1. State the problem: Solve the equation $$\frac{3}{2}x + 3 - \frac{1}{2}x + 1 = \frac{1}{x} + 1$$ for $$x$$. 2. Simplify the left side by combining like terms: $$\frac{3}{2}x - \frac{1}{2}x + 3 + 1 = \left(\frac{3}{2} - \frac{1}{2}\right)x + 4 = 1x + 4 = x + 4$$ 3. Rewrite the equation: $$x + 4 = \frac{1}{x} + 1$$ 4. Subtract 1 from both sides: $$x + 3 = \frac{1}{x}$$ 5. Multiply both sides by $$x$$ to eliminate the denominator (note that $$x \neq 0$$): $$x(x + 3) = 1$$ 6. Expand the left side: $$x^2 + 3x = 1$$ 7. Bring all terms to one side to set the quadratic to zero: $$x^2 + 3x - 1 = 0$$ 8. Use the quadratic formula to solve for $$x$$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=3$$, and $$c=-1$$. Calculate the discriminant: $$\Delta = 3^2 - 4(1)(-1) = 9 + 4 = 13$$ Thus, $$x = \frac{-3 \pm \sqrt{13}}{2}$$ 9. Final answer: $$x = \frac{-3 + \sqrt{13}}{2}$$ or $$x = \frac{-3 - \sqrt{13}}{2}$$