$\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{x+z}=\frac{133}{10}$
⇔ $19.(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z})=\frac{133}{10}$
⇒ $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=0,7$
$\frac{7x}{y+z}+\frac{7y}{x+z}+\frac{7z}{x+y}=\frac{133}{10}$
⇔ $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1,9$
⇔ $(x+y+z).(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z})=1,9+3$
⇔ $(x+y+z).0,7=4,9$
⇔ $x+y+z=7$