Solve {l}{2x-y+z=8}{(-3)x+4y-2z=(-12)}{x+2y-3z=(-11)}

$$\left. \begin{array}{l}{ 2 x - y + z = 8 }\\{ ( - 3 ) x + 4 y - 2 z = ( - 12 ) }\\{ x + 2 y - 3 z = ( - 11 ) }\end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=5$$

Short Steps Using Substitution

Solve 2x-y+z=8 for y.

$$y=2x+z-8$$

Substitute 2x+z-8 for y in the second and third equation.

$$-3x+4\left(2x+z-8\right)-2z=-12 x+2\left(2x+z-8\right)-3z=-11$$

Solve these equations for x and z respectively.

$$x=4-\frac{2}{5}z z=-5+5x$$

Substitute 4-\frac{2}{5}z for x in the equation z=-5+5x.

$$z=-5+5\left(4-\frac{2}{5}z\right)$$

Solve z=-5+5\left(4-\frac{2}{5}z\right) for z.

$$z=5$$

Substitute 5 for z in the equation x=4-\frac{2}{5}z.

$$x=4-\frac{2}{5}\times 5$$

Calculate x from x=4-\frac{2}{5}\times 5.

$$x=2$$

Substitute 2 for x and 5 for z in the equation y=2x+z-8.

$$y=2\times 2+5-8$$

Calculate y from y=2\times 2+5-8.

$$y=1$$

The system is now solved.

$$x=2 y=1 z=5$$