Problem 41 - Prealgebra and Introductory Algebra

Question

Two students are working with the equation \(A=P(1+i) .\) State whether the two students' answers are equivalent. a. When asked to solve the equation for \(i\), one student answered \(i=\frac{A}{P}-1\) and the other student answered \(i=\frac{A-P}{P}\) b. When asked to solve the equation for \(i\), one student answered \(i=-\frac{P-A}{P}\) and the other student answered \(i=\frac{A-P}{P}\)

Step-by-Step Solution

Answer

Yes, the answers provided by the students in both parts a and b of the exercise are equivalent. Both students have accurately solved the equation for 'i', showing a correct understanding of algebraic manipulation.

Step 1: Analyze Student Solutions for Part a

Start with the equation given, \(A=P(1+i)\), and solve for 'i'. Now, isolate 'i' on one side. By dividing both sides by P and subtracting 1, you get \(i=\frac{A}{P}-1\). Simplifying this expression, it turns into \(i=\frac{A-P}{P}\). Both students' solutions for part 'a' are the same as this result, so their answers are equivalent.

Step 2: Analyze Student Solutions for Part b

Starting again with the given equation \(A=P(1+i)\), solve for 'i'. The first student's solution was \(i=-\frac{P-A}{P}\), which when simplified, gives \(i=\frac{A-P}{P}\), equivalent to the second student's solution for 'i'. Hence, the solutions provided by both students in part 'b' are equivalent.

Key Concepts

Solving Equations

Solving equations is a foundational skill in algebra that often involves isolating the variable of interest. In this exercise, students are expected to solve for the variable \( i \) in the equation \( A=P(1+i) \). The goal here is to rearrange the equation so that \( i \) stands alone on one side.
Breaking it down into simpler steps involves:

• Transposing terms to move everything we do not want on the same side as \( i \) to the other side of the equation.
• Performing operations like addition, subtraction, multiplication, or division, hoping to clear unnecessary coefficients or terms away from \( i \).

For instance, dividing both sides by \( P \) and subtracting 1 provides a neat way to isolate \( i \) such that you arrive at the expressions \( i = \frac{A}{P} - 1 \). Regular practice in solving equations will enable one to quickly recognize equivalent forms or mistakes, making it a vital skill for students.

Mathematical Equivalence

Mathematical equivalence is all about determining whether different expressions or equations express the same value or concept. It is key in both simplifying expressions and verifying the correctness of different solutions.
In the problem, the challenge was to see if two student solutions were equivalent expressions for \( i \) after manipulations of the equation \( A = P(1+i) \).
By careful expansion and rearrangement:

• The solution \( i = \frac{A}{P} - 1 \) can be shown to be equivalent to \( i = \frac{A - P}{P} \) by simplifying.
• The insight comes from understanding that subtracting 1 is identical to subtracting a fraction with the same denominator: \( \frac{P}{P} \).

Recognizing these transformations helps in verifying if two seemingly different equations are actually the same, a critical skill for checking work and understanding deeper relationships in mathematics.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operations (like addition and multiplication). Understanding how to manipulate them is essential for solving equations and analyzing mathematical problems.
In this scenario, expressions for \( i \) were transformed: \( i = \frac{A}{P} - 1 \) and \( i = \frac{A - P}{P} \), both coming from the original algebraic expression \( A = P(1+i) \).

• Algebraic manipulation often involves combining like terms and using distribution laws, such as the distributive property \( a(b+c) = ab + ac \).
• Understanding how to rearrange and simplify algebraic expressions to other forms is key for flexible thinking. This flexibility is what allowed both students to arrive at equivalent solutions despite different initial transformations.

Practice with algebraic expressions enhances the ability to visualize and solve complex problems by breaking them into simpler pieces.