Respuesta :
We have the following system of linear equations:
[tex]\begin{gathered} 2x+y-5z=-2, \\ -4x+4y-6z=3, \\ 6x-3y+z=-7. \end{gathered}[/tex]1) We solve for x the first equation:
[tex]\begin{gathered} 2x+y-5z=-2, \\ 2x=5z-y-2, \\ x=\frac{1}{2}\cdot(5z-y-2)\text{.} \end{gathered}[/tex]2) We replace the last equation in the second and third equations:
[tex]\begin{gathered} -4\cdot\frac{1}{2}\cdot(5z-y-2)+4y-6z=3, \\ 6\cdot\frac{1}{2}\cdot(5z-y-2)-3y+z=-7. \end{gathered}[/tex]Simplifying these equations, we have:
[tex]\begin{gathered} 6y-16z+4=3, \\ -6y+16z-6=-7. \end{gathered}[/tex]3) Multiplying by -1 the last equation, we have:
[tex]\begin{gathered} 6y-16z+4=3, \\ 6y-16z+6=7. \end{gathered}[/tex]4) Isolating the term 6y - 16z in each equation, we have:
[tex]\begin{gathered} 6y-16z=-1, \\ 6y-16z=1. \end{gathered}[/tex]5) Equalling the equations, we get:
[tex]-1=1.[/tex]Which is absurd. So we conclude that the system of equations is incompatible and it has no solutions.
Answer: No Solution
