Solve {l}{x+y+z=6}{2x+y-z=1}{x+2y+z=8}

$$\left. \begin{array} { l } { x + y + z = 6 } \\ { 2 x + y - z = 1 } \\ { x + 2 y + z = 8 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=2$$

$$z=3$$

Short Steps Using Substitution

Solve x+y+z=6 for x.

$$x=-y-z+6$$

Substitute -y-z+6 for x in the second and third equation.

$$2\left(-y-z+6\right)+y-z=1 -y-z+6+2y+z=8$$

Solve these equations for z and y respectively.

$$z=\frac{11}{3}-\frac{1}{3}y y=2$$

Substitute 2 for y in the equation z=\frac{11}{3}-\frac{1}{3}y.

$$z=\frac{11}{3}-\frac{1}{3}\times 2$$

Calculate z from z=\frac{11}{3}-\frac{1}{3}\times 2.

$$z=3$$

Substitute 3 for z and 2 for y in the equation x=-y-z+6.

$$x=-2-3+6$$

Calculate x from x=-2-3+6.

$$x=1$$

The system is now solved.

$$x=1 y=2 z=3$$