Solve {c}{3x+2y=7}{2y+4z=3}{z+x=1}

$$\left. \begin{array} { c } { 3 x + 2 y = 7 } \\ { 2 y + 4 z = 3 } \\ { z + x = 1 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{8}{7} = 1\frac{1}{7} \approx 1.142857143$$

$$y = \frac{25}{14} = 1\frac{11}{14} \approx 1.785714286$$

$$z=-\frac{1}{7}\approx -0.142857143$$

Short Steps Using Substitution

Reorder the equations.

$$z+x=1 2y+4z=3 3x+2y=7$$

Solve z+x=1 for x.

$$x=-z+1$$

Substitute -z+1 for x in the equation 3x+2y=7.

$$3\left(-z+1\right)+2y=7$$

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

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

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

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

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

$$z=-\frac{1}{7}$$

Substitute -\frac{1}{7} for z in the equation y=-2z+\frac{3}{2}.

$$y=-2\left(-\frac{1}{7}\right)+\frac{3}{2}$$

Calculate y from y=-2\left(-\frac{1}{7}\right)+\frac{3}{2}.

$$y=\frac{25}{14}$$

Substitute -\frac{1}{7} for z in the equation x=-z+1.

$$x=-\left(-\frac{1}{7}\right)+1$$

Calculate x from x=-\left(-\frac{1}{7}\right)+1.

$$x=\frac{8}{7}$$

The system is now solved.

$$x=\frac{8}{7} y=\frac{25}{14} z=-\frac{1}{7}$$