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

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

Solve for x, y, z

$$x=\frac{1}{9}\approx 0.111111111$$

$$y=\frac{1}{3}\approx 0.333333333$$

$$z=\frac{1}{3}\approx 0.333333333$$

Short Steps Using Substitution

Solve 3x+2y=1 for x.

$$x=-\frac{2}{3}y+\frac{1}{3}$$

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

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

Solve the second equation for y and the third equation for z.

$$y=-2z+1 z=y$$

Substitute -2z+1 for y in the equation z=y.

$$z=-2z+1$$

Solve z=-2z+1 for z.

$$z=\frac{1}{3}$$

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

$$y=-2\times \frac{1}{3}+1$$

Calculate y from y=-2\times \frac{1}{3}+1.

$$y=\frac{1}{3}$$

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

$$x=-\frac{2}{3}\times \frac{1}{3}+\frac{1}{3}$$

Calculate x from x=-\frac{2}{3}\times \frac{1}{3}+\frac{1}{3}.

$$x=\frac{1}{9}$$

The system is now solved.

$$x=\frac{1}{9} y=\frac{1}{3} z=\frac{1}{3}$$