\(\frac{12}{35x + 1}\)
\(\frac{1}{35(x + 1)}\)
\(\frac{12x}{35(x + 7)}\)
\(\frac{12}{35x + 35}\)
Correct answer is D
\(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) = \(\frac{1}{5(x + 1)}\) + \(\frac{1}{7(x + 1)}\)
= \(\frac{7 + 5}{35(x + 1)}\)
= \(\frac{12}{35(x + 1)}\)