Solve {l}{15x+15y+15z=15}{15x+35y+25z=55}{15x+25y+30z=15}

$$\left. \begin{array} { l } { 15 x + 15 y + 15 z = 15 } \\ { 15 x + 35 y + 25 z = 55 } \\ { 15 x + 25 y + 30 z = 15 } \end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=3$$

$$z=-2$$

Short Steps Using Substitution

Solve 15x+15y+15z=15 for x.

$$x=1-y-z$$

Substitute 1-y-z for x in the second and third equation.

$$15\left(1-y-z\right)+35y+25z=55 15\left(1-y-z\right)+25y+30z=15$$

Solve these equations for y and z respectively.

$$y=2-\frac{1}{2}z z=-\frac{2}{3}y$$

Substitute 2-\frac{1}{2}z for y in the equation z=-\frac{2}{3}y.

$$z=-\frac{2}{3}\left(2-\frac{1}{2}z\right)$$

Solve z=-\frac{2}{3}\left(2-\frac{1}{2}z\right) for z.

$$z=-2$$

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

$$y=2-\frac{1}{2}\left(-2\right)$$

Calculate y from y=2-\frac{1}{2}\left(-2\right).

$$y=3$$

Substitute 3 for y and -2 for z in the equation x=1-y-z.

$$x=1-3-\left(-2\right)$$

Calculate x from x=1-3-\left(-2\right).

$$x=0$$

The system is now solved.

$$x=0 y=3 z=-2$$