Solve {r}{x+2y+5z=14}{2x+5y+3z=15}{5x+12y-5z=12}

$$\left. \begin{array} { r } { x + 2 y + 5 z = 14 } \\ { 2 x + 5 y + 3 z = 15 } \\ { 5 x + 12 y - 5 z = 12 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=2$$

Short Steps Using Substitution

Solve x+2y+5z=14 for x.

$$x=-2y-5z+14$$

Substitute -2y-5z+14 for x in the second and third equation.

$$2\left(-2y-5z+14\right)+5y+3z=15 5\left(-2y-5z+14\right)+12y-5z=12$$

Solve these equations for y and z respectively.

$$y=7z-13 z=\frac{29}{15}+\frac{1}{15}y$$

Substitute 7z-13 for y in the equation z=\frac{29}{15}+\frac{1}{15}y.

$$z=\frac{29}{15}+\frac{1}{15}\left(7z-13\right)$$

Solve z=\frac{29}{15}+\frac{1}{15}\left(7z-13\right) for z.

$$z=2$$

Substitute 2 for z in the equation y=7z-13.

$$y=7\times 2-13$$

Calculate y from y=7\times 2-13.

$$y=1$$

Substitute 1 for y and 2 for z in the equation x=-2y-5z+14.

$$x=-2-5\times 2+14$$

Calculate x from x=-2-5\times 2+14.

$$x=2$$

The system is now solved.

$$x=2 y=1 z=2$$