Intermediate Algebra (6th Edition)

$\dfrac{4}{x}$
RECALL: $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b} \cdot \dfrac{d}{c}$ Use the rule above to have: $\\=\dfrac{4}{x-1} \cdot \dfrac{x-1}{x}$ Cancel the common factors to have: $\require{cancel} \\=\dfrac{4}{\cancel{x-1}} \cdot \dfrac{\cancel{x-1}}{x} \\=\dfrac{4}{x}$