Which solution shown below contains an error? \( \frac{3}{x}+\frac{4}{x^{2}}=\frac{3 x}{x^{2}}+\frac{4}{x^{2}}=\frac{3 x+4}{x^{2}} \) \( \frac{1}{x+2}+\frac{1}{x+2}=\frac{2}{x+2}=\frac{1}{x+1} \) \( \frac{2}{x+1}+\frac{8}{x-6}=\frac{2 x-12}{(x+1)(x-6)}+\frac{8 x+8}{(x+1)(x-6)}=\frac{10 x-4}{(x+1)(x-6)} \)

1. For the first expression, write both fractions with the common denominator \(x^2\): \[ \frac{3}{x} = \frac{3x}{x^2}\quad \text{and} \quad \frac{4}{x^2} = \frac{4}{x^2}. \] Adding them gives: \[ \frac{3x}{x^2} + \frac{4}{x^2} = \frac{3x+4}{x^2}. \] This step is correct. 2. For the second expression, combine the two identical fractions: \[ \frac{1}{x+2}+\frac{1}{x+2} = \frac{2}{x+2}. \] However, the solution then incorrectly states: \[ \frac{2}{x+2} = \frac{1}{x+1}. \] There is no algebraic justification for replacing \(\frac{2}{x+2}\) with \(\frac{1}{x+1}\). This is an error. 3. For the third expression, first rewrite each fraction with the common denominator \((x+1)(x-6)\): \[ \frac{2}{x+1} = \frac{2(x-6)}{(x+1)(x-6)} = \frac{2x-12}{(x+1)(x-6)}, \] \[ \frac{8}{x-6} = \frac{8(x+1)}{(x+1)(x-6)} = \frac{8x+8}{(x+1)(x-6)}. \] Adding them gives: \[ \frac{2x-12}{(x+1)(x-6)} + \frac{8x+8}{(x+1)(x-6)} = \frac{2x-12+8x+8}{(x+1)(x-6)} = \frac{10x-4}{(x+1)(x-6)}. \] This step is correct. Thus, the solution with the error is the second one.