Solve the following systems of congruences. x≡2(mod 5)

Formula used:
1) Theorem: System of congruences:
Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences
$x\equiv a(modm)$
$x\equiv b(modn)$
Furthermore, any two solutions x and y are congruent modulo mn.
2) Theorem: Addition and Multiplication Properties:
If $a\equiv b(modn)$ and x is any integer, then $a+x-=b+x(modn)andax\equiv bx(modn)$.
Explanation:
Consider the system of congruences $x\equiv (mod5)$
$x\equiv 3(mod8)$
Since 5 and 8 are relatively prime, then $(5,8)=1$.
Then, by using a theorem, there exists an integer x that satisfies the system of congruences.
From the first congruence $x=2+5k$ for some integer k and substitute this expression for x into the second congruence.
$2+5k\equiv 3(mod8)$
By using addition property,
$2+5k+(—2)\equiv 3+(—2)(mod8)$
$=5k=1(mod8)$
By using multiplication property,
$\Rightarrow 2+5k+(-2)\equiv 3+(-2)(mod8)$
$\Rightarrow 5k\equiv 1(mod8)$
By using multiplication property,
$5\ast 5k-\equiv 5\ast 1(mod8)$
$\Rightarrow 25k\equiv 5(mod8)$
Since $25\equiv 1(mod8)$,
$\Rightarrow k\equiv 5(mod8)$
Thus, $x=2+5(5)=27$ satisfies the system and $x\equiv 27(mod5\ast 8)orx\equiv 27(mod40)$ gives all solutions to the given system of congruences.