Solve {r}{3x-2y+5z=1}{x+y-z=10}{2x-y+3z=5}

$$\left. \begin{array} { r } { 3 x - 2 y + 5 z = 1 } \\ { x + y - z = 10 } \\ { 2 x - y + 3 z = 5 } \end{array} \right.$$

Solve for x, y, z

$$x=-3$$

$$y=25$$

$$z=12$$

Short Steps Using Substitution

Reorder the equations.

$$x+y-z=10 3x-2y+5z=1 2x-y+3z=5$$

Solve x+y-z=10 for x.

$$x=-y+z+10$$

Substitute -y+z+10 for x in the second and third equation.

$$3\left(-y+z+10\right)-2y+5z=1 2\left(-y+z+10\right)-y+3z=5$$

Solve these equations for y and z respectively.

$$y=\frac{8}{5}z+\frac{29}{5} z=-3+\frac{3}{5}y$$

Substitute \frac{8}{5}z+\frac{29}{5} for y in the equation z=-3+\frac{3}{5}y.

$$z=-3+\frac{3}{5}\left(\frac{8}{5}z+\frac{29}{5}\right)$$

Solve z=-3+\frac{3}{5}\left(\frac{8}{5}z+\frac{29}{5}\right) for z.

$$z=12$$

Substitute 12 for z in the equation y=\frac{8}{5}z+\frac{29}{5}.

$$y=\frac{8}{5}\times 12+\frac{29}{5}$$

Calculate y from y=\frac{8}{5}\times 12+\frac{29}{5}.

$$y=25$$

Substitute 25 for y and 12 for z in the equation x=-y+z+10.

$$x=-25+12+10$$

Calculate x from x=-25+12+10.

$$x=-3$$

The system is now solved.

$$x=-3 y=25 z=12$$