I have solved a pair of linear congruence equations like $$x\equiv2\pmod {5}$$$$x\equiv4\pmod{7}$$
But now it is difficult for me to solve $$71x\equiv26\pmod {238}$$
Even I'm not able to think how can I start for this problem.
Please provide a hint so that I can begin it and proceed for its solution
Multiply both sides of the congruence by $71^{-1}$ (mod $238$)
HINT
In order to solve this linear congruence, you first need to find what is called a multiplicative inverse of the integer $71$ modulo $238$.
What is that? $ $ Well, it's an integer that when $71$ (mod $238$) is multiplied by it, you get $1$ (mod $238$).
So, if we say $ $ $u$ $ $ is a multiplicative inverse of $71$ (mod $238$), then $71u$ $\equiv 1$ (mod $238$).
$ $
Stop reading here if you want to give it a try yourself.
$ $
FULL ANSWER
Now, to find a value for $u$, I applied a method called Euclid's algorithm (you can google it to learn how it works), which gave the value $57$.
So, $57$ is a multiplicative inverse of $71$ modulo $238$. $ $ This makes sense because,
$71 \times 57 \equiv 4047 \equiv 17\times 238+1 \equiv 1$ (mod $238$).
Applying this to your congruence, we have
$71x \equiv 26$ (mod $238$)
$57\times 71x \equiv 57\times26$ (mod $238$)
$1\times x \equiv 1482$ (mod $238$)
$x \equiv 6\times 238 + 54$ (mod $238$)
$x \equiv 54$ (mod $238$),
as required.
Factorisation of $\quad 238=2\times 7\times 17$
$\begin{cases} 71\equiv 1\pmod 2\\ 71\equiv 1\pmod 7\\ 71^{16}\equiv 1\pmod {17} & \text{by little fermat theorem}\\ \end{cases}$
So by Chinese remainder theorem $71^{16}\equiv 1\pmod{238}$
It ensue that $71^{-1}\equiv 71^{15}\equiv 57 \pmod{238}$
Thus $x\equiv 57\times 26\equiv 54\pmod{238}$
