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 550: 15


$-\displaystyle \frac{8}{7(2x+1)}+\frac{4}{7(x-3)}$

Work Step by Step

First, factor the trinomial in the denominator (Two factors of $2\cdot(-3)=-6$ that add to $-5$ are $1$ and $-6$) $2x^{2}-5x-3=2x^{2}+x-6x-3$ $=x(2x+1)-3(2x+1)$ $=(2x+1)(x-3)$ (see p.541: Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.) $\displaystyle \frac{4}{(2x+1)(x-3)}=\frac{A}{(2x+1)}+\frac{B}{(x-3)}$ multiply both sides with $(2x-1)(x-3)\quad $(LCD) $4=A(x-3)+B(2x+1)$ $ 4=Ax-3A+2Bx+B\qquad$ /group like terms... $0x+4=x(A+2B)+(-3A+B)$ ... we now equate the coefficients of like powers... $\left\{\begin{array}{ll} I. & 0=A+2B\\ II. & 4=-3A+B \end{array}\right.\qquad $solve the system $(II+3\times I)$ eliminates $A...$ $4+0=-2A+2A+B+6B$ $4=7B$ $B=\displaystyle \frac{4}{7}$ back substitute: $0=A+2B$ $0=A+\displaystyle \frac{8}{7}$ $A=-\displaystyle \frac{8}{7}$ Rewrite the setup decomposition with $A=-\displaystyle \frac{8}{7}, B=\frac{4}{7}$ $\displaystyle \frac{4}{(2x+1)(x-3)}=\frac{-\frac{8}{7}}{(2x+1)}+\frac{\frac{4}{7}}{(x-3)}$ $\displaystyle \frac{4}{(2x+1)(x-3)}=-\frac{8}{7(2x+1)}+\frac{4}{7(x-3)}$
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