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

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

Solve for x, y, z

$$x=2$$

$$y=3$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=6 for x.

$$x=-y-z+6$$

Substitute -y-z+6 for x in the second and third equation.

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

Solve these equations for y and z respectively.

$$y=\frac{15}{4}-\frac{3}{4}z z=\frac{1}{3}y$$

Substitute \frac{15}{4}-\frac{3}{4}z for y in the equation z=\frac{1}{3}y.

$$z=\frac{1}{3}\left(\frac{15}{4}-\frac{3}{4}z\right)$$

Solve z=\frac{1}{3}\left(\frac{15}{4}-\frac{3}{4}z\right) for z.

$$z=1$$

Substitute 1 for z in the equation y=\frac{15}{4}-\frac{3}{4}z.

$$y=\frac{15}{4}-\frac{3}{4}$$

Calculate y from y=\frac{15}{4}-\frac{3}{4}.

$$y=3$$

Substitute 3 for y and 1 for z in the equation x=-y-z+6.

$$x=-3-1+6$$

Calculate x from x=-3-1+6.

$$x=2$$

The system is now solved.

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