1. The problem is to solve the equation $$\frac{2x+3}{x-1} = 4.$$\n\n2. We use the property that if $$\frac{a}{b} = c,$$ then $$a = bc,$$ provided $$b \neq 0.$$\n\n3. Multiply both sides of the equation by $$x-1$$ to eliminate the denominator:\n$$\cancel{\frac{2x+3}{x-1}} \times (x-1) = 4 \times (x-1)$$\nwhich simplifies to\n$$2x + 3 = 4(x - 1).$$\n\n4. Expand the right side:\n$$2x + 3 = 4x - 4.$$\n\n5. Rearrange terms to isolate $$x$$ on one side:\n$$2x + 3 - 4x = -4$$\n$$\Rightarrow -2x + 3 = -4.$$\n\n6. Subtract 3 from both sides:\n$$-2x + 3 - 3 = -4 - 3$$\n$$\Rightarrow -2x = -7.$$\n\n7. Divide both sides by $$-2$$ to solve for $$x$$:\n$$\cancel{-2x} \div \cancel{-2} = \frac{-7}{-2}$$\n$$x = \frac{7}{2}.$$\n\n8. Check the solution by substituting $$x = \frac{7}{2}$$ back into the original equation to ensure the denominator is not zero and the equality holds.\n\nFinal answer: $$x = \frac{7}{2}.$$