Solve the system of equations.

$\left\{\begin{array}{l}x+\frac{5}{2}y+z=-2\\ 2x+2y+\frac{1}{2}z=-4\\ \frac{1}{3}x-y-z=1\end{array}\right.$

The system of equations is,

$\left\{\begin{array}{l}x+\frac{5}{2}y+z=-2........\left(1\right)\\ 2x+2y+\frac{1}{2}z=-4.....\left(2\right)\\ \frac{1}{3}-y-z=1............\left(3\right)\end{array}\right.$

Simplifying equation (1), $x+\frac{5}{2}y+z=-2$by multiplying both sides by $2$,

$$\begin{gathered} 2(x+\frac{5}{2}y+z)=2(-2) \\ 2x+5y+2z=-4...........\left(4\right) \end{gathered}$$

Simplifying equation (2), $2x+2y+\frac{1}{2}z=-4$by multiplying both sides by $2$,

$$\begin{gathered} 2(2x+2y+\frac{1}{2}z)=2(-4) \\ 4x+4y+z=-8...........\left(5\right) \end{gathered}$$

Simplifying equation (3), $\frac{1}{3}x-y-z=1$by multiplying both sides by $3$,

$$\begin{gathered} 3(\frac{1}{3}x-y-z)=3\left(1\right) \\ x-3y-3z=3..........\left(6\right) \end{gathered}$$

Eliminating $x$from the equations (4) and (5).

$$\begin{gathered} -4x-10y-4z=8.........\left(4\right)\times -2 \\ 4x+4y+z=-8............\left(5\right) \end{gathered}$$

Solving the equations, we get,

$$\begin{gathered} -6y-3z=0 \\ 2y+z=0............\left(7\right) \end{gathered}$$

Eliminating $x$from the equations (5) and (6).

$$\begin{gathered} 4x+4y+z=-8..........\left(5\right) \\ -4x+12y+12z=-12.....\left(6\right)\times -4 \end{gathered}$$

Solving the equations, we get,

$16y+13z=-20........\left(8\right)$

Solving equations (7) and (8),

$$\begin{gathered} 16y+8z=0...........\left(7\right)\times 8 \\ 16y+13z=-20.....\left(8\right) \end{gathered}$$

Subtracting the equations, we get,

$$\begin{gathered} -5z=20 \\ z=-4 \end{gathered}$$

Substituting $z=-4$in the equation

$2y+z=0$,

$$\begin{gathered} 2y+(-4)=0 \\ 2y-4=0 \\ 2y=4 \\ y=2 \end{gathered}$$

Substituting $$\begin{gathered} y=2 \\ z=-4 \end{gathered}$$in the equation

$2x+5y+2z=-4$,

$$\begin{gathered} 2x+5\left(2\right)+2(-4)=-4 \\ 2x+10-8=-4 \\ 2x+2=-4 \\ 2x=-4-2 \\ 2x=-6 \\ x=-3 \end{gathered}$$

Substituting $$\begin{gathered} x=-3 \\ y=2 \\ z=-4 \end{gathered}$$in the equation

$x+\frac{5}{2}y+z=-2$,

$$\begin{gathered} -3+\frac{5}{2}\left(2\right)+(-4)=-2 \\ -3+5-4=-2 \\ 5-7=-2 \\ -2=-2 \end{gathered}$$

This is true.

Substituting $$\begin{gathered} x=-3 \\ y=2 \\ z=-4 \end{gathered}$$in the equation

$2x+2y+\frac{1}{2}z=-4$,

$$\begin{gathered} 2(-3)+2\left(2\right)+\frac{1}{2}(-4)=-4 \\ -6+4-2=-4 \\ -8+4=-4 \\ -4=-4 \end{gathered}$$

This is true.

Substituting $$\begin{gathered} x=-3 \\ y=2 \\ z=-4 \end{gathered}$$in the equation

$\frac{1}{3}x-y-z=1$,

$$\begin{gathered} \frac{1}{3}(-3)-2-(-4)=1 \\ -1-2+4=1 \\ 4-3=1 \\ 1=1 \end{gathered}$$

This is true.