Problem 45 - Precalculus Mathematics for Calculus

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 = (a+b)/2; B = (a-b)/2.

Step 1: Understand the Expression

We have the expression \(\frac{a x+b}{x^2-1}\) and need to express it as \(\frac{A}{x-1}+\frac{B}{x+1}\). The denominator \(x^2-1\) can be factored as \((x-1)(x+1)\). We will use partial fraction decomposition to express the given fraction in the desired form.

Step 2: Set Up the Partial Fractions

Set the equation \(\frac{a x+b}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\). To solve for \(A\) and \(B\), combine the right side into a single fraction: \(\frac{A(x+1) + B(x-1)}{(x-1)(x+1)}\).

Step 3: Equate Numerators

With the denominators the same, equate the numerators: \(a x+b = A(x+1) + B(x-1)\). Expand the right side: \(A(x+1) + B(x-1) = Ax + A + Bx - B = (A + B)x + (A - B)\).

Step 4: Solve for A and B

From the equation: \(a x + b = (A + B)x + (A - B)\), compare the coefficients. For \( x\): we have \(a = A + B\). For the constant term: \(b = A - B\). Solve these equations simultaneously:1. \(A + B = a\) 2. \(A - B = b\).

Step 5: Solve System of Equations

Add the two equations to eliminate \(B\):\( (A + B) + (A - B) = a + b \), yielding \(2A = a + b\). Thus, \(A = \frac{a + b}{2}\).Now, subtract the second equation from the first to eliminate \(A\):\( (A + B) - (A - B) = a - b \), giving \(2B = a - b\). Thus, \(B = \frac{a - b}{2}\).

Step 6: Final Expression for A and B

The values of \(A\) and \(B\) in terms of \(a\) and \(b\) are:\[ A = \frac{a + b}{2} \]\[ B = \frac{a - b}{2} \]

Key Concepts

Polynomial Fractions

Polynomial fractions are expressions in which one polynomial is divided by another polynomial. This kind of algebraic fraction can sometimes look complex, but is helpful for modeling various real-world problems and solving advanced equations.
In the original exercise, the fraction \( \frac{a x + b}{x^2 - 1} \) is a polynomial fraction, because both the numerator, \( ax + b \), and the denominator, \( x^2 - 1 \), are polynomials.
Polynomial fractions often need to be simplified or transformed in calculations, especially in integration and other calculus operations. Through a method known as partial fraction decomposition, we can break down complex polynomial fractions into simpler components.
By decomposing \( \frac{a x + b}{x^2 - 1} \) into \( \frac{A}{x-1} + \frac{B}{x+1} \), we make it easier to work with, because each simpler fraction has a denominator that's just a linear term \((x-?1)\).
This technique is vital in calculus because it simplifies the integration process.

Algebraic Expressions

An algebraic expression is a mathematical phrase that includes numbers, variables, and operations. They are fundamental components in algebra and are used to describe real-world scenarios mathematically.
In the given exercise, we see an algebraic expression in the form of \( a x + b \). This is a linear expression as it contains variables raised to the power of one and constants. Algebraic expressions help us perform mathematical operations, such as addition, subtraction, and multiplication, on variables.
Transforming the expression \( a x + b \) into its partial fraction form shows how variables and constants interact in an equation. The given expression is comparable to a template that we fill in as needed with the appropriate values of the variables.
The expressions within the partial fraction setup: \( A(x+1) + B(x-1) \), demonstrate how algebraic expressions can be equated and manipulated to solve for unknowns using systems of equations. This understanding improves our ability to solve mathematical problems and can be applied to both theoretical and practical problems.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same variables. The aim is to find values for those variables that satisfy all the equations in the system.
In the exercise, we have the system:\( A + B = a \) and \( A - B = b \). Solving this system helps us find the values of \( A \) and \( B \).
To do this, combine the equations strategically to eliminate one variable, enabling easier solution for the other. By adding the equations, we cancel out \( B \) and calculate \( A \): \( 2A = a + b \), so \( A = \frac{a + b}{2} \).
By subtracting the equations, we cancel out \( A \) and calculate \( B \): \( 2B = a - b \), so \( B = \frac{a - b}{2} \). This is a fundamental algebraic technique, often used in various branches of mathematics and science.
Understanding how to solve systems of linear equations empowers you to tackle a broad array of problems, from balancing chemical equations to analyzing financial forecasts.