## College Algebra (10th Edition)

$\dfrac{3}{5x}; x \ne 0. \frac{9}{2}$
Factor each polynomial completely to obtain: $=\dfrac{3(2x-9)}{5(x)} \cdot \dfrac{2}{2(2x-9)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{3\cancel{(2x-9)}}{5(x)} \cdot \dfrac{\cancel{2}}{\cancel{2}\cancel{(2x-9)}} \\=\dfrac{3}{5x}; x \ne 0, \frac{9}{2}$