Solve {l}{2x+10y-3z=-1}{5x+6y+z=3}{6x+3y+2z=8}

$$\left. \begin{array} { l } { 2 x + 10 y - 3 z = - 1 } \\ { 5 x + 6 y + z = 3 } \\ { 6 x + 3 y + 2 z = 8 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{128}{41} = 3\frac{5}{41} \approx 3.12195122$$

$$y = -\frac{66}{41} = -1\frac{25}{41} \approx -1.609756098$$

$$z = -\frac{121}{41} = -2\frac{39}{41} \approx -2.951219512$$

Short Steps Using Substitution

Reorder the equations.

$$5x+6y+z=3 2x+10y-3z=-1 6x+3y+2z=8$$

Solve 5x+6y+z=3 for z.

$$z=-5x-6y+3$$

Substitute -5x-6y+3 for z in the second and third equation.

$$2x+10y-3\left(-5x-6y+3\right)=-1 6x+3y+2\left(-5x-6y+3\right)=8$$

Solve these equations for y and x respectively.

$$y=-\frac{17}{28}x+\frac{2}{7} x=-\frac{9}{4}y-\frac{1}{2}$$

Substitute -\frac{17}{28}x+\frac{2}{7} for y in the equation x=-\frac{9}{4}y-\frac{1}{2}.

$$x=-\frac{9}{4}\left(-\frac{17}{28}x+\frac{2}{7}\right)-\frac{1}{2}$$

Solve x=-\frac{9}{4}\left(-\frac{17}{28}x+\frac{2}{7}\right)-\frac{1}{2} for x.

$$x=\frac{128}{41}$$

Substitute \frac{128}{41} for x in the equation y=-\frac{17}{28}x+\frac{2}{7}.

$$y=-\frac{17}{28}\times \frac{128}{41}+\frac{2}{7}$$

Calculate y from y=-\frac{17}{28}\times \frac{128}{41}+\frac{2}{7}.

$$y=-\frac{66}{41}$$

Substitute -\frac{66}{41} for y and \frac{128}{41} for x in the equation z=-5x-6y+3.

$$z=-5\times \frac{128}{41}-6\left(-\frac{66}{41}\right)+3$$

Calculate z from z=-5\times \frac{128}{41}-6\left(-\frac{66}{41}\right)+3.

$$z=-\frac{121}{41}$$

The system is now solved.

$$x=\frac{128}{41} y=-\frac{66}{41} z=-\frac{121}{41}$$