Solve {l}{x+2y+5z=6}{2x-2y-z=1}{3x+4y-2z=4}

$$\left. \begin{array} { l } { x + 2 y + 5 z = 6 } \\ { 2 x - 2 y - z = 1 } \\ { 3 x + 4 y - 2 z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{13}{10} = 1\frac{3}{10} = 1.3$$

$$y=\frac{33}{80}=0.4125$$

$$z=\frac{31}{40}=0.775$$

Short Steps Using Substitution

Solve x+2y+5z=6 for x.

$$x=-2y-5z+6$$

Substitute -2y-5z+6 for x in the second and third equation.

$$2\left(-2y-5z+6\right)-2y-z=1 3\left(-2y-5z+6\right)+4y-2z=4$$

Solve these equations for y and z respectively.

$$y=\frac{11}{6}-\frac{11}{6}z z=\frac{14}{17}-\frac{2}{17}y$$

Substitute \frac{11}{6}-\frac{11}{6}z for y in the equation z=\frac{14}{17}-\frac{2}{17}y.

$$z=\frac{14}{17}-\frac{2}{17}\left(\frac{11}{6}-\frac{11}{6}z\right)$$

Solve z=\frac{14}{17}-\frac{2}{17}\left(\frac{11}{6}-\frac{11}{6}z\right) for z.

$$z=\frac{31}{40}$$

Substitute \frac{31}{40} for z in the equation y=\frac{11}{6}-\frac{11}{6}z.

$$y=\frac{11}{6}-\frac{11}{6}\times \frac{31}{40}$$

Calculate y from y=\frac{11}{6}-\frac{11}{6}\times \frac{31}{40}.

$$y=\frac{33}{80}$$

Substitute \frac{33}{80} for y and \frac{31}{40} for z in the equation x=-2y-5z+6.

$$x=-2\times \frac{33}{80}-5\times \frac{31}{40}+6$$

Calculate x from x=-2\times \frac{33}{80}-5\times \frac{31}{40}+6.

$$x=\frac{13}{10}$$

The system is now solved.

$$x=\frac{13}{10} y=\frac{33}{80} z=\frac{31}{40}$$