Answer
$\frac{3x+8}{3x+10}; x \ne -3, -\frac{10}{3};$
Work Step by Step
$\frac{3-\frac{1}{x+3}}{3+\frac{1}{x+3}}$
Take LCD in the numerator and denominator.
$= \frac{\frac{3(x+3)-1}{x+3}}{\frac{3(x+3)+1}{x+3}}$
$= \frac{\frac{3x+9-1}{x+3}}{\frac{3x+9+1}{x+3}}$
$= \frac{\frac{3x+8}{x+3}}{\frac{3x+10}{x+3}}; x \ne -3, -\frac{10}{3};$
$= \frac{3x+8}{x+3} \times \frac{x+3}{3x+10}; x \ne -3, -\frac{10}{3};$
$=\frac{3x+8}{3x+10}; x \ne -3, -\frac{10}{3};$
