Show that $6x + 5 \equiv 7 \pmod 5$ has infinitely many solutions. Let $u$ be a solution to $6x + 5 \equiv 7 \pmod 5$ so that $6u + 5 \equiv 7 \pmod 5$. Then there's some $t$ such that $u \equiv t \pmod 5$.
Then
$$6u \equiv t \pmod 5,$$
$$6u + 5 \equiv t + 5 \pmod 5,$$
$$7 \equiv t + 5 \pmod 5,$$
$$t + 5 \equiv 6t + 5 \pmod 5,$$
$$0 \equiv 5t \pmod 5.$$
Can we say that since $7 \equiv 6t + 5 \pmod 5$ and $0 \equiv 5t \pmod 5$ are equivalent(?), then $7 \equiv 6t + 5 \pmod 5$ has infinite number of solutions?
 A: There seem to be a lot of misunderstandings here. Let me adress a few:


*

*The statement 
"Then there's some $t$ such that $u=t\pmod5$."
is completely void; for every $u$ we can simply take $t=u$ to get $u=t\pmod5$.

*The implication
$$6u + 5 = t + 5 \pmod 5\qquad\to\qquad 7 = t + 5 \pmod 5,$$
is simply false, even assuming $u\equiv t\pmod5$. Take $u=t=0$, for example.

*The implication
$$7 = t + 5 \pmod 5\qquad\to\qquad t + 5 = 6t + 5 \pmod 5,$$
is also completely void; the latter is true regardless of the former.

*The same goes for the implication
$$t + 5 = 6t + 5 \pmod 5\qquad\to\qquad0 = 5t \pmod 5.$$

*The statements $7 = 6t + 5 \pmod 5$ and $0=5t\pmod5$ are not equivalent. Take $t=0$ for example.

*And from the previous point, I cannot see any reason to believe that it follows that there are infinitely many solutions to $7=6t+5\pmod5$.
Suggestion: Look up what $a\equiv b\pmod c$ means for integers $a$, $b$ and $c$ with $c>0$. Then try proving that if $x$ is a solution to $6x+5=7\pmod5$, also $x+5k$ is a solution for any integer $k$.
