Solve {l}{3x+3y+2z=130-0}{3x+2y=85-6}{x+3y+z=60-(3)}

$$\left. \begin{array} { l } { 3 x + 3 y + 2 z = 130 - 0 } \\ { 3 x + 2 y = 85 - 6 } \\ { x + 3 y + z = 60 - ( 3 ) } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{269}{11} = 24\frac{5}{11} \approx 24.454545455$$

$$y = \frac{31}{11} = 2\frac{9}{11} \approx 2.818181818$$

$$z = \frac{265}{11} = 24\frac{1}{11} \approx 24.090909091$$

Short Steps Using Substitution

Reorder the equations.

$$x+3y+z=60-3 3x+2y=85-6 3x+3y+2z=130-0$$

Solve x+3y+z=60-3 for x.

$$x=-3y-z+57$$

Substitute -3y-z+57 for x in the second and third equation.

$$3\left(-3y-z+57\right)+2y=85-6 3\left(-3y-z+57\right)+3y+2z=130-0$$

Solve these equations for y and z respectively.

$$y=-\frac{3}{7}z+\frac{92}{7} z=-6y+41$$

Substitute -\frac{3}{7}z+\frac{92}{7} for y in the equation z=-6y+41.

$$z=-6\left(-\frac{3}{7}z+\frac{92}{7}\right)+41$$

Solve z=-6\left(-\frac{3}{7}z+\frac{92}{7}\right)+41 for z.

$$z=\frac{265}{11}$$

Substitute \frac{265}{11} for z in the equation y=-\frac{3}{7}z+\frac{92}{7}.

$$y=-\frac{3}{7}\times \frac{265}{11}+\frac{92}{7}$$

Calculate y from y=-\frac{3}{7}\times \frac{265}{11}+\frac{92}{7}.

$$y=\frac{31}{11}$$

Substitute \frac{31}{11} for y and \frac{265}{11} for z in the equation x=-3y-z+57.

$$x=-3\times \frac{31}{11}-\frac{265}{11}+57$$

Calculate x from x=-3\times \frac{31}{11}-\frac{265}{11}+57.

$$x=\frac{269}{11}$$

The system is now solved.

$$x=\frac{269}{11} y=\frac{31}{11} z=\frac{265}{11}$$