Solve {l}{4x+y-3z=11}{2x-3y+2z=9}{z+y+z=-3}

$$\left\{ \begin{array} { l } { 4 x + y - 3 z = 11 } \\ { 2 x - 3 y + 2 z = 9 } \\ { z + y + z = - 3 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$y = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$z=-\frac{2}{3}\approx -0.666666667$$

Short Steps Using Substitution

Solve 4x+y-3z=11 for y.

$$y=-4x+3z+11$$

Substitute -4x+3z+11 for y in the second and third equation.

$$2x-3\left(-4x+3z+11\right)+2z=9 z-4x+3z+11+z=-3$$

Solve these equations for x and z respectively.

$$x=3+\frac{1}{2}z z=-\frac{14}{5}+\frac{4}{5}x$$

Substitute 3+\frac{1}{2}z for x in the equation z=-\frac{14}{5}+\frac{4}{5}x.

$$z=-\frac{14}{5}+\frac{4}{5}\left(3+\frac{1}{2}z\right)$$

Solve z=-\frac{14}{5}+\frac{4}{5}\left(3+\frac{1}{2}z\right) for z.

$$z=-\frac{2}{3}$$

Substitute -\frac{2}{3} for z in the equation x=3+\frac{1}{2}z.

$$x=3+\frac{1}{2}\left(-\frac{2}{3}\right)$$

Calculate x from x=3+\frac{1}{2}\left(-\frac{2}{3}\right).

$$x=\frac{8}{3}$$

Substitute \frac{8}{3} for x and -\frac{2}{3} for z in the equation y=-4x+3z+11.

$$y=-4\times \frac{8}{3}+3\left(-\frac{2}{3}\right)+11$$

Calculate y from y=-4\times \frac{8}{3}+3\left(-\frac{2}{3}\right)+11.

$$y=-\frac{5}{3}$$

The system is now solved.

$$x=\frac{8}{3} y=-\frac{5}{3} z=-\frac{2}{3}$$