In the following exercises, solve each system of equations. Use either substitution or elimination.

$\left\{\begin{array}{l}x+y-z=-1\\ 2x-y+2z=8\\ -3x+2y+z=-9\end{array}\right.$

The system of equations

$\left\{\begin{array}{l}x+y-z=-1\\ 2x-y+2z=8\\ -3x+2y+z=-9\end{array}\right.$

To find the solution of the system of equations.

Consider the system of equations:

$\begin{array}{l}x+y-z=-1...........\left(1\right)\\ 2x-y+2z=8...........\left(2\right)\\ -3x+2y+z=-9......\left(3\right)\end{array}$

Multiply equation (1) with 2

$$\begin{gathered} 2(x+y-z)=2(-1) \\ 2x+2y-2z=-2..........\left(4\right) \end{gathered}$$

From equation (2) and (4):

$$\begin{gathered} \frac{2x-y+2z=8 \\ 2x+2y-2z=-2}{4x+y=6} \\ 4x+y=6.........\left(5\right) \end{gathered}$$

From equation (1) and (3):

$$\begin{gathered} \frac{x+y-z=-1 \\ -3x+2y+z=-9}{-2x+3y=-10} \\ -2x+3y=-10..........\left(6\right) \end{gathered}$$

Multiply equation (5) by -3

$$\begin{gathered} -3(4x+y)=-3\left(6\right) \\ -\text{'}12x-3y=-18..........\left(7\right) \end{gathered}$$

from equation (6) and (7)

role="math" localid="1644663547310" $$\begin{gathered} \frac{-12x-3y=-18 \\ -2x+3y=-10}{14x=-28} \\ x=2 \end{gathered}$$

Put the value of $x=2$in $-2x+3y=-10$

$$\begin{gathered} -2x+3y=-10 \\ -4+3y=-10 \\ 3y=-6 \\ y=-2 \end{gathered}$$

Put $x=2,y=-2$in the equation role="math" localid="1644663746866" $$\begin{gathered} x+y-z=-1 \\ 2-2-z=-1 \\ z=1 \end{gathered}$$

Put the values of $x=2,y=-2,z=1$in the equation $x+y-z=-1$

$$\begin{gathered} x+y-z=-1 \\ 2-2-1=-1 \\ -1=-1 \end{gathered}$$

which holds true.