Solve {l}{4x+2y-3z=24}{5x+3y-8z=10}{x-4y+5z=3}

$$\left. \begin{array} { l } { 4 x + 2 y - 3 z = 24 } \\ { 5 x + 3 y - 8 z = 10 } \\ { x - 4 y + 5 z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{409}{65} = 6\frac{19}{65} \approx 6.292307692$$

$$y = \frac{511}{65} = 7\frac{56}{65} \approx 7.861538462$$

$$z = \frac{366}{65} = 5\frac{41}{65} \approx 5.630769231$$

Short Steps Using Substitution

Reorder the equations.

$$x-4y+5z=3 5x+3y-8z=10 4x+2y-3z=24$$

Solve x-4y+5z=3 for x.

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

Substitute 4y-5z+3 for x in the second and third equation.

$$5\left(4y-5z+3\right)+3y-8z=10 4\left(4y-5z+3\right)+2y-3z=24$$

Solve these equations for y and z respectively.

$$y=\frac{33}{23}z-\frac{5}{23} z=-\frac{12}{23}+\frac{18}{23}y$$

Substitute \frac{33}{23}z-\frac{5}{23} for y in the equation z=-\frac{12}{23}+\frac{18}{23}y.

$$z=-\frac{12}{23}+\frac{18}{23}\left(\frac{33}{23}z-\frac{5}{23}\right)$$

Solve z=-\frac{12}{23}+\frac{18}{23}\left(\frac{33}{23}z-\frac{5}{23}\right) for z.

$$z=\frac{366}{65}$$

Substitute \frac{366}{65} for z in the equation y=\frac{33}{23}z-\frac{5}{23}.

$$y=\frac{33}{23}\times \frac{366}{65}-\frac{5}{23}$$

Calculate y from y=\frac{33}{23}\times \frac{366}{65}-\frac{5}{23}.

$$y=\frac{511}{65}$$

Substitute \frac{511}{65} for y and \frac{366}{65} for z in the equation x=4y-5z+3.

$$x=4\times \frac{511}{65}-5\times \frac{366}{65}+3$$

Calculate x from x=4\times \frac{511}{65}-5\times \frac{366}{65}+3.

$$x=\frac{409}{65}$$

The system is now solved.

$$x=\frac{409}{65} y=\frac{511}{65} z=\frac{366}{65}$$