Solve {l}{x+y+z=6}{x+2y+3z=14}{x+3y+9z=34}

$$\left\{ \begin{array} { l } { x + y + z = 6 } \\ { x + 2 y + 3 z = 14 } \\ { x + 3 y + 9 z = 34 } \end{array} \right.$$

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.

$$-y-z+6+2y+3z=14 -y-z+6+3y+9z=34$$

Solve these equations for y and z respectively.

$$y=8-2z z=\frac{7}{2}-\frac{1}{4}y$$

Substitute 8-2z for y in the equation z=\frac{7}{2}-\frac{1}{4}y.

$$z=\frac{7}{2}-\frac{1}{4}\left(8-2z\right)$$

Solve z=\frac{7}{2}-\frac{1}{4}\left(8-2z\right) for z.

$$z=3$$

Substitute 3 for z in the equation y=8-2z.

$$y=8-2\times 3$$

Calculate y from y=8-2\times 3.

$$y=2$$

Substitute 2 for y and 3 for z 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.

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