1. The problem is to solve the equation $$\frac{2x+3}{x-1} = 4.$$\n\n2. We use the property that if $$\frac{a}{b} = c$$ and $$b \neq 0$$, then $$a = bc$$. Here, $$a = 2x+3$$, $$b = x-1$$, and $$c = 4$$.\n\n3. Multiply both sides by $$x-1$$ to eliminate the denominator:\n$$\cancel{\frac{2x+3}{\cancel{x-1}}} \times \cancel{x-1} = 4 \times (x-1)$$\nwhich simplifies to\n$$2x + 3 = 4(x - 1).$$\n\n4. Expand the right side:\n$$2x + 3 = 4x - 4.$$\n\n5. Bring all terms to one side to isolate $$x$$:\n$$2x + 3 - 4x + 4 = 0$$\nwhich simplifies to\n$$-2x + 7 = 0.$$\n\n6. Solve for $$x$$:\n$$-2x = -7$$\n$$x = \frac{-7}{-2} = \frac{7}{2}.$$\n\n7. Check that $$x \neq 1$$ to avoid division by zero in the original equation. Since $$\frac{7}{2} \neq 1$$, the solution is valid.\n\nFinal answer: $$x = \frac{7}{2}.$$