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: 9


The partial fraction function is $\frac{3}{\left( x-3 \right)}-\frac{2}{\left( x-2 \right)}$.

Work Step by Step

Finding the partial fraction of the provided expression is given below, $\frac{x}{\left( x-3 \right)\left( x-2 \right)}=\frac{A}{\left( x-3 \right)}+\frac{B}{\left( x-2 \right)}$ Then, multiply $\left( x-3 \right)\left( x-2 \right)$ on both sides, $\begin{align} & \left( x-3 \right)\left( x-2 \right)\frac{x}{\left( x-3 \right)\left( x-2 \right)}=\left( x-3 \right)\left( x-2 \right)\frac{A}{\left( x-3 \right)}+\left( x-3 \right)\left( x-2 \right)\frac{B}{\left( x-2 \right)} \\ & x=\left( x-2 \right)A+\left( x-3 \right)B \\ & =Ax-2A+Bx-3B \\ & x=x\left( A+B \right)-2A-3B \end{align}$ …… (I) Now, compare the coefficient of equation (I), $A+B=1$ …… (II) $-2A-3B=0$ …… (III) And multiply equation (II) by 2 and add with equation (III), Then $\begin{align} & -B=2 \\ & B=-2 \end{align}$ By putting the value of B in equation (II), $\begin{align} & A-2=1 \\ & A=3 \end{align}$ Therefore, the partial fraction is given below, $\frac{x}{\left( x-3 \right)\left( x-2 \right)}=\frac{3}{\left( x-3 \right)}-\frac{2}{\left( x-2 \right)}$ Thus, the partial fraction of the given expression is, $\frac{3}{\left( x-3 \right)}-\frac{2}{\left( x-2 \right)}$.
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