1. **State the problem:** We are given the function $$y = \frac{ax + b}{cx + d}$$ where $$a, b, c, d$$ are constants and $$c \neq 0$$. We want to show that for $$c \neq 0$$, $$\frac{ax + b}{cx + d} = \frac{a}{c} + \frac{bc - ad}{c(cx + d)}$$ and then determine the equation of the horizontal asymptote. 2. **Start with the given function:** $$y = \frac{ax + b}{cx + d}$$ 3. **Rewrite the right side expression:** We want to express $$\frac{ax + b}{cx + d}$$ as a sum of two terms: $$\frac{a}{c} + \frac{bc - ad}{c(cx + d)}$$ 4. **Find a common denominator to combine the right side:** $$\frac{a}{c} + \frac{bc - ad}{c(cx + d)} = \frac{a(cx + d)}{c(cx + d)} + \frac{bc - ad}{c(cx + d)}$$ 5. **Expand the numerator of the first fraction:** $$\frac{a(cx + d)}{c(cx + d)} = \frac{acx + ad}{c(cx + d)}$$ 6. **Add the two fractions:** $$\frac{acx + ad}{c(cx + d)} + \frac{bc - ad}{c(cx + d)} = \frac{acx + ad + bc - ad}{c(cx + d)}$$ 7. **Simplify the numerator:** $$acx + ad + bc - ad = acx + bc$$ 8. **So the sum is:** $$\frac{acx + bc}{c(cx + d)} = \frac{c(ax + b)}{c(cx + d)}$$ 9. **Cancel the common factor $$c$$ in numerator and denominator:** $$\frac{\cancel{c}(ax + b)}{\cancel{c}(cx + d)} = \frac{ax + b}{cx + d}$$ This matches the original function, so the equality is proven. 10. **Determine the horizontal asymptote:** As $$x \to \pm \infty$$, the term $$\frac{bc - ad}{c(cx + d)} \to 0$$ because the denominator grows without bound. Therefore, the horizontal asymptote is: $$y = \frac{a}{c}$$ **Final answer:** $$\frac{ax + b}{cx + d} = \frac{a}{c} + \frac{bc - ad}{c(cx + d)}$$ Horizontal asymptote: $$y = \frac{a}{c}$$