Đề bài không cho 2x + 3y - 5z bằng bao nhiêu nên đặt 2x + 3y - 5z = k
Ta có:
\(6x=4y=3z\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{30}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{5z}{30}=\dfrac{2x+3y-5z}{6+12-30}=\dfrac{k}{-12}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{k}{-12}.6:2\Leftrightarrow x=\dfrac{k}{-4}\\y=\dfrac{k}{-12}.12:3\Leftrightarrow y=\dfrac{k}{-3}\\z=\dfrac{k}{-12}.30:5\Leftrightarrow z=\dfrac{k}{-2}\end{matrix}\right.\)
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