Question

\(2x=3y=-2z\) và \(2x-3y+4z=48\) 

Answer 1

Có: \(2x=3y=-2z\)

=> \(2x=3y\) và \(3y=-2z\) 

=> \(\frac{x}{3}=\frac{y}{2}\) và \(\frac{y}{-2}=\frac{z}{3}\)

=> \(\frac{x}{3}=\frac{y}{2}\) và \(\frac{y}{2}=\frac{-z}{3}\)

=> \(\frac{x}{3}=\frac{y}{2}=-\frac{z}{3}\)

=>\(\frac{2x}{6}=\frac{-3y}{-6}=\frac{4z}{-12}=\frac{2x-3y+4z}{6-6-12}=\frac{48}{-12}=-4\)

+) \(2x=6\cdot-4=-24\Rightarrow x=-12\)

+)\(-3y=-6\cdot-4=24\Rightarrow y=-8\)

+)\(4z=-12\cdot-4=48\Rightarrow x=12\)

Answer 2

Từ 2x=3y= - 2 z

\(\Rightarrow\frac{2x}{6}=\frac{3y}{6}=-\frac{2z}{6}\)

\(\Rightarrow\frac{2x}{6}=\frac{3y}{6}=\frac{4z}{-12}=\frac{2x-3y+4z}{6-6+12}=\frac{48}{-12}=-4\)

\(\Rightarrow\begin{cases}x=-24\\y=-8\\z=12\end{cases}\)