Solve {l}{3x+2y+3z=0}{5x+4y-3z=3}{6x+5y-4z=12}

$$\left\{ \begin{array} { l } { 3 x + 2 y + 3 z = 0 } \\ { 5 x + 4 y - 3 z = 3 } \\ { 6 x + 5 y - 4 z = 12 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{147}{4} = -36\frac{3}{4} = -36.75$$

$$y = \frac{99}{2} = 49\frac{1}{2} = 49.5$$

$$z = \frac{15}{4} = 3\frac{3}{4} = 3.75$$

Short Steps Using Substitution

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

$$x=-\frac{2}{3}y-z$$

Substitute -\frac{2}{3}y-z for x in the second and third equation.

$$5\left(-\frac{2}{3}y-z\right)+4y-3z=3 6\left(-\frac{2}{3}y-z\right)+5y-4z=12$$

Solve these equations for y and z respectively.

$$y=\frac{9}{2}+12z z=-\frac{6}{5}+\frac{1}{10}y$$

Substitute \frac{9}{2}+12z for y in the equation z=-\frac{6}{5}+\frac{1}{10}y.

$$z=-\frac{6}{5}+\frac{1}{10}\left(\frac{9}{2}+12z\right)$$

Solve z=-\frac{6}{5}+\frac{1}{10}\left(\frac{9}{2}+12z\right) for z.

$$z=\frac{15}{4}$$

Substitute \frac{15}{4} for z in the equation y=\frac{9}{2}+12z.

$$y=\frac{9}{2}+12\times \frac{15}{4}$$

Calculate y from y=\frac{9}{2}+12\times \frac{15}{4}.

$$y=\frac{99}{2}$$

Substitute \frac{99}{2} for y and \frac{15}{4} for z in the equation x=-\frac{2}{3}y-z.

$$x=-\frac{2}{3}\times \frac{99}{2}-\frac{15}{4}$$

Calculate x from x=-\frac{2}{3}\times \frac{99}{2}-\frac{15}{4}.

$$x=-\frac{147}{4}$$

The system is now solved.

$$x=-\frac{147}{4} y=\frac{99}{2} z=\frac{15}{4}$$