Description for Correct answer:
Given that \( \Large 8x \equiv 6 \left(mod\ 14\right) \)

=> \( \Large 8x-6 = 14 P,\ \left(x\ \epsilon\ z\right) \)

=> \( \Large x = \frac{1}{8} \left(14P+6\right),\ \left(x\ \epsilon\ z\right) \)

=> \( \Large x = \frac{1}{4} \left(7P+3\right) \)

=> \( \Large x = 6,\ 13,\ 20,\ 27, 34,\ 41,\ 48... \)

Solution Set = = \( \Large x:x\ is\ a\ multiple\ of\ 12 \)

\( \Large \{ 6,\ 20,\ 34,\ 48... \} \cup\ \{ 13,\ 27,\ 41 \} \)

\( \Large \{ 6 \} \cup\ \{ 13 \} \)

Where \( \Large \left[ 6 \right],\ \left[ 13 \right] \) are equivalence classes of 6 and d13 respectively.