Solve the following systems of congruences. 2xequiv 5(mod 3) 3x+2equiv 3(mod 8) 5x+4equiv 5(mod 7)

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 $a=b(modn)$ 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, $2x\equiv 5(mod3)$
Since $5\equiv 2(mod3)$,
$\Rightarrow 2x\equiv 2(mod3)$
By using multiplication property,
$2\ast 2x\equiv 2\ast 2(mod3)$
$\Rightarrow 4x\equiv 4(mod3)$
Since $4\equiv 1(mod3)$,
$\Rightarrow x=1(mod3)$
Now, consider $5x+4\equiv 5(mod7)$
By using addition property,
$5x+4+(—4)\equiv 5+(-4)(mod7)$
$\Rightarrow 5x=1(mod7)$
By using multiplication property,
$\Rightarrow 5\ast 3x\equiv 1\ast 3(mod7)$
$\Rightarrow 15x\equiv 3(mod7)$
Since $15\equiv 1(mod7)$,
$\Rightarrow x\equiv 3(mod7)$
Therefore, the system of congruences is
$x\equiv 1(mod3)$
$x\equiv 3(mod7)$
Since 3 and 7 are relatively prime then $(3,7)=1$.
From the first congruence $x=1+3k$ for some integer k and substitute this expression for x into the second congruence.
$1+3k=3(mod7)$
By using addition property,
$1+3k+(-1)\equiv 3+(-1)(mod7)$
$\Rightarrow 3k\equiv 2(mod7)$
By using multiplication property,
$3\ast 3k\equiv 2\ast 3(mod7)$
$\Rightarrow 9k\equiv 6(mod7)$
Since $9\equiv 2(mod7)$
$\Rightarrow 2k\equiv 6(mod7)$
Since $(2,7)=1$ then by using cancellation law,
$\Rightarrow k\equiv (mod7)$
Thus, $x=1+3(3)=10$ satisfies the system and $x\equiv 10(mod3\ast 7)orx\equiv (mod21)$ gives all solutions to the given systems of congruences.