1. **State the problem:** We have two separate equations involving fractions: Equation 1: $$\frac{x - 2}{2} = \frac{1 - 9}{2x}$$ Equation 2: $$\frac{x - 1}{3x} = \frac{x - 3}{9x}$$ 2. **Solve Equation 1:** Start by simplifying the right side numerator: $$1 - 9 = -8$$ So the equation becomes: $$\frac{x - 2}{2} = \frac{-8}{2x}$$ 3. Cross-multiply to eliminate denominators: $$ (x - 2) \cdot 2x = 2 \cdot (-8) $$ Simplify: $$ 2x(x - 2) = -16 $$ 4. Expand the left side: $$ 2x^2 - 4x = -16 $$ 5. Bring all terms to one side: $$ 2x^2 - 4x + 16 = 0 $$ Divide entire equation by 2 to simplify: $$ x^2 - 2x + 8 = 0 $$ 6. Use the quadratic formula to solve for $x$: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=-2$, $c=8$. Calculate discriminant: $$ \Delta = (-2)^2 - 4 \cdot 1 \cdot 8 = 4 - 32 = -28 $$ Since $\Delta < 0$, there are no real solutions for Equation 1. 7. **Solve Equation 2:** $$ \frac{x - 1}{3x} = \frac{x - 3}{9x} $$ 8. Cross-multiply: $$ (x - 1) \cdot 9x = (x - 3) \cdot 3x $$ Simplify both sides: $$ 9x(x - 1) = 3x(x - 3) $$ 9. Expand both sides: $$ 9x^2 - 9x = 3x^2 - 9x $$ 10. Bring all terms to one side: $$ 9x^2 - 9x - 3x^2 + 9x = 0 $$ Simplify: $$ 6x^2 = 0 $$ 11. Divide both sides by 6: $$ x^2 = 0 $$ 12. Solve for $x$: $$ x = 0 $$ 13. **Check for restrictions:** In Equation 2, denominators are $3x$ and $9x$, so $x \neq 0$ to avoid division by zero. Therefore, $x=0$ is not a valid solution. 14. **Final answers:** - Equation 1 has no real solutions. - Equation 2 has no valid solutions because $x=0$ is excluded. Hence, both equations have no real solutions under the given conditions.