The denominator can't equal 0.
[tex]11(x-7) \not= 0 \\ x-7 \not=0 \\ x \not= 7 \\ \\ \\ \frac{(2x+1)(x-7)}{11(x-7)}=\frac{2x+1}{11} \\ \\
(2x+1)(x-7) \cdot 11= 11(x-7) \cdot (2x+1) \\
11(2x+1)(x-7)=11(2x+1)(x-7) \\ \\
x \in R \setminus \{7 \}[/tex]
The answer is D.