Solve {l}{x+y+z=2}{x+2y+4z=1}{x+3y+9z=6}

$$\left. \begin{array} { l } { x + y + z = 2 } \\ { x + 2 y + 4 z = 1 } \\ { x + 3 y + 9 z = 6 } \end{array} \right.$$

Solve for x, y, z

$$x=9$$

$$y=-10$$

$$z=3$$

Short Steps Using Substitution

Solve x+y+z=2 for x.

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

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

$$-y-z+2+2y+4z=1 -y-z+2+3y+9z=6$$

Solve these equations for y and z respectively.

$$y=-3z-1 z=\frac{1}{2}-\frac{1}{4}y$$

Substitute -3z-1 for y in the equation z=\frac{1}{2}-\frac{1}{4}y.

$$z=\frac{1}{2}-\frac{1}{4}\left(-3z-1\right)$$

Solve z=\frac{1}{2}-\frac{1}{4}\left(-3z-1\right) for z.

$$z=3$$

Substitute 3 for z in the equation y=-3z-1.

$$y=-3\times 3-1$$

Calculate y from y=-3\times 3-1.

$$y=-10$$

Substitute -10 for y and 3 for z in the equation x=-y-z+2.

$$x=-\left(-10\right)-3+2$$

Calculate x from x=-\left(-10\right)-3+2.

$$x=9$$

The system is now solved.

$$x=9 y=-10 z=3$$