Showing that two binomials are relatively prime for all positive integers (Euclidean ?) Show that $3x+11$ and $5x+18$ are relatively prime for all positive integers $x$. 
Hi everyone I've looked around a lot and found similar questions like this here but when trying some of the tips I feel like whenever I get close it doesn't match. 
As I understand it there are two approaches I can take for a proof strategy and I can't get them to work (or am doing something wrong and not able to find the next logical step). 
For example via the Euclidean Algorithm 
So $3x+11$ and $5x+18$ are relatively prime. That means they are not both 0 and $$gcd((3x+11), (5x+18)) = 1$$
Also two integers $a$ and $b$ are relatively prime if and only if there exist integers $s$ and $t$ such that $$as+bt = 1$$
So I first tried dividing and cannot arrive at a way that I get this nice 1 sitting alone at the end.
$$5x+18 = (1) (3x+11) + (2x+7)$$
$$3x+11 = (1) (2x+7) + (x+4)$$
$$2x+7 = (1)(x+4)+(x+3)$$
$$x+4 = (1)(x+3)+1$$
$$x+3 = 1(1)+(x+2)$$ 
By that last step I think I'm lost... 
Another common solution I'm seeing which is no help is people just saying
well you know that if $(3x+11)$ and $(5x+18)$ are relatively prime then there there are two integers s, t such that $$as+bt =1$$ and they pull out like $(3x+11)(5) + (-3)(5x+18) = 1 $ which is great and fine and all but I have no idea how to get those two numbers by any sort of method besides just guessing. There must be a way I'm missing or a fundamental step in the Euclidean Algorithm or definitions of Linear Combinations that I'm missing. 
 A: One simple way to prove these numbers are not relative primes is to show that the following fraction is irriducible:
$$\frac{5x+18}{3x+11}=1+\frac{2x+7}{3x+11}$$ Hence, $\,2x+7\,$ and $\,3x+11\,$ can be co-primes iff (if and only if) the following fraction is reducible:
$$\frac{3x+11}{2x+7}=1+\frac{x+4}{2x+7}.$$
Now for $\,x+4\,$ and $\,2x+7\,$ to be co-primes this fraction must be reducible:
$$\frac{2x+7}{x+4}=1+\frac{x+3}{x+4}\,$$
So, if $\,x+3\,$ and $\,x+4\,$ are co-primes, then the next fraction is reducible:
$$\frac{x+4}{x+3}=1+\frac{1}{x+3}\,$$
Now finally, we come to the following fraction which must also be reducible $$\frac{1}{x+3}$$ but it is not. So, the original numbers are not co-primes. Q.E.D.
In order for two numbers to be co-primes, the fraction must be reducible i.e. it must be in the form: $$\frac{xk}{xp}$$ where some number, say, $\,x\,$ is a common factor. If there are no common factor, the fraction is irreducible and its numerator and denominator are co-primes.
$$............$$
There's even a simpler way to do it.
$$\frac{15x+55}{15x+54}=1+\frac{1}{15x+54}$$
The fraction $$\frac{1}{15x+54}$$ is irreducible.
So the initial numbers are co-primes because it is trivial to show that $\,15x+54\,$ is not divisible by $\,5\,$ but divisible by $\,3,$ and $\,15x+55\,$ is conversely not divisible by $\,3\,$but divisible by $\,5$. In other words multiplying one number by $\,5$ and the other by $\,3$ doesn't change the situation because after that we know that the initial numbers will still not be both (at the same time) divisible by $\,5\,$ or by $\,3\,$.
