1. The problem asks to solve the equation $$\frac{2x+3}{x-1} = 3$$ for $x$. 2. The formula used here is to eliminate the denominator by multiplying both sides of the equation by $(x-1)$, assuming $x \neq 1$ to avoid division by zero. 3. Multiply both sides by $(x-1)$: $$\frac{2x+3}{x-1} \times (x-1) = 3 \times (x-1)$$ which simplifies to: $$2x + 3 = 3(x - 1)$$ 4. Expand the right side: $$2x + 3 = 3x - 3$$ 5. Rearrange terms to isolate $x$: $$2x + 3 - 3x = -3$$ $$-x + 3 = -3$$ 6. Subtract 3 from both sides: $$-x = -6$$ 7. Multiply both sides by $-1$: $$x = 6$$ 8. Check the solution by substituting $x=6$ back into the original equation: $$\frac{2(6)+3}{6-1} = \frac{12+3}{5} = \frac{15}{5} = 3$$ which is true. Therefore, the solution is $x = 6$.