1. We are asked to solve the inequality $$2 < \frac{2x + 3}{x + 1} < \frac{5}{2}$$ for real numbers $x$ such that $x > 1$. 2. Important: Since $x > 1$, the denominator $x + 1 > 0$, so we can multiply inequalities by $x + 1$ without reversing inequality signs. 3. Solve the left inequality: $$2 < \frac{2x + 3}{x + 1}$$ Multiply both sides by $x + 1$: $$2(x + 1) < 2x + 3$$ Simplify: $$2x + 2 < 2x + 3$$ Subtract $2x$ from both sides: $$2 < 3$$ This is always true. 4. Solve the right inequality: $$\frac{2x + 3}{x + 1} < \frac{5}{2}$$ Multiply both sides by $x + 1$: $$2x + 3 < \frac{5}{2}(x + 1)$$ Multiply out the right side: $$2x + 3 < \frac{5}{2}x + \frac{5}{2}$$ Multiply both sides by 2 to clear fractions: $$4x + 6 < 5x + 5$$ Subtract $4x$ from both sides: $$6 < x + 5$$ Subtract 5 from both sides: $$1 < x$$ 5. Since $x > 1$ is given, the right inequality holds for all $x > 1$. 6. Conclusion: For all real $x$ such that $x > 1$, the inequality $$2 < \frac{2x + 3}{x + 1} < \frac{5}{2}$$ holds true. Final answer: $$\boxed{\text{For all } x > 1, \quad 2 < \frac{2x + 3}{x + 1} < \frac{5}{2}}$$