Solve {l}{2x+y+2z=3}{x-3y-2z=2}{4x+y+3z=1}

$$\left\{ \begin{array} { l } { 2 x + y + 2 z = 3 } \\ { x - 3 y - 2 z = 2 } \\ { 4 x + y + 3 z = 1 } \end{array} \right.$$

Solve for x, y, z

$$x=-19$$

$$y=-31$$

$$z=36$$

Short Steps Using Substitution

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

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

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

$$x-3\left(-2x-2z+3\right)-2z=2 4x-2x-2z+3+3z=1$$

Solve these equations for x and z respectively.

$$x=-\frac{4}{7}z+\frac{11}{7} z=-2-2x$$

Substitute -\frac{4}{7}z+\frac{11}{7} for x in the equation z=-2-2x.

$$z=-2-2\left(-\frac{4}{7}z+\frac{11}{7}\right)$$

Solve z=-2-2\left(-\frac{4}{7}z+\frac{11}{7}\right) for z.

$$z=36$$

Substitute 36 for z in the equation x=-\frac{4}{7}z+\frac{11}{7}.

$$x=-\frac{4}{7}\times 36+\frac{11}{7}$$

Calculate x from x=-\frac{4}{7}\times 36+\frac{11}{7}.

$$x=-19$$

Substitute -19 for x and 36 for z in the equation y=-2x-2z+3.

$$y=-2\left(-19\right)-2\times 36+3$$

Calculate y from y=-2\left(-19\right)-2\times 36+3.

$$y=-31$$

The system is now solved.

$$x=-19 y=-31 z=36$$