91Ó°ÊÓ

The system of equations is,

$\left\{\begin{array}{l}x+\frac{1}{3}y-2z=-1...........\left(1\right)\\ \frac{1}{3}x+y+\frac{1}{2}z=0.............\left(2\right)\\ \frac{1}{2}x+\frac{1}{3}y-\frac{1}{2}z=-1.........\left(3\right)\end{array}\right.$

Simplifying equation (1), $x+\frac{1}{3}y-2z=-1$by multiplying both sides by $3$,

$3(x+\frac{1}{3}y-2z)=3(-1)$

localid="1644514308075" $3x+y-6z=-3.........\left(4\right)$

Simplifying equation (2), $\frac{1}{3}x+y+\frac{1}{2}z=0$by multiplying both sides by $6$,

$6(\frac{1}{3}x+y+\frac{1}{2}z)=6\left(0\right)$

localid="1644514451288" $2x+6y+3z=0........\left(5\right)$

Simplifying equation (3)$\frac{1}{2}x+\frac{1}{3}y-\frac{1}{2}z=-1$by multiplying both sides by $6$,

$6(\frac{1}{2}x+\frac{1}{3}y-\frac{1}{2}z)=6(-1)$

$3x+2y-3z=-6.......\left(6\right)$

Eliminating $z$from the equations (4) and (5),

Multiplying $2$in the equation (5),

$4x+12y+6z=0$

$3x+y-6z=-3$

Solving the equations we get,

$7x+13y=-3........\left(7\right)$

Eliminating $z$from the equations (5) and (6),

role="math" localid="1644517953992" $2x+6y+3z=0$

$3x+2y-3z=-6$

Solving the two equations, we get,

$5x+8y=-6........\left(8\right)$

Solving the equations (7) and (8),

$$\begin{gathered} 35x+65y=-15.......\left(7\right)×5 \\ 35x+56y=-42......\left(8\right)×7 \end{gathered}$$

Subtracting the equations we get,

$$\begin{gathered} 9y=27 \\ y=3 \end{gathered}$$

Substituting the value of $y$in the equation $7x+13y=-3$,

$7x+13\left(3\right)=-3$

$$\begin{gathered} 7x=-3-39 \\ 7x=-42 \\ x=-6 \end{gathered}$$

Substituting $$\begin{gathered} x=-6 \\ y=3 \end{gathered}$$in the equation

$3x+y-6z=-3$

$3(-6)+3-6z=-3$

$-18+3-6z=-3$

$-15-6z=-3$

$$\begin{gathered} -6z=-3+15 \\ -6z=12 \\ z=-2 \end{gathered}$$

Substituting $$\begin{gathered} x=-6 \\ y=3 \\ z=-2 \end{gathered}$$in the equation

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

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

$-6+1+4=-1$

$-6+5=-1$

$-1=-1$

This is true.

Substituting $$\begin{gathered} x=-6 \\ y=3 \\ z=-2 \end{gathered}$$in the equation

$\frac{1}{3}x+y+\frac{1}{2}z=0$

$$\begin{gathered} \frac{1}{3}(-6)+3+\frac{1}{2}(-2)=0 \\ -2+3-1=0 \\ -3+3=0 \\ 0=0 \end{gathered}$$

This is true.

Substituting $$\begin{gathered} x=-6 \\ y=3 \\ z=-2 \end{gathered}$$in the equation

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

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

This is true.