## Algebra and Trigonometry 10th Edition

$[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$
We must first find the Partial Fraction Decomposition: $\frac{A}{x-3} + \frac{B}{x+2} = \frac{x+1}{(x-3)(x+2)}$ We must then solve for the constants: $A(x+2) + B(x-3) = x+1$ $Ax + 2A + Bx - 3B = x +1$ $Ax + Bx = x$ $2A - 3B = 1$ This can be represented with the following matrix: $\begin{bmatrix} 1 & 1 & |1\\ 2 & -3 & |1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & |1\\ 0 & -5 & |-1\\ \end{bmatrix}$ $-5B = -1$ $B = \frac{1}{5}$ $A + B = 1$ $A + \frac{1}{5} = 1$ $A = \frac{4}{5}$ The Partial Fraction is: $[\frac{1}{5}][\frac{4}{x-3} + \frac{1}{x+2}]$ Checking the Result: $[\frac{1}{5}]\frac{4x+8+x-3}{(x-3)(x+2)} = [\frac{1}{5}]\frac{5x + 5}{(x-3)(x+2)} = \frac{x+1}{(x-3)(x+2)}$ Therefore, the answer is correct.