91Ó°ÊÓ

The system of equations is,

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

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

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

localid="1644481254464" $x-3y-3z=3.......\left(4\right)$

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

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

localid="1644481067950" $2x+5y+2z=-4..........\left(5\right)$

Simplifying equation (3), $2x+2y+\frac{1}{2}z=-4$by multiplying both sides by $2$,

$2(2x+2y+\frac{1}{2}y)=2(-4)$

localid="1644480818136" $4x+4y+x=-8.......\left(6\right)$

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

Multiplying equation (4) by $-2$,

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

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

Solving we get,

role="math" localid="1644481121965" $11y+8z=-10........\left(7\right)$

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

Multiplying equation (5) by $-2$,

$-4x-10y-4z=8$

$4x+4y+z=-8$

Solving we get,

$-6y-3z=0$

$2y+z=0........\left(8\right)$

Solving equations (7) and (8),

$11y+8z=-10$

$16y+8z=0$

Subtracting we get,

$-5y=-10$

$y=2$

Substituting the value of $y=2$in the equation

$2y+z=0$

$2\left(2\right)+z=0$

$4+z=0$

$z=-4$

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

$x-3y-3z=3$

$x-3\left(2\right)-3(-4)=3$

$x-6+12=3$

$x+6=3$

$x=3-6$

$x=-3$

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

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

$\frac{1}{3}(-3)-2-(-4)=1$

$-1-2+4=1$

$-3+4=1$

$1=1$

This is true.

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

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

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

$-3+5-4=-2$

$-7+5=-2$

$-2=-2$

This is true.

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

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

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

$-6+4-2=-4$

$-8+4=-4$

$-4=-4$

This is true.