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=(-5-3z)/(2)`
From first equation count x:

`4x-y+2z=-6 \\ 4x=-6-2z+y \qquad /:4 \\ x=(-6-2z+y)/(4)`
Substitute "y", which is counted from third equation:

`x=(-6-2z+(-5-3z)/(2))/(4)=(-(12)/(2)-(4z)/(2)+(-5-3z)/(2))/(4)=(-12-4z-5-3z)/(8)=(-7z-17)/(8)`
So now you've got:

`y=(-5-3z)/(2) \\ x=(-z-7)/(8) \ \\ \hbox{Substitute this to second equation:} \\ -2 \cdot (-7z-17)/(8)+3 \cdot (-5-3z)/(2)-z=8 \\ (7z+17)/(4)+(-15-9z)/(2)-z=8 \qquad /\cdot 4 \\ 7z+17-30-18z-4z=32 \\ -15z-13=32 \\ -15z=45 \qquad /:(-15) \\ z=-3 \\ x=(3-7)/(8)=(4)/(8)=(1)/(2) \\ y=(-5+9)/(2)=(4)/(2)=2`
So the solution is:

`\begin{cases} x=(1)/(2) \\ y=2 \\ z=-3 \end{cases}`