1. **State the problem:** We are given the system of equations: $$2x - 2y = 3$$ and $$xy = 1$$ We need to solve for $x$. 2. **Rewrite the first equation:** Divide both sides of the first equation by 2: $$x - y = \frac{3}{2}$$ 3. **Express $y$ in terms of $x$:** From the above, $$y = x - \frac{3}{2}$$ 4. **Substitute $y$ into the second equation:** Given $xy = 1$, substitute $y$: $$x \left(x - \frac{3}{2}\right) = 1$$ 5. **Simplify and form a quadratic equation:** $$x^2 - \frac{3}{2}x = 1$$ Multiply both sides by 2 to clear the fraction: $$2x^2 - 3x = 2$$ Bring all terms to one side: $$2x^2 - 3x - 2 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-3$, and $c=-2$. Calculate the discriminant: $$\Delta = (-3)^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$ 7. **Calculate the roots:** $$x = \frac{3 \pm 5}{4}$$ 8. **Find the two possible values for $x$:** - When $+$: $$x = \frac{3 + 5}{4} = \frac{8}{4} = 2$$ - When $-$: $$x = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}$$ 9. **Final answer:** $$x = 2 \quad \text{or} \quad x = -\frac{1}{2}$$