Difficulties with Modulus - Chinese Remainder Theorem, GCD and the Extended Euclidean Algorithm I am finding it quite difficult to deconstruct the steps for seemingly simple questions and decide which mathematical process to use. For example

If $x \equiv 2$ (mod $11$) and $ x \equiv 1$ (mod $17$), what is the value of $x$ (mod $187$)?

There are many online resources that patiently walk through questions such as this, but I find that some of the numbers suggested in the steps are either random guesses to get closer to an objective (without saying so) or there is no explanation as to how they were calculated or why they were chosen during a step. Then trying to follows the process onwards after that is unproductive because one is trying to puzzle out unknown step calculations instead of the objective. I'd appreciate any advice please. As in all of mathematics, it's more rewarding to solve it oneself, but sometimes productive learning hours turn to despair.
Saying that x is congruent to 2 modulo 11 means that x= 2+ 11m for some integer m.  Saying that x is congruent to 1 modulo 17 means that x= 1+ 17n for some integer n.
So 2+ 11m= 1+ 17n which is the same as 17n- 11m= 1.
Now 11 divides into 17 once with remaider 6: 17- 11= 6.
6 divides into 11  once with remainder 5:  11- 6= 5.
And, of course, 6- 5= 1.
6- 5= 6- (11- 6)= 2(6)- 11= 1.
2(17- 11)- 11= 2(17)- 3(11)= 1.
So one solution to this is n= 2, m= 3. x= 1+ 17(2)= 35 or x= 2+ 11(3)= 35.   Of course any number of the forM x= 35+ (11)(17)k=  35+ 187k is also a solution. So x= 35 (mod 187).
Saying that x is congruent to 2 modulo 11 means that x= 2+ 11m for some integer m.  Saying that x is congruent to 1 modulo 17 means that x= 1+ 17n for some integer n.
So 2+ 11m= 1+ 17n which is the same as 17n- 11m= 1.
Now 11 divides into 17 once with remaider 6: 17- 11= 6.
6 divides into 11  once with remainder 5:  11- 6= 5.
And, of course, 6- 5= 1.
6- 5= 6- (11- 6)= 2(6)- 11= 1.
2(17- 11)- 11= 2(17)- 3(11)= 1.
So one solution to this is n= 2, m= 3. x= 1+ 17(2)= 35 or x= 2+ 11(3)= 35.   Of course any number of the forM x= 35+ (11)(17)k=  35+ 187k is also a solution. So x= 35 (mod 187).