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

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

Solve for x, y, z

$$x=2$$

$$y=-5$$

$$z=10$$

Short Steps Using Substitution

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

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

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

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

Solve the second equation for y and the third equation for z.

$$y=-z+5 z=-\frac{3}{2}y+\frac{5}{2}$$

Substitute -z+5 for y in the equation z=-\frac{3}{2}y+\frac{5}{2}.

$$z=-\frac{3}{2}\left(-z+5\right)+\frac{5}{2}$$

Solve z=-\frac{3}{2}\left(-z+5\right)+\frac{5}{2} for z.

$$z=10$$

Substitute 10 for z in the equation y=-z+5.

$$y=-10+5$$

Calculate y from y=-10+5.

$$y=-5$$

Substitute -5 for y and 10 for z in the equation x=-2y-z+2.

$$x=-2\left(-5\right)-10+2$$

Calculate x from x=-2\left(-5\right)-10+2.

$$x=2$$

The system is now solved.

$$x=2 y=-5 z=10$$