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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 16

In Exercises 16–24, write the partial fraction decomposition of each rational expression. x/(x - 3)(x + 2)

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Step 1: Recognize that the given rational expression \( \frac{x}{(x - 3)(x + 2)} \) is a proper fraction because the degree of the numerator (1) is less than the degree of the denominator (2). This means we can proceed with partial fraction decomposition.
Step 2: Set up the partial fraction decomposition. Since the denominator \((x - 3)(x + 2)\) consists of two distinct linear factors, the decomposition will take the form: \( \frac{x}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2} \), where \(A\) and \(B\) are constants to be determined.
Step 3: Multiply through by the common denominator \((x - 3)(x + 2)\) to eliminate the fractions. This gives: \( x = A(x + 2) + B(x - 3) \).
Step 4: Expand and simplify the right-hand side. Distribute \(A\) and \(B\) to get: \( x = A \cdot x + 2A + B \cdot x - 3B \). Combine like terms: \( x = (A + B)x + (2A - 3B) \).
Step 5: Equate coefficients of like terms from both sides of the equation. For the \(x\)-terms: \( A + B = 1 \). For the constant terms: \( 2A - 3B = 0 \). Solve this system of linear equations to find \(A\) and \(B\). Substitute these values back into the partial fraction decomposition.

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