Solving Linear congruence using backward substitution Use the method of back substitution to find all integers $x$ such that $x ≡ 1 \mod 5$,
$x ≡ 2 \mod 6 $, and $x ≡ 3 \mod 7$.
Solution: The first congruence can be rewritten as an equality,
$x = 5t + 1$, where $t$ is an integer. Substituting this expression for $x$ into the second congruence
tells us that
$5t + 1 ≡ 2 \mod 6$,
which can be solved to show that $t ≡ 5 $ (mod $6$) (as the reader should verify).
Can someone verify that and show me how to do it?

To solve $$5t+1\equiv2\pmod6,$$
subtract $1$ from both sides:  $$5t\equiv1\pmod6.$$
To get  $t\equiv5\pmod6$ from that, multiply both sides by $5$.
In general, if $a\equiv b\pmod n$, then $a+c\equiv b+c\pmod n$ and $a\times c\equiv b\times c\pmod n$.
