Solve {l}{3x-y+2z=3}{2x+y-3z=11}{x+y+z=12}

$$\left. \begin{array} { l } { 3 x - y + 2 z = 3 } \\ { 2 x + y - 3 z = 11 } \\ { x + y + z = 12 } \end{array} \right.$$

Solve for x, y, z

$$x=3$$

$$y=8$$

$$z=1$$

Short Steps Using Substitution

Solve 3x-y+2z=3 for y.

$$y=3x+2z-3$$

Substitute 3x+2z-3 for y in the second and third equation.

$$2x+3x+2z-3-3z=11 x+3x+2z-3+z=12$$

Solve these equations for x and z respectively.

$$x=\frac{14}{5}+\frac{1}{5}z z=-\frac{4}{3}x+5$$

Substitute \frac{14}{5}+\frac{1}{5}z for x in the equation z=-\frac{4}{3}x+5.

$$z=-\frac{4}{3}\left(\frac{14}{5}+\frac{1}{5}z\right)+5$$

Solve z=-\frac{4}{3}\left(\frac{14}{5}+\frac{1}{5}z\right)+5 for z.

$$z=1$$

Substitute 1 for z in the equation x=\frac{14}{5}+\frac{1}{5}z.

$$x=\frac{14}{5}+\frac{1}{5}\times 1$$

Calculate x from x=\frac{14}{5}+\frac{1}{5}\times 1.

$$x=3$$

Substitute 3 for x and 1 for z in the equation y=3x+2z-3.

$$y=3\times 3+2\times 1-3$$

Calculate y from y=3\times 3+2\times 1-3.

$$y=8$$

The system is now solved.

$$x=3 y=8 z=1$$