Respuesta :
Given,
The system of equations are,
[tex]\begin{gathered} 2x+y-3z=1\ldots\ldots\ldots\ldots......\ldots.\mathrm{\cdot}..(1) \\ 3x-y-4z=7\ldots\ldots\ldots\ldots\ldots\ldots\ldots.(2) \\ 5x+2y-6z=5\ldots\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(3) \end{gathered}[/tex]Adding equation (1) and (2)
[tex]\begin{gathered} 5x-7z=8\ldots.\ldots\ldots\ldots\ldots\ldots\text{.}(4) \\ 7z=5x-8 \\ x=\frac{7z+8}{5} \end{gathered}[/tex]Substracting equation (4) from (3)
[tex]\begin{gathered} 5x+2y-6z-5x+7z=5-8 \\ 2y+z=-3 \\ y=\frac{-z-3}{2} \end{gathered}[/tex]Substituting the value of x and y then,
[tex]\begin{gathered} 2x+y-3z=1 \\ 2(\frac{7z+8}{5})+(\frac{-z-3}{2})-3z=1 \\ \frac{14z+16}{5}-\frac{z-3}{2}-3z=1 \\ \frac{28z+32-5z-15-30z}{10}=1 \\ \frac{-7z+17}{10}=1 \\ -7z=10-17 \\ z=1 \end{gathered}[/tex]Substituting the value of z to get the value of x and y,
[tex]\begin{gathered} x=\frac{7z+8}{5}=3 \\ y=\frac{-z-3}{2}=-2 \end{gathered}[/tex]Hence, the value of x is 3, y is -2 and z is 1.
