Problem 45 - Algebra and Trigonometry

Question

Determine \(A\) and \(B\) in terms of \(a\) and \(b\) $$ \frac{a x+b}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1} $$

Step-by-Step Solution

Answer

\( A = \frac{a+b}{2} \) and \( B = \frac{a-b}{2} \).

Step 1: Clear denominators

To determine the values of \(A\) and \(B\), start by eliminating the denominators in the given equation. Multiply both sides of the equation by \((x-1)(x+1)\). This gives us the equation: \\[ a x + b = A(x + 1) + B(x - 1) \]

Step 2: Expand the right-hand side

Next, expand the right-hand side of the equation: \\[ A(x+1) + B(x-1) = Ax + A + Bx - B = (A+B)x + (A-B) \]

Step 3: Compare coefficients

Now compare the coefficients of like terms on both sides of the equation \(ax + b = (A+B)x + (A-B)\). This gives us two separate equations: \ 1. \(a = A + B\) \ 2. \(b = A - B\)

Step 4: Solve the system of equations

Use the method of substitution or addition to solve for \(A\) and \(B\). First, add the two equations: \\(a = A + B\) \ \(b = A - B\) \ Adding these: \\[ a + b = (A + B) + (A - B) = 2A \] \So, \(A = \frac{a+b}{2}\).\Substituting \(A\) back into the equation \(a = A + B\), we get: \\[ a = \frac{a+b}{2} + B \rightarrow B = a - \frac{a+b}{2} \rightarrow B = \frac{2a - a - b}{2} = \frac{a-b}{2} \]

Step 5: Substitute solved values

Substitute the values obtained into the terms: \\( A = \frac{a+b}{2} \) \\( B = \frac{a-b}{2} \)

Key Concepts

Rational Functions

Rational functions are mathematical expressions that represent the ratio of two polynomials. In simpler terms, these are functions of form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). These functions can appear in a variety of mathematical contexts and are essential in calculus, algebra, and even in real-world applications like physics and engineering.
One common task with rational functions is expressing them in different forms, which can help in understanding their behavior or in solving specific problems. For example, in the exercise given, we aim to express the rational function in terms of partial fraction decomposition. This involves simplifying the function into a sum of simpler fractions, making it easier to integrate or differentiate if needed.
Partial fraction decomposition is particularly useful when dealing with integrals or solving differential equations, and it's a powerful tool for breaking down complex rational expressions into more manageable pieces.

Algebraic Manipulation

Algebraic manipulation involves a series of operations that help simplify or solve equations and expressions. In the context of partial fraction decomposition, it often includes clearing denominators, expanding expressions, and comparing coefficients. Let's break these steps down.

• Clearing Denominators: The first step involves multiplying both sides by the common denominator of all terms. This eliminates fractions and leaves us with a polynomial equation. In our exercise, multiplying by \( (x-1)(x+1) \) cleared the denominators, simplifying the equation to polynomial format.
• Expanding Expressions: Once the denominators are cleared, the next step is to expand the expressions if necessary. This involves distributing terms to write each polynomial in standard form, making it easier to compare like terms later on.
• Comparing Coefficients: By aligning terms on both sides of an equation, you can form a system of equations containing the coefficients. This allows you to isolate unknowns and solve for them, as done when identifying \(A\) and \(B\) in the exercise.

These techniques are foundational in algebra and are crucial for solving various types of mathematical problems. Mastering them provides a strong base for more advanced mathematical studies.

System of Equations

A system of equations is a set of two or more equations with common variables. Solving such systems is crucial in many areas of mathematics and science, often leading to solutions that describe how variables interrelate.
In our exercise, solving a simple system of two equations allowed us to find the values for \( A \) and \( B \). Here's a quick breakdown of the steps involved:

• Formulating Equations: Once coefficients are compared, equations relate each unknown to the given parameters. In our case, the system was \( a = A + B \) and \( b = A - B \).
• Solving Through Addition or Substitution: We used addition to eliminate one variable. Adding both equations gave a direct solution for \( A \). Substituting \( A \) back into one of the original equations then provided the value of \( B \).
• Validation: Finally, it's always a good idea to substitute the found values back into the original problem to ensure they satisfy all equations. Mistake-free solutions are key to confidence in mathematical manipulations.

Solving systems of equations is a vital skill, used widely in solving real-world problems like optimization, economics, and engineering scenarios. With practice, finding such relationships becomes second nature.