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: 10


$\displaystyle \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$

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: $\displaystyle \frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\qquad /\times x(x-1)\quad$(LCD) $1=A(x-1)+Bx$ $1=Ax-A+Bx \qquad$ /group like terms... $0\cdot x+1=x(A+B)+(-A)$ (Emphasize the 0 linear constant on the LHS, as we now equate the coefficients of like powers...) $\left\{\begin{array}{l} 0=A+B\\ 1=-A \end{array}\right.\qquad $solve the system The second equation gives us: $A=-1$ back substitute: $0=-1+B$ $B=1$ Rewrite the setup decomposition with $A=-1, B=1$ $\displaystyle \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$
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