1. The problem is to solve the equation $$\frac{2x+3}{x-1} = 4.$$\n\n2. We use the formula for solving rational equations: multiply both sides by the denominator to eliminate the fraction. Important rule: check for values that make the denominator zero, as they are excluded from the solution. Here, $x \neq 1$.\n\n3. Multiply both sides by $x-1$: $$\cancel{\frac{2x+3}{x-1}} \times (x-1) = 4 \times (x-1) \implies 2x+3 = 4(x-1).$$\n\n4. Expand the right side: $$2x + 3 = 4x - 4.$$\n\n5. Rearrange terms to isolate $x$: $$2x + 3 - 4x = -4 \implies -2x + 3 = -4.$$\n\n6. Subtract 3 from both sides: $$-2x + \cancel{3} - \cancel{3} = -4 - 3 \implies -2x = -7.$$\n\n7. Divide both sides by $-2$: $$\cancel{-2}x / \cancel{-2} = \frac{-7}{-2} \implies x = \frac{7}{2}.$$\n\n8. Check that $x=\frac{7}{2}$ does not make the denominator zero (it doesn't), so this is the valid solution.\n\nFinal answer: $$x = \frac{7}{2}.$$