Problem 70 - Algebra for College Students

Question

Solve \(i=\operatorname{Prt}\) for \(r\), given that \(i=\$ 356.50, P=\$ 1550\), and \(t=2\) years. Express \(r\) as a percent.

Step-by-Step Solution

Answer

The interest rate \( r \) is 11.5\%.

Step 1: Write Down the Given Formula

The formula given is \( i = P \, r \, t \). We are tasked to solve for \( r \).

Step 2: Substitute the Known Values

Substitute the known values into the equation. We have \( i = 356.50 \), \( P = 1550 \), and \( t = 2 \) years. The equation becomes \( 356.50 = 1550 \, r \, 2 \).

Step 3: Simplify the Equation

Simplify the right side of the equation: \( 1550 \, r \, 2 = 3100r \). The equation now is \( 356.50 = 3100r \).

Step 4: Solve for r

To solve for \( r \), divide both sides by 3100. This gives: \( r = \frac{356.50}{3100} \).

Step 5: Calculate the Value of r

Calculate \( r \) by performing the division: \( r = \frac{356.50}{3100} \approx 0.115 \).

Step 6: Convert r to a Percentage

Convert \( r \) to a percentage by multiplying by 100: \( r = 0.115 \times 100 = 11.5\% \).

Key Concepts

Interest Rate Calculation

The concept of interest rate calculation is fundamental in finance and mathematics. It helps in understanding how much interest one can earn or pay over a period of time. In the exercise provided, we use the formula for simple interest:

• \( i = P \cdot r \cdot t \)

where:

• \( i \) stands for the interest earned or paid.
• \( P \) is the principal amount or the initial amount of money.
• \( r \) represents the rate of interest.
• \( t \) is the time period for which the money is borrowed or invested, usually measured in years.

To calculate the interest rate, we aim to isolate \( r \) in the equation. Here, we are given the total interest \( i \), the principal \( P \), and the time \( t \). By substituting the known values into the formula, we can determine \( r \), the interest rate.

Solving for Variables in Equations

In mathematics, solving for a variable in an equation involves finding the value of the unknown that makes the equation true. When dealing with simple interest, you often need to solve for one variable when the others are known. For this exercise, we solve for \( r \). The original equation is:\( 356.50 = 1550 \cdot r \cdot 2 \)First, simplify the equation by multiplying the principal \( P \) and time \( t \), then isolate \( r \) by dividing both sides by the product of \( P \times t \). The action of isolating \( r \) is typical when solving equations where you need to express one variable in terms of others. This action reveals \( r \) as \( r = \frac{356.50}{3100} \). The goal is to simplify the problem by breaking it down into stepwise operations.

Algebraic Manipulation

Algebraic manipulation is a crucial skill for transforming equations, like converting a formula to solve for a different variable. This is important when applying mathematical formulas to real-world scenarios. It involves using operations such as addition, subtraction, division, or multiplication to rearrange the formula. In our exercise, we manipulate the simple interest formula to solve for \( r \). Initially, we rearrange:\( i = P \cdot r \cdot t \) into:\( r = \frac{i}{P \cdot t} \).We do this by dividing both sides by \( P \cdot t \), leaving \( r \) on one side of the equation. This type of rearrangement is fundamental in algebra, enabling you to handle various algebraic expressions effectively.

Elementary Algebra

Elementary algebra serves as the foundation for understanding more complicated math topics. It focuses on representing numbers with symbols and solving simple equations. Key concepts include:

• Identifying and using variables.
• Performing arithmetic operations.
• Simplifying expressions.
• Solving equations.

In the context of our exercise, elementary algebra aids in simplifying the equation \( 1550 \cdot r \cdot 2 = 3100r \) to \( 356.50 = 3100r \).These basic skills help in approaching more complex problems, guiding you from the initial equation to the final solution where \( r \) is expressed as a percentage, ensuring clarity and precision in financial calculations.