Solve {c}{2x-3y+4z=2}{5x+y-3z=0}{x+2y-z=3}

$$\left. \begin{array} { c } { 2 x - 3 y + 4 z = 2 } \\ { 5 x + y - 3 z = 0 } \\ { x + 2 y - z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{5}{8}=0.625$$

$$y = \frac{41}{20} = 2\frac{1}{20} = 2.05$$

$$z = \frac{69}{40} = 1\frac{29}{40} = 1.725$$

Short Steps Using Substitution

Reorder the equations.

$$5x+y-3z=0 2x-3y+4z=2 x+2y-z=3$$

Solve 5x+y-3z=0 for y.

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

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

$$2x-3\left(-5x+3z\right)+4z=2 x+2\left(-5x+3z\right)-z=3$$

Solve these equations for x and z respectively.

$$x=\frac{5}{17}z+\frac{2}{17} z=\frac{3}{5}+\frac{9}{5}x$$

Substitute \frac{5}{17}z+\frac{2}{17} for x in the equation z=\frac{3}{5}+\frac{9}{5}x.

$$z=\frac{3}{5}+\frac{9}{5}\left(\frac{5}{17}z+\frac{2}{17}\right)$$

Solve z=\frac{3}{5}+\frac{9}{5}\left(\frac{5}{17}z+\frac{2}{17}\right) for z.

$$z=\frac{69}{40}$$

Substitute \frac{69}{40} for z in the equation x=\frac{5}{17}z+\frac{2}{17}.

$$x=\frac{5}{17}\times \frac{69}{40}+\frac{2}{17}$$

Calculate x from x=\frac{5}{17}\times \frac{69}{40}+\frac{2}{17}.

$$x=\frac{5}{8}$$

Substitute \frac{5}{8} for x and \frac{69}{40} for z in the equation y=-5x+3z.

$$y=-5\times \frac{5}{8}+3\times \frac{69}{40}$$

Calculate y from y=-5\times \frac{5}{8}+3\times \frac{69}{40}.

$$y=\frac{41}{20}$$

The system is now solved.

$$x=\frac{5}{8} y=\frac{41}{20} z=\frac{69}{40}$$