College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.3 - Page 549: 12


$ \displaystyle \frac{3}{x-2}+\frac{2}{x+1}$

Work Step by Step

Both factors of the denominator are distinct linear factors... see p.541. Decomposition with Distinct Linear Factors In the Denominator: Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator. Setup: $\displaystyle \frac{5x-1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\qquad $ multiply both sides with $(x-2)(x+1)\quad$(LCD) $5x-1=A(x+1)+B(x-2)$ $ 5x-1=Ax+A+Bx-2B\qquad$ /group like terms... $5x-1=x(A+B)+(A-2B)$ ... we now equate the coefficients of like powers... $\left\{\begin{array}{l} 5=A+B\\ -1=A-2B \end{array}\right.\qquad $solve the system Subtract the second equation from the first: $5+1=A-A+B+2B$ $6=3B$ $B=2$ back substitute:$5=A+B$ $5=A+2$ $A=3$ Rewrite the setup decomposition with $A=3, B=2$ $\displaystyle \frac{5x-1}{(x-2)(x+1)}= \frac{3}{x-2}+\frac{2}{x+1}$
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