Algebra: A Combined Approach (4th Edition)

$\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}=\dfrac{2x}{x-3}$
$\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}$ Evaluate the sum indicated in the numerator and the substraction indicated in the denominator: $\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}=\dfrac{\dfrac{2x+6}{x}}{\dfrac{x^{2}-9}{x^{2}}}=...$ Evaluate the division: $...=\dfrac{2x+6}{x}\div\dfrac{x^{2}-9}{x^{2}}=\dfrac{x^{2}(2x+6)}{x(x^{2}-9)}=...$ Take out common factor $2$ from the parentheses in the numerator and factor the parentheses in the denominator. After that, simplify: $...=\dfrac{2x^{2}(x+3)}{x(x-3)(x+3)}=\dfrac{2x^{2}}{x(x-3)}=\dfrac{2x}{x-3}$