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.
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 $\,3\,$.