Solve {l}{x+y+z=12}{x+2y-z=6}{3x-y+z=10}

$$\left\{ \begin{array} { l } { x + y + z = 12 } \\ { x + 2 y - z = 6 } \\ { 3 x - y + z = 10 } \end{array} \right.$$

Solve for x, y, z

$$x=3$$

$$y=4$$

$$z=5$$

Short Steps Using Substitution

Solve x+y+z=12 for x.

$$x=-y-z+12$$

Substitute -y-z+12 for x in the second and third equation.

$$-y-z+12+2y-z=6 3\left(-y-z+12\right)-y+z=10$$

Solve these equations for y and z respectively.

$$y=-6+2z z=13-2y$$

Substitute -6+2z for y in the equation z=13-2y.

$$z=13-2\left(-6+2z\right)$$

Solve z=13-2\left(-6+2z\right) for z.

$$z=5$$

Substitute 5 for z in the equation y=-6+2z.

$$y=-6+2\times 5$$

Calculate y from y=-6+2\times 5.

$$y=4$$

Substitute 4 for y and 5 for z in the equation x=-y-z+12.

$$x=-4-5+12$$

Calculate x from x=-4-5+12.

$$x=3$$

The system is now solved.

$$x=3 y=4 z=5$$