Đáp án:
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Có : `x/6 = y/(-7)` `(1)`
Có : `x/3 = z/8`
`↔ x/6 = z/16` `(2)`
Từ `(1), (2)`
`-> x/6 = y/(-7) = z/16`
`↔ x/6= (2y)/(-14) = (3z)/48`
Áp dụng tính chất dãy tỉ số bằng nhau có :
`x/6 = (2y)/(-14) = (3z)/48 = (x - 2y+3z)/(6 - (-14) + 48) = 56/68 = 14/17`
`↔` \(\left\{ \begin{array}{l}\dfrac{x}{6}=\dfrac{14}{17}\\ \dfrac{2y}{-14}=\dfrac{14}{17}\\ \dfrac{3z}{48}=\dfrac{14}{17}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=6×\dfrac{14}{17}\\ 2y = -14 × \dfrac{14}{17}\\3z=48×\dfrac{14}{17}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{84}{17}\\2y=\dfrac{-196}{17}\\3z=\dfrac{672}{17}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{84}{17}\\y=\dfrac{-196}{17}÷2\\z=\dfrac{672}{17}÷3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=\dfrac{84}{17}\\y=\dfrac{-98}{17}\\z=\dfrac{224}{17}\end{array} \right.\)
Vậy `(x;y;z) = (84/17; (-98)/17; 224/17)`