Question

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

Explanation:

Step1: Check first expression

To add $\frac{3}{x}+\frac{4}{x^2}$, get common denominator $x^2$:
$\frac{3}{x} = \frac{3x}{x^2}$, so $\frac{3x}{x^2}+\frac{4}{x^2}=\frac{3x+4}{x^2}$. This is correct.

Step2: Check second expression

Add $\frac{1}{x+2}+\frac{1}{x+2}$:
Sum of numerators is $1+1=2$, so $\frac{2}{x+2}$. Simplifying $\frac{2}{x+2}$ to $\frac{1}{x+1}$ is wrong, as you cannot cancel 2 from numerator and denominator (denominator is $x+2$, not $2(x+1)$).

Step3: Check third expression

To add $\frac{2}{x+1}+\frac{8}{x-6}$, get 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 gives $\frac{(2x-12)+(8x+8)}{(x+1)(x-6)}=\frac{10x-4}{(x+1)(x-6)}$. This is correct.

Answer:

$\boldsymbol{\frac{1}{x+2}+\frac{1}{x+2} = \frac{2}{x+2} = \frac{1}{x+1}}$ (this solution contains an error)