Respuesta :

Answer:

Adding 5 times the third equation to the first one we get:

[tex]\begin{gathered} 0x+22y+39z=-27, \\ -4x+2y-3z=-1, \\ -x+5y+7z=-6. \end{gathered}[/tex]

Multiplying the third equation by -1 we get:

[tex]\begin{gathered} 0x+22y+39z=-27, \\ -4x+2y-3z=-1, \\ x-5y-7z=6. \end{gathered}[/tex]

Adding 4 times the third equation to the second one we get:

[tex]\begin{gathered} 0x+22y+39z=-27, \\ 0x-18y-31z=23, \\ x-5y-7z=6. \end{gathered}[/tex]

Dividing the first equation by 22 we get:

[tex]\begin{gathered} 0x+y+\frac{39}{22}z=-\frac{27}{22}, \\ 0x-18y-31z=23, \\ x-5y-7z=6. \end{gathered}[/tex]

Adding 18 times the first equation to the second one we get:

[tex]\begin{gathered} 0x+y+\frac{39}{22}z=-\frac{27}{22}, \\ 0x+0y+\frac{10}{11}z=\frac{10}{11}, \\ x-5y-7z=6. \end{gathered}[/tex]

Adding 5 times the first equation to the third one we get:

[tex]\begin{gathered} 0x+y+\frac{39}{22}z=-\frac{27}{22}, \\ 0x+0y+\frac{10}{11}z=\frac{10}{11}, \\ x+0y+\frac{41}{22}z=-\frac{3}{22}. \end{gathered}[/tex]

Multiplying the second equation by 11/10 we get:

[tex]\begin{gathered} 0x+y+\frac{39}{22}z=-\frac{27}{22}, \\ 0x+0y+z=1, \\ x+0y+\frac{41}{22}z=-\frac{3}{22}. \end{gathered}[/tex]

Subtracting 39/22 times the second equation from the first one we get:

[tex]\begin{gathered} 0x+y+0z=-3, \\ 0x+0y+z=1, \\ x+0y+\frac{41}{22}z=-\frac{3}{22}. \end{gathered}[/tex]

Subtracting 41/22 times the second equation from the third one we get:

[tex]\begin{gathered} 0x+y+0z=-3, \\ 0x+0y+z=1, \\ x+0y+0z=-2. \end{gathered}[/tex]

Therefore, the solution to the given system of equations is x=-2, y=-3, and z=1.

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