Sub-topic: Solve Quad Solve for \( x \) : a) \( 2 x+\frac{8}{x+1}=8 \) b) \( \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} \) c) \( \frac{6}{x+1}-\frac{3 x}{1-x^{2}}=\frac{2 x}{x-1} \)

To solve \( 2 x+\frac{8}{x+1}=8 \), first isolate the fraction: \( 2x + \frac{8}{x+1} = 8 \) leads to \( \frac{8}{x+1} = 8 - 2x \). Multiply through by \( (x+1) \) to eliminate the denominator, yielding: \( 8 = (8 - 2x)(x + 1) \). Expanding and rearranging leads to a quadratic equation you can solve using the quadratic formula or factoring. For the second equation \( \frac{x}{x-2}=\frac{1}{x-3}-\frac{2}{2-x} \), first simplify the right side. Note that \( \frac{2}{2-x} = -\frac{2}{x-2} \), so you rewrite it as \( \frac{1}{x-3} + \frac{2}{x-2} \). Combine the fractions on the right side and find a common denominator. This will give you a rational equation that you can cross-multiply to solve for \( x \). Remember, checking your solutions in the original equations is always a good practice to avoid extraneous results!