1. The problem is to solve the equation $$\frac{2x+3}{x-1} = 4$$ for $x$. 2. We use the formula for solving rational equations: multiply both sides by the denominator to eliminate the fraction. 3. Multiply both sides by $x-1$: $$\cancel{(x-1)} \cdot \frac{2x+3}{\cancel{x-1}} = 4(x-1)$$ which simplifies to $$2x + 3 = 4(x - 1)$$ 4. Expand the right side: $$2x + 3 = 4x - 4$$ 5. Rearrange terms to isolate $x$: $$2x + 3 - 4x = -4$$ $$\cancel{2x} + 3 - \cancel{4x} = -4$$ $$-2x + 3 = -4$$ 6. Subtract 3 from both sides: $$-2x + 3 - 3 = -4 - 3$$ $$-2x = -7$$ 7. Divide both sides by $-2$: $$\frac{-2x}{\cancel{-2}} = \frac{-7}{\cancel{-2}}$$ $$x = \frac{7}{2}$$ 8. Check that $x=\frac{7}{2}$ does not make the denominator zero. Since $x-1 = \frac{7}{2} - 1 = \frac{5}{2} \neq 0$, the solution is valid. Final answer: $$x = \frac{7}{2}$$