have three equations:

• Total amount borrowed: \( x + y + z = 12,500 \)
• Relation between amounts: \( y = \frac{1}{2}x + 1000 \)
• Interest total: \( 0.02x + 0.03y + 0.025z = 305 \)

These equations are interconnected through the variables \(x\), \(y\), and \(z\). The goal is to find values for these variables that satisfy all the equations at once.
To solve a system like this, you often start by expressing one variable in terms of another, as done here. This helps transform the set of equations into ones with fewer variables, making it easier to solve. Often, you will substitute the variables back into the other equations to gradually reduce the complexity until you solve for all the unknowns.

Loan Interest Calculation

Loan interest calculation in algebra involves understanding how interest accrues over time. In this specific problem, different parts of the loan are charged different interest rates:

• \(2\%\) for \(x\)
• \(3\%\) for \(y\)
• \(2.5\%\) for \(z\)

Each portion of the loan contributes a part to the total interest of \$305.
The corresponding interest equation formed is \(0.02x + 0.03y + 0.025z = 305\).
The coefficients \(0.02\), \(0.03\), and \(0.025\) are the interest rates converted into decimal form. They multiply their respective loan amounts to calculate the interest cost for each part.
Understanding these calculations helps manage loans better and ensures you know how much you are paying in terms of interest.

Linear Equations

Linear equations are equations that involve terms that are either constants or the product of a constant and a variable. They graph as straight lines. In solving the loan amounts, we dealt with equations in linear form:

• The total amount equation: \(1.5x + z = 11,500\)
• The interest equation: \(0.035x + 0.025z = 275\)

Linear equations are simpler because they maintain a proportional relationship. When you substitute and simplify like we've done, the equations reflect this simplicity by eventually resolving to basic arithmetic.
The process usually involves rearranging terms to isolate variables, gradually simplifying the system so each unknown can be directly calculated. Here, expressing \(z\) in terms of \(x\) allowed for substitution, which broke down the equations further to solve for each variable accurately.