Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 11


The partial fraction is $\frac{7}{\left( x-9 \right)}-\frac{4}{\left( x+2 \right)}$.

Work Step by Step

We find the partial fraction of the provided expression as given below, $\frac{3x+50}{\left( x-9 \right)\left( x+2 \right)}=\frac{A}{\left( x-9 \right)}+\frac{B}{\left( x+2 \right)}$ Then, multiply $\left( x-9 \right)\left( x+2 \right)$ on both sides, $\begin{align} & \left( x-9 \right)\left( x+2 \right)\frac{3x+50}{\left( x-9 \right)\left( x+2 \right)}=\left( x-9 \right)\left( x+2 \right)\frac{A}{\left( x-9 \right)}+\left( x-9 \right)\left( x+2 \right)\frac{B}{\left( x+2 \right)} \\ & 3x+50=\left( x+2 \right)A+\left( x-9 \right)B \\ & =Ax+2A+Bx-9B \\ & 3x+50=x\left( A+B \right)+2A-9B \end{align}$ …… (I) Now, compare the coefficient of equation (I), $A+B=3$ …… (II) $2A-9B=50$ …… (III) Then, multiply equation (II) by 2 and subtract from equation (III), Then, $11B=44$ $\begin{align} & B=\frac{44}{-11} \\ & =-4 \end{align}$ Now, put the value in equation (II), $\begin{align} & A-4=3 \\ & =7 \end{align}$ Therefore, $\frac{3x+50}{\left( x-9 \right)\left( x+2 \right)}=\frac{7}{\left( x-9 \right)}-\frac{4}{\left( x+2 \right)}$ Thus, the partial fraction of the given expression is $\frac{7}{\left( x-9 \right)}-\frac{4}{\left( x+2 \right)}$.
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