Solve {c}{2x+z=30}{3x+y-z=8}{x+3y+2z=3}

$$\left\{ \begin{array} { c } { 2 x + z = 30 } \\ { 3 x + y - z = 8 } \\ { x + 3 y + 2 z = 3 } \end{array} \right.$$

Solve for x, z, y

$$x = \frac{19}{2} = 9\frac{1}{2} = 9.5$$

$$y = -\frac{19}{2} = -9\frac{1}{2} = -9.5$$

$$z=11$$

Short Steps Using Substitution

Solve 2x+z=30 for z.

$$z=30-2x$$

Substitute 30-2x for z in the second and third equation.

$$3x+y-\left(30-2x\right)=8 x+3y+2\left(30-2x\right)=3$$

Solve these equations for x and y respectively.

$$x=-\frac{1}{5}y+\frac{38}{5} y=-19+x$$

Substitute -\frac{1}{5}y+\frac{38}{5} for x in the equation y=-19+x.

$$y=-19-\frac{1}{5}y+\frac{38}{5}$$

Solve y=-19-\frac{1}{5}y+\frac{38}{5} for y.

$$y=-\frac{19}{2}$$

Substitute -\frac{19}{2} for y in the equation x=-\frac{1}{5}y+\frac{38}{5}.

$$x=-\frac{1}{5}\left(-\frac{19}{2}\right)+\frac{38}{5}$$

Calculate x from x=-\frac{1}{5}\left(-\frac{19}{2}\right)+\frac{38}{5}.

$$x=\frac{19}{2}$$

Substitute \frac{19}{2} for x in the equation z=30-2x.

$$z=30-2\times \frac{19}{2}$$

Calculate z from z=30-2\times \frac{19}{2}.

$$z=11$$

The system is now solved.

$$x=\frac{19}{2} z=11 y=-\frac{19}{2}$$