Solve {l}{5x-10y-5z=-15}{4x+10y-4z=-39}{x+2y+2z=0}{text{Findtheuniquesolutiontothissystemofequatic}}

$$\left. \begin{array}{l}{ 5 x - 10 y - 5 z = - 15 }\\{ 4 x + 10 y - 4 z = - 39 }\\{ x + 2 y + 2 z = 0 }\\{ \text { Find the unique solution to this system of equatic } }\end{array} \right.$$

Solve for x, y, z

$$x=-3$$

$$y = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$z=3$$

Short Steps Using Substitution

Reorder the equations.

$$x+2y+2z=0 4x+10y-4z=-39 5x-10y-5z=-15$$

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

$$x=-2y-2z$$

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

$$4\left(-2y-2z\right)+10y-4z=-39 5\left(-2y-2z\right)-10y-5z=-15$$

Solve these equations for y and z respectively.

$$y=-\frac{39}{2}+6z z=1-\frac{4}{3}y$$

Substitute -\frac{39}{2}+6z for y in the equation z=1-\frac{4}{3}y.

$$z=1-\frac{4}{3}\left(-\frac{39}{2}+6z\right)$$

Solve z=1-\frac{4}{3}\left(-\frac{39}{2}+6z\right) for z.

$$z=3$$

Substitute 3 for z in the equation y=-\frac{39}{2}+6z.

$$y=-\frac{39}{2}+6\times 3$$

Calculate y from y=-\frac{39}{2}+6\times 3.

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

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

$$x=-2\left(-\frac{3}{2}\right)-2\times 3$$

Calculate x from x=-2\left(-\frac{3}{2}\right)-2\times 3.

$$x=-3$$

The system is now solved.

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