Solve {l}{x-2y-3z+18=0}{x+3y-3z-8=0}{x+y+2z-24=0}

$$\left\{ \begin{array} { l } { x - 2 y - 3 z + 18 = 0 } \\ { x + 3 y - 3 z - 8 = 0 } \\ { x + y + 2 z - 24 = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{206}{25} = 8\frac{6}{25} = 8.24$$

$$y = \frac{26}{5} = 5\frac{1}{5} = 5.2$$

$$z = \frac{132}{25} = 5\frac{7}{25} = 5.28$$

Short Steps Using Substitution

Solve x-2y-3z+18=0 for x.

$$x=2y+3z-18$$

Substitute 2y+3z-18 for x in the second and third equation.

$$2y+3z-18+3y-3z-8=0 2y+3z-18+y+2z-24=0$$

Solve these equations for y and z respectively.

$$y=\frac{26}{5} z=\frac{42}{5}-\frac{3}{5}y$$

Substitute \frac{26}{5} for y in the equation z=\frac{42}{5}-\frac{3}{5}y.

$$z=\frac{42}{5}-\frac{3}{5}\times \frac{26}{5}$$

Calculate z from z=\frac{42}{5}-\frac{3}{5}\times \frac{26}{5}.

$$z=\frac{132}{25}$$

Substitute \frac{26}{5} for y and \frac{132}{25} for z in the equation x=2y+3z-18.

$$x=2\times \frac{26}{5}+3\times \frac{132}{25}-18$$

Calculate x from x=2\times \frac{26}{5}+3\times \frac{132}{25}-18.

$$x=\frac{206}{25}$$

The system is now solved.

$$x=\frac{206}{25} y=\frac{26}{5} z=\frac{132}{25}$$