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

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

Solve for x, y, z

$$x=1$$

$$y=1$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=3 for x.

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

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

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

Solve these equations for z and y respectively.

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

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

$$z=-\frac{1}{2}+\frac{3}{2}$$

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

$$z=1$$

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

$$x=-1-1+3$$

Calculate x from x=-1-1+3.

$$x=1$$

The system is now solved.

$$x=1 y=1 z=1$$