Solve {l}{x+y+z=6}{x+2y+3z=10}{x+2y+1z=4}

$$\left. \begin{array} { l } { x + y + z = 6 } \\ { x + 2 y + 3 z = 10 } \\ { x + 2 y + 1 z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x=5$$

$$y=-2$$

$$z=3$$

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.

$$-y-z+6+2y+3z=10 -y-z+6+2y+1z=4$$

Solve these equations for z and y respectively.

$$z=-\frac{1}{2}y+2 y=-2$$

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

$$z=-\frac{1}{2}\left(-2\right)+2$$

Calculate z from z=-\frac{1}{2}\left(-2\right)+2.

$$z=3$$

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

$$x=-\left(-2\right)-3+6$$

Calculate x from x=-\left(-2\right)-3+6.

$$x=5$$

The system is now solved.

$$x=5 y=-2 z=3$$