Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{y+z-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+y+z-3}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2$
$⇒x+y+z=\dfrac{1}{2}$
Ta có: $\dfrac{y+z+1}{x}=2⇒\dfrac{y+z+1}{x}+1=3$
$⇒\dfrac{x+y+z+1}{x}=3$
$⇒\dfrac{\dfrac{1}{2}+1}{x}=3$
$⇒x=\dfrac{1}{2}$
$\dfrac{x+z+2}{y}=2⇒\dfrac{x+z+2}{y}+1=3$
$⇒\dfrac{x+y+z+2}{y}=3$
$⇒\dfrac{\dfrac{1}{2}+2}{y}=3$
$⇒y=\dfrac{5}{6}$
$⇒z=\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{5}{6}=-\dfrac{5}{6}$