a) Solve the unknown in each of the following: 8) \( \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} \)

To solve the equation \( \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} \), first simplify the right side. Notice that \( \frac{2}{2-x} \) can be rewritten as \( -\frac{2}{x-2} \) since they are negatives. Thus, the equation becomes: \[ \frac{x}{x-2} = \frac{1}{x-3} + \frac{2}{x-2} \] Now, combine the fractions on the right-hand side under a common denominator \( (x-3)(x-2) \): \[ \frac{x}{x-2} = \frac{(x-2) + 2(x-3)}{(x-3)(x-2)} \] This simplifies to: \[ \frac{x}{x-2} = \frac{x-2+2x-6}{(x-3)(x-2)} = \frac{3x-8}{(x-3)(x-2)} \] Now, cross-multiply to eliminate the fractions: \[ x(x-3) = (x-2)(3x-8) \] Expanding both sides gives: \[ x^2 - 3x = 3x^2 - 8x - 6x + 16 \] \[ x^2 - 3x = 3x^2 - 14x + 16 \] Rearranging the terms leads to: \[ 0 = 2x^2 - 11x + 16 \] Now, apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 16}}{2 \cdot 2} = \frac{11 \pm \sqrt{121 - 128}}{4} = \frac{11 \pm \sqrt{-7}}{4} \] Since the discriminant is negative, we end up with two complex solutions: \[ x = \frac{11 \pm i\sqrt{7}}{4} \] Thus, the solutions are: \[ x = \frac{11}{4} + \frac{i\sqrt{7}}{4} \quad \text{and} \quad x = \frac{11}{4} - \frac{i\sqrt{7}}{4} \]