Chapter 7.5, Problem 4E

To determine

To find the value of A and B.

Answer to Problem 4E

The value is, $A=\frac{1}{2}\text{ and }B=-\frac{1}{2}$

Explanation of Solution

Given information:The equation is , $\frac{3}{x^2-9}=\frac{A}{x-3}+\frac{B}{x+3}$ .

Calculation: The required solution is obtained as,

  $\begin{array}{l}\frac{3}{x^2-9}=\frac{A}{x-3}+\frac{B}{x+3}\\ \frac{3}{x^2-9}=\frac{A\left(x+3\right)+B\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ \frac{3}{x^2-9}=\frac{A\left(x+3\right)+B\left(x-3\right)}{x^2-9}\\ 3=A\left(x+3\right)+B\left(x-3\right)\left(\text{cancel denominator both side}\right)\end{array}$

Set $x=-3$ to solve for B.

  $\begin{array}{l}3=A\left(-3+3\right)+B\left(-3-3\right)\\ 3=A\left(0\right)+B\left(-6\right)\\ B=-\frac{1}{2}\end{array}$

Set $x=3$ to solve for A.

  $\begin{array}{l}3=A\left(3+3\right)+B\left(3-3\right)\\ 3=A\left(6\right)+B\left(0\right)\\ A=\frac{1}{2}\end{array}$

Hence, the value is, $A=\frac{1}{2}\text{ and }B=-\frac{1}{2}$