Solve {l}{x+2y-z=-1}{3x+8y+2z=28}{4x+9y-z=12}

$$\left. \begin{array} { l } { x + 2 y - z = - 1 } \\ { 3 x + 8 y + 2 z = 28 } \\ { 4 x + 9 y - z = 12 } \end{array} \right.$$

Solve for x, y, z

$$x=-26$$

$$y=13$$

$$z=1$$

Short Steps Using Substitution

Solve x+2y-z=-1 for x.

$$x=-2y+z-1$$

Substitute -2y+z-1 for x in the second and third equation.

$$3\left(-2y+z-1\right)+8y+2z=28 4\left(-2y+z-1\right)+9y-z=12$$

Solve these equations for y and z respectively.

$$y=-\frac{5}{2}z+\frac{31}{2} z=-\frac{1}{3}y+\frac{16}{3}$$

Substitute -\frac{5}{2}z+\frac{31}{2} for y in the equation z=-\frac{1}{3}y+\frac{16}{3}.

$$z=-\frac{1}{3}\left(-\frac{5}{2}z+\frac{31}{2}\right)+\frac{16}{3}$$

Solve z=-\frac{1}{3}\left(-\frac{5}{2}z+\frac{31}{2}\right)+\frac{16}{3} for z.

$$z=1$$

Substitute 1 for z in the equation y=-\frac{5}{2}z+\frac{31}{2}.

$$y=-\frac{5}{2}+\frac{31}{2}$$

Calculate y from y=-\frac{5}{2}+\frac{31}{2}.

$$y=13$$

Substitute 13 for y and 1 for z in the equation x=-2y+z-1.

$$x=-2\times 13+1-1$$

Calculate x from x=-2\times 13+1-1.

$$x=-26$$

The system is now solved.

$$x=-26 y=13 z=1$$