Consider a sum $$\frac{a}{b}+\frac{c}{d} = \frac{x}{y}$$ where each fraction is reduced. Alternatively using the familiar process of lowest common denominators, we have $$\frac{a}{b}+\frac{c}{d} = \frac{a\cdot\frac{d}{(b,d)}+c\cdot\frac{b}{(b,d)}}{[b,d]}$$ where $(b,d)$ denotes the gcd and $[b,d]$ denotes the lcm. My question is, when is it true that $y = [b,d]$? For example, this does not hold for $$\frac{5}{6}+\frac{1}{14} = \frac{38}{42} = \frac{19}{21}$$ where $21\neq [14, 6]$.
Are there any simple necessary and sufficient conditions for $y = [b,d]$?
Edit It has been suggested that
$$y = [b,d] \iff (b, d) = 1$$
is a necessary and sufficient condition. I'm interested in either a proof or a counterexample if possible.
Edit 2 After a bit of searching, I've found $$\frac{1}{24} + \frac{1}{16} = \frac{5}{48}$$ as a counterexample.
I am still looking for nice conditions for this to be satisfied and I feel that I should give an explanation of exactly what type of condition I am seeking. Angela has provided a necessary and sufficient condition, but it does not seem to be "simpler" than simply adding the fraction and seeing if it reduces. This is perhaps ambitious, but I am looking for a condition which is simple enough to use by inspection for simple fractions.
let $x=\gcd(b,d)$
$b=xe$
$d=xf$
$\frac{a}{b}+\frac{c}{d}=\frac{af+ce}{efx}$
$efx=\operatorname{lcm}(b,d)$
$e,f$ are relative primes
The fraction simplifies when $\gcd(af+ce,x)\not=1$, which is when $\gcd(ad+bc,bd)\not=\gcd(b,d)$.