1. **State the problem:** Determine the domain and range of the function $$y=\frac{x+3}{x-2}$$. 2. **Domain:** The domain is all real numbers except where the denominator is zero because division by zero is undefined. Set denominator equal to zero: $$x-2=0$$ $$x=2$$ So, the domain is all real numbers except $$x=2$$. 3. **Range:** To find the range, solve for $$x$$ in terms of $$y$$: $$y=\frac{x+3}{x-2}$$ Multiply both sides by $$x-2$$: $$y(x-2)=x+3$$ $$yx - 2y = x + 3$$ Bring all terms involving $$x$$ to one side: $$yx - x = 2y + 3$$ Factor out $$x$$: $$x(y - 1) = 2y + 3$$ If $$y-1=0$$, then $$y=1$$, which we must check separately. Otherwise, $$x = \frac{2y + 3}{y - 1}$$ Since $$x$$ must be real, the expression is defined for all $$y$$ except where denominator is zero: $$y - 1 = 0 \Rightarrow y = 1$$ Check if $$y=1$$ is in the range by substituting back: $$1 = \frac{x+3}{x-2}$$ $$1(x-2) = x + 3$$ $$x - 2 = x + 3$$ $$-2 = 3$$ (False) So, $$y=1$$ is not in the range. 4. **Final answers:** - Domain: $$\{x \in \mathbb{R} : x \neq 2\}$$ - Range: $$\{y \in \mathbb{R} : y \neq 1\}$$