Solve {l}{x+y+z=3}{2x+y+z=0}{2x+z=-1}

$$\left. \begin{array} { l } { x + y + z = 3 } \\ { 2 x + y + z = 0 } \\ { 2 x + z = - 1 } \end{array} \right.$$

Solve for x, y, z

$$x=-3$$

$$y=1$$

$$z=5$$

Short Steps Using Substitution

Solve x+y+z=3 for x.

$$x=-y-z+3$$

Substitute -y-z+3 for x in the second and third equation.

$$2\left(-y-z+3\right)+y+z=0 2\left(-y-z+3\right)+z=-1$$

Solve these equations for y and z respectively.

$$y=-z+6 z=-2y+7$$

Substitute -z+6 for y in the equation z=-2y+7.

$$z=-2\left(-z+6\right)+7$$

Solve z=-2\left(-z+6\right)+7 for z.

$$z=5$$

Substitute 5 for z in the equation y=-z+6.

$$y=-5+6$$

Calculate y from y=-5+6.

$$y=1$$

Substitute 1 for y and 5 for z in the equation x=-y-z+3.

$$x=-1-5+3$$

Calculate x from x=-1-5+3.

$$x=-3$$

The system is now solved.

$$x=-3 y=1 z=5$$