\(\dfrac{x+1}{3}=\dfrac{3x+3}{9}\)
\(\dfrac{y+2}{-4}=\dfrac{2y+4}{-8}\)
\(\dfrac{t+3}{5}=\dfrac{4t+12}{20}\)
Do đó : \(\dfrac{3x+3}{9}=\dfrac{2y+4}{-8}=\dfrac{4t+12}{20}=\dfrac{3x+2y+4t+19}{21}=\dfrac{47+19}{21}=\dfrac{22}{7}\)
Suy ra : x = \(\dfrac{59}{7}\); y = \(-\dfrac{102}{7}\); z = \(\dfrac{89}{7}\)