Solve {l}{x+y+z=11}{3x+y=20}{x=2z}

$$\left\{ \begin{array} { l } { x + y + z = 11 } \\ { 3 x + y = 20 } \\ { x = 2 z } \end{array} \right.$$

Solve for x, y, z

$$x=6$$

$$y=2$$

$$z=3$$

Short Steps Using Substitution

Reorder the equations.

$$x=2z 3x+y=20 x+y+z=11$$

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

$$3\times 2z+y=20 2z+y+z=11$$

Solve these equations for y and z respectively.

$$y=20-6z z=-\frac{1}{3}y+\frac{11}{3}$$

Substitute 20-6z for y in the equation z=-\frac{1}{3}y+\frac{11}{3}.

$$z=-\frac{1}{3}\left(20-6z\right)+\frac{11}{3}$$

Solve z=-\frac{1}{3}\left(20-6z\right)+\frac{11}{3} for z.

$$z=3$$

Substitute 3 for z in the equation y=20-6z.

$$y=20-6\times 3$$

Calculate y from y=20-6\times 3.

$$y=2$$

Substitute 3 for z in the equation x=2z.

$$x=2\times 3$$

Calculate x from x=2\times 3.

$$x=6$$

The system is now solved.

$$x=6 y=2 z=3$$