## College Algebra 7th Edition

(a) $\frac{x +3}{x - 3}$ (b) $\frac{x - 2}{x - 1}$ (c) $\frac{1}{x - 3}$ (d) $\frac{y - x}{2}$
These questions are solved using the same methods used in Question 2a-f, the only difference is they are divided instead of multiplied. (a) $\frac{x^2 + 4x + 3}{x^2 - 2x -3}$ = $\frac{(x +3)(x + 1)}{(x - 3)(x + 1)}$ = $\frac{x + 3}{x - 3}$ When factored like terms, in this case ($\mathcal{x}$ + 1), cancel out leaving $\frac{x + 3}{x - 3}$. (b) $\frac{2x^2 -3x -2}{x^2 - 1}$$\times$$\frac{x + 1}{2x + 1}$ = $\frac{(2x + 1)(x - 2)}{(x - 1)(x + 1)}$$\times$$\frac{x + 1}{2x + 1}$ = $\frac{x - 2}{x - 1}$ The like terms cancel as in (a). (c) $\frac{x^2 - x}{x^2 - 9}$ - $\frac{x + 1}{x + 3}$ = $\frac{x^2-x}{(x+3)(x-3)}$ + $\frac{-(x + 1)}{-(x +3)}$ = $\frac{1}{x-3}$ Factor, then cancel the like terms. (d) $\frac{\frac{1}{x}-{\frac{1}{y}}}{\frac{2}{xy}}$ = ($\frac{\frac{1}{x}-{\frac{1}{y}}}{\frac{2}{xy}}$)($\mathcal{xy}$)=$\frac{\frac{1\times y}{x \times y}-{\frac{1 \times x}{y \times x}}}{\frac{2}{xy} \times xy }$ = $\frac{y-x}{2}$ To subtract the numerator, it must have a common denominator. Which cancels with the bottom equation leaving the solution.