Question
Solve.$$\begin{aligned}&\frac{x+2}{3}-\frac{y+4}{2}+\frac{z+1}{6}=0,\\&\frac{x-4}{3}+\frac{y+1}{4}+\frac{z-2}{2}=-1,\\&\frac{x+1}{2}+\frac{y}{2}+\frac{z-1}{4}=\frac{3}{4}\end{aligned} $$
Step 1
This gives us: $$\begin{aligned} &2x-3y+z=7,\\ &4x+3y+6z=13,\\ &2x+2y+z=2. \end{aligned} $$ Show more…
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