Write the partial fraction decomposition of the rational expression. Check your result algebraically. \[ \frac{9}{x^{2}-9 x} \]

To decompose the given rational expression, we first factor the denominator: \[ x^2 - 9x = x(x - 9) \] Now, we express the rational function in terms of its partial fractions: \[ \frac{9}{x(x - 9)} = \frac{A}{x} + \frac{B}{x - 9} \] Next, we need to find the values of \(A\) and \(B\). To do this, we'll clear the fractions by multiplying the entire equation by the denominator \(x(x - 9)\): \[ 9 = A(x - 9) + Bx \] Now, we can expand the right side: \[ 9 = Ax - 9A + Bx \] Combine like terms: \[ 9 = (A + B)x - 9A \] Now, we can set up the system of equations by equating coefficients. The constant term gives us: 1. \(-9A = 9\) and from the coefficients of \(x\): 2. \(A + B = 0\) From the first equation, we solve for \(A\): \[ A = -1 \] Substituting \(A\) into the second equation gives: \[ -1 + B = 0 \implies B = 1 \] Thus, we have: \[ A = -1 \quad \text{and} \quad B = 1 \] The partial fraction decomposition is: \[ \frac{9}{x(x - 9)} = \frac{-1}{x} + \frac{1}{x - 9} \] Now, let's check our result algebraically by recombining the fractions: \[ \frac{-1}{x} + \frac{1}{x - 9} = \frac{-1(x - 9) + 1(x)}{x(x - 9)} = \frac{-x + 9 + x}{x(x - 9)} = \frac{9}{x(x - 9)} \] Since we have verified that the left-hand side equals the right-hand side, our partial fraction decomposition is indeed correct.