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

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

Solve for x, y, z

$$x = \frac{40}{29} = 1\frac{11}{29} \approx 1.379310345$$

$$y = -\frac{41}{29} = -1\frac{12}{29} \approx -1.413793103$$

$$z = \frac{32}{29} = 1\frac{3}{29} \approx 1.103448276$$

Short Steps Using Substitution

Solve -2x-4y+z=4 for z.

$$z=4+2x+4y$$

Substitute 4+2x+4y for z in the second and third equation.

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

Solve these equations for y and x respectively.

$$y=-\frac{1}{13}x-\frac{17}{13} x=-\frac{16}{13}-\frac{24}{13}y$$

Substitute -\frac{1}{13}x-\frac{17}{13} for y in the equation x=-\frac{16}{13}-\frac{24}{13}y.

$$x=-\frac{16}{13}-\frac{24}{13}\left(-\frac{1}{13}x-\frac{17}{13}\right)$$

Solve x=-\frac{16}{13}-\frac{24}{13}\left(-\frac{1}{13}x-\frac{17}{13}\right) for x.

$$x=\frac{40}{29}$$

Substitute \frac{40}{29} for x in the equation y=-\frac{1}{13}x-\frac{17}{13}.

$$y=-\frac{1}{13}\times \frac{40}{29}-\frac{17}{13}$$

Calculate y from y=-\frac{1}{13}\times \frac{40}{29}-\frac{17}{13}.

$$y=-\frac{41}{29}$$

Substitute -\frac{41}{29} for y and \frac{40}{29} for x in the equation z=4+2x+4y.

$$z=4+2\times \frac{40}{29}+4\left(-\frac{41}{29}\right)$$

Calculate z from z=4+2\times \frac{40}{29}+4\left(-\frac{41}{29}\right).

$$z=\frac{32}{29}$$

The system is now solved.

$$x=\frac{40}{29} y=-\frac{41}{29} z=\frac{32}{29}$$