Show that $(a) + (b)= R$ for $\gcd(a,b) = 1$ The question I am trying to solve it:

Let $R$ be a principal ideal domain, $a,b\in R$. Suppose $\gcd(a,b) = 1$. Show that $(a)+(b)=R$.

First I have tried to show that $(a)+(b)$ is in R:
  $\gcd(a,b)=1$ means there exists $s$, $t$ in R such that $as+bt=1$
  where $as$ is in $(a)$ and $bt$ is in $(b)$
  so $as+bt$ is in $(a) + (b)$ and since $1$ is in R, we have that $as+bt$ in R implies $(a)+(b)$ is a subset of R.
Then I want to show the other inclusion that R is in $(a)+(b)$ but I am not sure how to do this?

It's obvious that $(a)+(b)\subseteq R$.
The other inclusion in the point of the question, and you (almost) proved it! From $as+bt=1$ we get $asr+btr=r$ for any $r\in R$, so $r\in (a)+(b)$. 
