addition with a variable (mod) Given $2+x \equiv 7 \pmod 3$.

$2 + 0 = 2$

$2 + 1 = 3$

$2 + 2 = 4$ 

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$2 + 5 = 7$

so, the answer will be $x = 5, 8, 11, 14, 17,\dots$ 

Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$
Hint: $\ x\equiv 7-2 \equiv \color{#c00}5 \pmod 3 = \color{#c00}5 + 3\,\Bbb Z = \{\,\ldots,-4, -1, 2, \color{#c00}5, 8, 11,14,\ldots\}$

Or: $\ x\equiv 5\pmod{3}\!\iff\! 3\mid x\!-\!5\!\iff\! 3n = x\!-\!5 \!\iff\! x = 5\! +\! 3n,\ $ for some $\, n\in \Bbb Z$
We are given that:

$(2+x) = 7 (\mod 3)$

Thus, $2+x-7 = 3k$ where $k$ is an integer.

Simplifying, yields:

$x-5 = 3k$

Therefore, $x\in (5,8,..)$