91Ó°ÊÓ

Rational expressions are a type of fraction where both the numerator and the denominator are polynomials. In this exercise, we have a rational expression given by \( \frac{x}{(x-3)(x-2)} \). Understanding rational expressions is crucial as they form the basis of solving various mathematical problems, including partial fraction decomposition.

In general, rational expressions can be simplified, added, subtracted, multiplied, and divided, but special attention is needed to handle their domains. The domain consists of all real numbers except those that make the denominator zero because division by zero is undefined.

For the rational expression \( \frac{x}{(x-3)(x-2)} \), the domain excludes \( x = 3 \) and \( x = 2 \) because these values make the denominator zero.

To eliminate denominators in partial fraction decomposition, we multiply both sides of the equation by the least common denominator. This allows us to simplify the equation and focus solely on the numerators, which makes solving for unknown constants easier. In our example, the least common denominator is \((x-3)(x-2)\).

By multiplying both sides of the equation \( \frac{x}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2} \) by this common denominator, we obtain the simplified equation: \[ x = A(x-2) + B(x-3) \].

This step is crucial as it clears the fractions, allowing us to handle a pure algebraic expression that's ready for substitution and simplification.

Solving linear equations involves finding the values of unknown variables that make the equation true. After eliminating denominators, the expression becomes \( x = A(x-2) + B(x-3) \). Here, we have a linear equation in terms of \( A \) and \( B \).

To find \( A \) and \( B \), we can choose specific values for \( x \) that simplify the calculations. Setting \( x = 2 \), we have: \( 2 = A(2-2) + B(2-3) \), simplifying to \( 2 = -B \), so \( B = -2 \).

Similarly, setting \( x = 3 \), the equation becomes: \( 3 = A(3-2) + B(3-3) \), leading to \( 3 = A \). Thus, \( A = 3 \). These steps show how solving linear equations directly leads us to determine the unknown constants.

Determining the constants in partial fraction decomposition is the final step after simplifying equations. In our example, once the denominators are eliminated, we solve for \( A \) and \( B \) with selected values of \( x \) that make one term zero and the other term easy to solve.

With \( x = 2 \), we isolate \( B \) and find \( B = -2 \). With \( x = 3 \), we isolate \( A \) and find \( A = 3 \). These values are crucial because they provide us with the numeric coefficients that completely describe the initial rational expression in terms of simpler fractions.

Substituting these back into the decomposed equation gives us the partial fraction decomposition of \( \frac{x}{(x-3)(x-2)} = \frac{3}{x-3} - \frac{2}{x-2} \).