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Answer :

The given system of linear equations is

[tex]\mleft\{\begin{aligned}2x-y-5z=27 \\ -2x-4y+7z=-27 \\ -2x-6y+3z=-3\end{aligned}\mright.[/tex]

Let's sum the first and the second equation to eliminate x, and to get an equation with y and z only.

[tex]\begin{gathered} 2x-2x-y-4y-5z+7z=27-27 \\ -5y+2z=0 \end{gathered}[/tex]

Then, we sum the first and the third equation to eliminate x, and to get an equation with y and z only.

[tex]\begin{gathered} 2x-2x-y-6y-5z+3z=27-3 \\ -7y-2z=24 \end{gathered}[/tex]

Now, we form a system with the new two equations.

[tex]\mleft\{\begin{aligned}-5y+2z=0 \\ -7y-2z=24\end{aligned}\mright.[/tex]

Let's sum the equations to eliminate z and find y.

[tex]\begin{gathered} -5y-7y+2z-2z=0+24 \\ -12y=24 \\ y=\frac{24}{-12} \\ y=-2 \end{gathered}[/tex]

y is equal to -2.

Then, we use the y-value to find z.

[tex]\begin{gathered} -5y+2z=0 \\ -5(-2)+2z=0 \\ 10+2z=0 \\ 2z=-10 \\ z=-\frac{10}{2} \\ z=-5 \end{gathered}[/tex]

z is equal to -5.

Now, we use y and z values to find x.

[tex]\begin{gathered} 2x-y-5z=27 \\ 2x-(-2)-5(-5)=27 \\ 2x+2+25=27 \\ 2x=27-25-2 \\ 2x=0 \\ x=\frac{0}{2} \\ x=0 \end{gathered}[/tex]

x is equal to zero.

Therefore, the solution to the system is (0, -2, -5).