Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9

[tex]\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}[/tex]

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

[tex]5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)[/tex]

Subtracting (3) from 3 times (1), we have

[tex]3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)[/tex]

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

[tex]\begin{gathered} 13z=-48 \\ \\ z=-\frac{48}{13} \end{gathered}[/tex]

Using the value of z obtained in (5), we have

[tex]\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\ \\ 3y+\frac{240}{13}=-3 \\ \\ 3y=-3-\frac{240}{13} \\ \\ 3y=-\frac{279}{13} \\ \\ y=-\frac{279}{39} \end{gathered}[/tex]

Using the values obtained for y and z in (1), we have

[tex]\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\ \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\ \\ x-\frac{231}{13}=-4 \\ \\ x=-4+\frac{231}{13} \\ \\ x=\frac{179}{13} \end{gathered}[/tex]