Solve the following systems of congruences. x≡4(mod 5) x≡3(mod 8) x≡2(mod 3)

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
$x\equiv (modm)$
$x\equiv b(modn)$
Furthermore, any two solutions x and y are congruent modulo mn.
2) Theorem: Addition and Multiplication Properties:
If $a=b(modn)$ and x is any integer, then $a+x\equiv b+x(modn)andax\equiv bx(modn)$.
3) Theorem: Cancellation Law:
If $ax\equiv ay(modn)and(a,n)=1,thenx\equiv y(modn)$.
Explanation:
Consider the system of congruences
$x\equiv 4(mod5)$
$x\equiv 2(mod3)$
Since 5 and 3 are relatively prime, then $(5,3)=1$.
Then, by using theorem there exists an integer x that satisfies the system of congruences.
From the first congruence $x=4+5k$ for some integer k and substitute this expression for x into the second congruence.
$4+5k\equiv 2(mod3)$
By using addition property,
$4+5k+(—4)\equiv 2+(—4)(mod3)$
$\Rightarrow 5k=—2(mod3)$
Since $5\equiv 2(mod3)$,
$\Rightarrow 2k\equiv -2(mod3)$
Since $(2,3)=1$ then by using cancellation law,
$\Rightarrow k\equiv -1(mod3)$
Now, $-1\equiv 2(mod3)$,
Therefore, $k=2(mod3)$
Thus, $x=4+5(2)=14$ satisfies the system and $x=14(mod5\ast 3)$ or $x=14(mod15)$ gives all solutions to the given system of congruences.