To find: The smallest positive integer that solves the congruences x≡3(b mod 7),x≡4(b mod 5)

Given information:
The congruences ${\displaystyle x\equiv 3\left(bmod7\right),x\equiv 4\left(bmod5\right)}$
Consider the given congruences
${\displaystyle x\equiv 3\left(bmod7\right),x\equiv 4\left(bmod5\right)}$
‘The congruence ${\displaystyle x\equiv 3\left(bmod7\right)}$ means if x is divided by 7, the remainder is 3.
So the number x is one of the numbers in the following list:
3, 10, 17, 24, 31, 38, 45, -
Similarly, the congruence ${\displaystyle x\equiv 4\left(bmod5\right)}$ means if x is divided by 5, the remainder is 4.
So the number x is one of the numbers in the following list:
4,9, 14, 19,24, 29,34, 39, 44, -
The smallest number that is found in both the lists is 24, so the
smallest number that solves the congruences
${\displaystyle x\equiv 3\left(bmod7\right),x\equiv 4\left(bmod5\right)}$ is 24.
${\displaystyle \Rightarrow x=24}$
Final Statement:
The smallest positive integer that solves the congruences
${\displaystyle x\equiv 3\left(bmod7\right),x\equiv 4\left(bmod5\right)}$ is x = 24.