Solve {l}{2x-2y+z=-10}{3x+y-2z=4}{5x+2y-3z=7}

$$\left. \begin{array} { l } { 2 x - 2 y + z = - 10 } \\ { 3 x + y - 2 z = 4 } \\ { 5 x + 2 y - 3 z = 7 } \end{array} \right.$$

Solve for x, y, z

$$x=-1$$

$$y=3$$

$$z=-2$$

Short Steps Using Substitution

Solve 2x-2y+z=-10 for z.

$$z=-10-2x+2y$$

Substitute -10-2x+2y for z in the second and third equation.

$$3x+y-2\left(-10-2x+2y\right)=4 5x+2y-3\left(-10-2x+2y\right)=7$$

Solve these equations for y and x respectively.

$$y=\frac{7}{3}x+\frac{16}{3} x=\frac{4}{11}y-\frac{23}{11}$$

Substitute \frac{7}{3}x+\frac{16}{3} for y in the equation x=\frac{4}{11}y-\frac{23}{11}.

$$x=\frac{4}{11}\left(\frac{7}{3}x+\frac{16}{3}\right)-\frac{23}{11}$$

Solve x=\frac{4}{11}\left(\frac{7}{3}x+\frac{16}{3}\right)-\frac{23}{11} for x.

$$x=-1$$

Substitute -1 for x in the equation y=\frac{7}{3}x+\frac{16}{3}.

$$y=\frac{7}{3}\left(-1\right)+\frac{16}{3}$$

Calculate y from y=\frac{7}{3}\left(-1\right)+\frac{16}{3}.

$$y=3$$

Substitute 3 for y and -1 for x in the equation z=-10-2x+2y.

$$z=-10-2\left(-1\right)+2\times 3$$

Calculate z from z=-10-2\left(-1\right)+2\times 3.

$$z=-2$$

The system is now solved.

$$x=-1 y=3 z=-2$$