Solve {l}{6(x-3)+6(y-1)+3(z+4)=0}{x+3/2-y+1/2-2(z-4)+9=0}{x+3+y+1/2-(z-4)=0}

$$\left. \begin{array} { l } { 6 ( x - 3 ) + 6 ( y - 1 ) + 3 ( z + 4 ) = 0 } \\ { \frac { x + 3 } { 2 } - \frac { y + 1 } { 2 } - 2 ( z - 4 ) + 9 = 0 } \\ { x + 3 + \frac { y + 1 } { 2 } - ( z - 4 ) = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=-17$$

$$y=19$$

$$z=0$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$6\left(x-3\right)+6\left(y-1\right)+3\left(z+4\right)=0 36+x-y-4z=0 15+2x+y-2z=0$$

Reorder the equations.

$$36+x-y-4z=0 6\left(x-3\right)+6\left(y-1\right)+3\left(z+4\right)=0 15+2x+y-2z=0$$

Solve 36+x-y-4z=0 for x.

$$x=-36+y+4z$$

Substitute -36+y+4z for x in the second and third equation.

$$6\left(-36+y+4z-3\right)+6\left(y-1\right)+3\left(z+4\right)=0 15+2\left(-36+y+4z\right)+y-2z=0$$

Solve these equations for y and z respectively.

$$y=19-\frac{9}{4}z z=\frac{19}{2}-\frac{1}{2}y$$

Substitute 19-\frac{9}{4}z for y in the equation z=\frac{19}{2}-\frac{1}{2}y.

$$z=\frac{19}{2}-\frac{1}{2}\left(19-\frac{9}{4}z\right)$$

Solve z=\frac{19}{2}-\frac{1}{2}\left(19-\frac{9}{4}z\right) for z.

$$z=0$$

Substitute 0 for z in the equation y=19-\frac{9}{4}z.

$$y=19-\frac{9}{4}\times 0$$

Calculate y from y=19-\frac{9}{4}\times 0.

$$y=19$$

Substitute 19 for y and 0 for z in the equation x=-36+y+4z.

$$x=-36+19+4\times 0$$

Calculate x from x=-36+19+4\times 0.

$$x=-17$$

The system is now solved.

$$x=-17 y=19 z=0$$