The solution set of 8x equiv 6 pmod{14},x in | vedclass

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(C) Given the linear congruence $8x \equiv 6 \pmod{14}$.This is equivalent to $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we ge

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$[6] \cup [13]$

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(C) Given the linear congruence $8x \equiv 6 \pmod{14}$.This is equivalent to $8x - 6 = 14k$ for some integer $k \in \mathbb{Z}$.Dividing by $2$,we get $4x - 3 = 7k$,which implies $4x \equiv 3 \pmod{7}$.Multiplying by the modular inverse of $4$ modulo $7$,which is $2$ (since $4 	imes 2 = 8 \equiv 1 \pmod{7}$):$2(4x) \equiv 2(3) \pmod{7} \implies 8x \equiv 6 \pmod{7} \implies x \equiv 6 \pmod{7}$.Thus,$x$ can be written as $x = 7n + 6$ for $n \in \mathbb{Z}$.For $n = 0, x = 6$. For $n = 1, x = 13$. For $n = 2, x = 20$,and so on.The solutions are $x \in \{ \dots, -1, 6, 13, 20, 27, \dots \}$.This set can be represented as the union of two equivalence classes modulo $14$: $[6] = \{ \dots, -8, 6, 20, 34, \dots \}$ and $[13] = \{ \dots, -1, 13, 27, 41, \dots \}$.Therefore,the solution set is $[6] \cup [13]$.

The solution set of $8x \equiv 6 \pmod{14}$,$x \in \mathbb{Z}$,is

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