Respuesta :
We are given the following system of equations:
[tex]\begin{gathered} x+2y+z=8,(1) \\ 2x+y-z=1,(2) \\ x+y-2z=-3,(3) \end{gathered}[/tex]To solve the system we will add equations (1) and (2):
[tex]x+2y+z+2x+y-z=8+1[/tex]Adding like terms:
[tex]\begin{gathered} 3x+3y=9 \\ x+y=3,(4) \end{gathered}[/tex]Now we multiply equation (2) by -2:
[tex]-4x-2y+2z=-2[/tex]Now we add this equation to equation (3):
[tex]x+y-2z-4x-2y+2z=-3-2[/tex]Adding like terms:
[tex]-3x-y=-5,(5)[/tex]Now we add equations (4) and (5):
[tex]x+y-3x-y=3-5[/tex]Adding like terms:
[tex]-2x=-2[/tex]Dividing both sides by -2:
[tex]x=-\frac{2}{-2}=1[/tex]Now we replace this value of "x" in equation (4):
[tex]\begin{gathered} x+y=3 \\ 1+y=3 \end{gathered}[/tex]Subtracting 1 to both sides:
[tex]\begin{gathered} 1-1+y=3-1 \\ y=2 \end{gathered}[/tex]Now we replace the values of "x" and "y" in equation (1):
[tex]\begin{gathered} x+2y+z=8 \\ 1+2(2)+z=8 \end{gathered}[/tex]Adding like terms:
[tex]\begin{gathered} 1+4+z=8 \\ 5+z=8 \end{gathered}[/tex]Subtracting 5 to both sides:
[tex]\begin{gathered} 5-5+z=8-5 \\ z=3 \end{gathered}[/tex]Therefore, the solution of the system is:
[tex]x=1,\text{ y=2, z=3}[/tex]