Solve {r}{x+z=1}{2x-y-2z=4}{x+3y+z=3}

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

Solve for x, z, y

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$y=\frac{2}{3}\approx 0.666666667$$

$$z=-\frac{2}{3}\approx -0.666666667$$

Short Steps Using Substitution

Solve x+z=1 for x.

$$x=-z+1$$

Substitute -z+1 for x in the second and third equation.

$$2\left(-z+1\right)-y-2z=4 -z+1+3y+z=3$$

Solve these equations for z and y respectively.

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

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

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

Calculate z from z=-\frac{1}{2}-\frac{1}{4}\times \frac{2}{3}.

$$z=-\frac{2}{3}$$

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

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

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

$$x=\frac{5}{3}$$

The system is now solved.

$$x=\frac{5}{3} z=-\frac{2}{3} y=\frac{2}{3}$$