Solve {l}{3x+y+2z=13}{x+2y+z=7}{x-y+3z=10}

$$\left\{ \begin{array} { l } { 3 x + y + 2 z = 13 } \\ { x + 2 y + z = 7 } \\ { x - y + 3 z = 10 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=3$$

Short Steps Using Substitution

Solve 3x+y+2z=13 for y.

$$y=-3x-2z+13$$

Substitute -3x-2z+13 for y in the second and third equation.

$$x+2\left(-3x-2z+13\right)+z=7 x-\left(-3x-2z+13\right)+3z=10$$

Solve these equations for x and z respectively.

$$x=-\frac{3}{5}z+\frac{19}{5} z=-\frac{4}{5}x+\frac{23}{5}$$

Substitute -\frac{3}{5}z+\frac{19}{5} for x in the equation z=-\frac{4}{5}x+\frac{23}{5}.

$$z=-\frac{4}{5}\left(-\frac{3}{5}z+\frac{19}{5}\right)+\frac{23}{5}$$

Solve z=-\frac{4}{5}\left(-\frac{3}{5}z+\frac{19}{5}\right)+\frac{23}{5} for z.

$$z=3$$

Substitute 3 for z in the equation x=-\frac{3}{5}z+\frac{19}{5}.

$$x=-\frac{3}{5}\times 3+\frac{19}{5}$$

Calculate x from x=-\frac{3}{5}\times 3+\frac{19}{5}.

$$x=2$$

Substitute 2 for x and 3 for z in the equation y=-3x-2z+13.

$$y=-3\times 2-2\times 3+13$$

Calculate y from y=-3\times 2-2\times 3+13.

$$y=1$$

The system is now solved.

$$x=2 y=1 z=3$$