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

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

Solve for x, y, z

$$x=\frac{4}{5}=0.8$$

$$y=\frac{3}{10}=0.3$$

$$z=\frac{9}{10}=0.9$$

Short Steps Using Substitution

Solve 3x-y+z=3 for y.

$$y=3x+z-3$$

Substitute 3x+z-3 for y in the second and third equation.

$$2x+3x+z-3-z=1 x+3x+z-3+z=2$$

Solve these equations for x and z respectively.

$$x=\frac{4}{5} z=-2x+\frac{5}{2}$$

Substitute \frac{4}{5} for x in the equation z=-2x+\frac{5}{2}.

$$z=-2\times \frac{4}{5}+\frac{5}{2}$$

Calculate z from z=-2\times \frac{4}{5}+\frac{5}{2}.

$$z=\frac{9}{10}$$

Substitute \frac{4}{5} for x and \frac{9}{10} for z in the equation y=3x+z-3.

$$y=3\times \frac{4}{5}+\frac{9}{10}-3$$

Calculate y from y=3\times \frac{4}{5}+\frac{9}{10}-3.

$$y=\frac{3}{10}$$

The system is now solved.

$$x=\frac{4}{5} y=\frac{3}{10} z=\frac{9}{10}$$