Problem 1 - Thomas Calculus

Question

Expand the quotients in Exercises \(1-8\) by partial fractions. $$\frac{5 x-13}{(x-3)(x-2)}$$

Step-by-Step Solution

Answer

The expanded partial fractions are \( \frac{2}{x-3} + \frac{3}{x-2} \).

Step 1: Identify the Structure

We have the fraction \( \frac{5x-13}{(x-3)(x-2)} \). The denominator has two linear factors: \((x-3)\) and \((x-2)\). We will break this into partial fractions where each fraction has one of these linear factors as its denominator.

Step 2: Set Up the Partial Fractions

We express \( \frac{5x-13}{(x-3)(x-2)} \) as \( \frac{A}{x-3} + \frac{B}{x-2} \). Here, \( A \) and \( B \) are constants that we need to determine.

Step 3: Clear the Denominator

Multiply every term by \((x-3)(x-2)\) to clear the denominators:\[5x - 13 = A(x-2) + B(x-3)\] This eliminates the fraction and sets up an equation to solve for \(A\) and \(B\).

Step 4: Expand and Collect Like Terms

Expand the right side:\[A(x - 2) + B(x - 3) = Ax - 2A + Bx - 3B = (A + B)x - (2A + 3B)\]Now, we equate coefficients from both sides of the equation \( 5x - 13 = (A+B)x - (2A + 3B) \).

Step 5: Form and Solve the System of Equations

From equating coefficients, we get two equations:1. \(A + B = 5\)2. \(-2A - 3B = -13\)Solve this system of equations. First, multiply the first equation by 2:\(2A + 2B = 10\)Add this to the second equation:\(-2A - 3B + 2A + 2B = -13 + 10\)\(-B = -3\)\(B = 3\).Substitute \( B = 3 \) into \( A + B = 5 \):\(A + 3 = 5\)\(A = 2\).

Step 6: Write the Expanded Expression

Substitute \( A = 2 \) and \( B = 3 \) back into the partial fraction expansion:\[\frac{5x - 13}{(x-3)(x-2)} = \frac{2}{x-3} + \frac{3}{x-2}\] This gives the expression in its expanded form using partial fractions.

Key Concepts

Linear Factors

In the world of algebra, linear factors are essential building blocks. When we talk about linear factors, we are referring to expressions of the form \((x - a)\), where \(a\) is a constant.
This means the expression represents a straight line when plotted on a graph. They are called "linear" because they involve a first-degree term—\(x\)—with no exponents greater than one. Linear factors are crucial when we decompose fractions into simpler parts through partial fraction decomposition.
For instance, in the problem \(\frac{5x-13}{(x-3)(x-2)}\), the denominator contains two linear factors: \((x-3)\) and \((x-2)\). These linear factors tell us that the fraction is composed of parts related to these factors. By recognizing and isolating these factors, we can break down complex fractions, making them far easier to understand and solve.
Without identifying linear factors, solving such expressions would be cumbersome and complex.

System of Equations

To find values for unknowns, like \(A\) and \(B\) in our fraction decomposition, we often need to solve a system of equations. Here, a system of equations is a set of two or more equations with the same variables.
In our exercise, equating coefficients from both sides of the equation \(5x - 13 = (A+B)x - (2A + 3B)\), we derive two equations. These are:

• \(A + B = 5\)
• \(-2A - 3B = -13\)

Solving these equations simultaneously gives us the values of \(A\) and \(B\).
First, by manipulating the equations, such as multiplying or adding them together, you can isolate one variable and subsequently find the other. In this problem:
- Multiply the first equation by 2: \(2A + 2B = 10\).
- Then, add it to the second equation to eliminate \(A\), simplifying to \(-B = -3\), giving \(B = 3\).
- Substitute \(B\) back into \(A + B = 5\) to find \(A = 2\).
This strategy makes handling complex algebraic expressions quite approachable.

Fraction Expansion

Fraction expansion through partial fractions involves expressing a complex algebraic fraction as a sum of simpler fractions. This technique is particularly useful in integration and algebra problem-solving.
In our exercise, the fraction \(\frac{5x-13}{(x-3)(x-2)}\) is expanded into \(\frac{2}{x-3} + \frac{3}{x-2}\). Here's how it works:

• Identify and express the fraction in terms of its linear factors, assigning unknown values to each partial fraction.
• Clear the denominator by multiplying all parts by the denominator of the original fraction.
• This results in an equation involving terms without denominators, which makes it much simpler to work with.

After solving for the unknowns using a system of equations, you substitute back to find the expanded expression.
This technique of fraction expansion reduces complex division problems into manageable segments, streamlining the process of solving or integrating complex expressions. It's an invaluable tool for students tackling both high school and college-level algebra and calculus.