Problem 86 - Algebra for College Students

Question

Solve each problem. The Friendly Bob Loan Company gives each applicant a rating, \(t,\) from 0 to 10 according to the applicant's ability to repay, a higher rating indicating higher risk. The interest rate, \(r\), is then determined by the function \(r=0.02 t+0.15\) a) If your rating were \(8,\) then what would be your interest rate? b) Sketch the graph of the equation for \(t\) ranging from 0 to 10

Step-by-Step Solution

Answer

a) The interest rate is 0.31. b) Sketch the graph of \(r = 0.02t + 0.15\) from \(t = 0\) to \(t = 10\).

Step 1: Understand the Given Function

The interest rate function is given by the equation: \[ r = 0.02t + 0.15 \] Here, \(r\) is the interest rate and \(t\) is the rating.

Step 2: Substitute the Given Rating

For part a), we need to find the interest rate when the rating \(t\) is 8. Substitute \(t = 8\) into the equation: \[ r = 0.02(8) + 0.15 \]

Step 3: Calculate the Interest Rate

Now, perform the calculation: \[ r = 0.16 + 0.15 = 0.31 \] The interest rate when the rating is 8 is 0.31.

Step 4: Sketch the Graph

For part b), sketch the graph of the equation \(r = 0.02t + 0.15\) for \(t\) ranging from 0 to 10. Start by plotting several points: - When \(t = 0\), \(r = 0.15\) - When \(t = 5\), \(r = 0.25\) - When \(t = 10\), \(r = 0.35\) Draw a straight line through these points.

Key Concepts

Solving Equations

Understanding how to solve equations is essential for tackling mathematical problems effectively. An equation is a mathematical statement that asserts the equality of two expressions. In our example, we have the linear equation: \( r = 0.02t + 0.15 \). Here’s a simple guide to solving an equation:

- Identify the variables and constants: In our function, \( r \) is the interest rate (dependent variable), and \( t \) is the rating (independent variable). The numbers 0.02 and 0.15 are constants.
- Substitute the given values into the equation: For part a), we replace \( t \) with 8 in the equation: \( r = 0.02(8) + 0.15 \).

- Perform the calculation: Multiply 0.02 by 8 and then add 0.15 to get \( r = 0.16 + 0.15 = 0.31 \).
This step-by-step approach helps you systematically solve almost any equation.

Graphing Linear Equations

Graphing linear equations gives a visual representation of the relationship between two variables. A linear equation, like \( r = 0.02t + 0.15 \), forms a straight line when plotted on a graph. Follow these steps to sketch a graph:

- Create a table of values: Choose several values for the independent variable (\( t \)), and calculate the corresponding values of the dependent variable (\( r \)).

- Plot the points: For example, when \( t = 0 \), \( r = 0.15 \); when \( t = 5 \), \( r = 0.25 \); and when \( t = 10 \), \( r = 0.35 \).

- Draw the line: Connect these points with a straight line extending across the range of \( t \) from 0 to 10.
This visual aid is particularly helpful for understanding and interpreting the relationship between the two variables in real-life applications.

Substitution Method

The substitution method is a fundamental technique in algebra used to solve equations. It involves replacing one variable with an equivalent expression containing another variable. Let’s explore this method using our example:

- Identify the equation and the values: We start with \( r = 0.02t + 0.15 \) and a given value of \( t \), which is 8.

- Substitute the given value: Replace \( t \) with 8: \( r = 0.02(8) + 0.15 \).
- Simplify the equation: Perform the arithmetic to find \( r \): Multiply 0.02 by 8 to get 0.16, then add the constant 0.15, resulting in \( r = 0.31 \).
The substitution method is straightforward and helps to translate problems into easily solvable algebraic expressions, making it a powerful tool for solving real-world problems.