Question

4x-y+2z=-6,-2x+3y-z=8,2y+3z=-5

Answer 1

From third equation count y:
$2y+3z=-5 \\ 2y=-5-3z \qquad /:2 \\ y=\frac{-5-3z}{2}$
From first equation count x:
$4x-y+2z=-6 \\ 4x=-6-2z+y \qquad /:4 \\ x=\frac{-6-2z+y}{4}$
Substitute "y", which is counted from third equation:
$x=\frac{-6-2z+\frac{-5-3z}{2}}{4}=\frac{-\frac{12}{2}-\frac{4z}{2}+\frac{-5-3z}{2}}{4}=\frac{-12-4z-5-3z}{8}=\frac{-7z-17}{8}$
So now you've got:
$y=\frac{-5-3z}{2} \\ x=\frac{-z-7}{8} \ \\ \hbox{Substitute this to second equation:} \\ -2 \cdot \frac{-7z-17}{8}+3 \cdot \frac{-5-3z}{2}-z=8 \\ \frac{7z+17}{4}+\frac{-15-9z}{2}-z=8 \qquad /\cdot 4 \\ 7z+17-30-18z-4z=32 \\ -15z-13=32 \\ -15z=45 \qquad /:(-15) \\ z=-3 \\ x=\frac{3-7}{8}=\frac{4}{8}=\frac{1}{2} \\ y=\frac{-5+9}{2}=\frac{4}{2}=2$
So the solution is:
$\begin{cases} x=\frac{1}{2} \\ y=2 \\ z=-3 \end{cases}$