1. We are asked to solve the equation $gf(x) = 1$ where $f(x) = \frac{x+1}{2}$ and $g(x) = \frac{2x+3}{x-1}$ with $x \in \mathbb{R}$ and $x \neq 1$. 2. First, find the expression for $gf(x)$, which means $g(f(x))$. 3. Substitute $f(x)$ into $g$: $$gf(x) = g\left(\frac{x+1}{2}\right) = \frac{2\left(\frac{x+1}{2}\right) + 3}{\left(\frac{x+1}{2}\right) - 1}$$ 4. Simplify numerator: $$2 \times \frac{x+1}{2} = x+1$$ So numerator becomes: $$x + 1 + 3 = x + 4$$ 5. Simplify denominator: $$\frac{x+1}{2} - 1 = \frac{x+1}{2} - \frac{2}{2} = \frac{x+1-2}{2} = \frac{x-1}{2}$$ 6. So, $$gf(x) = \frac{x+4}{\frac{x-1}{2}} = (x+4) \times \frac{2}{x-1} = \frac{2(x+4)}{x-1}$$ 7. The equation $gf(x) = 1$ becomes: $$\frac{2(x+4)}{x-1} = 1$$ 8. Multiply both sides by $x-1$ (noting $x \neq 1$): $$2(x+4) = 1 \times (x-1)$$ 9. Expand both sides: $$2x + 8 = x - 1$$ 10. Bring all terms to one side: $$2x + 8 - x + 1 = 0$$ $$x + 9 = 0$$ 11. Solve for $x$: $$x = -9$$ 12. Check domain restrictions: $x \neq 1$, and $-9 \neq 1$, so solution is valid. Final answer: $$\boxed{x = -9}$$