[tex]5x-2y+z=-1 \\ 2x+y+2z=6 \\ x-3y-z=-5[/tex]
firat we need to get two equation with two varibles let us work on x,y
so adding the first and last one will yield
[tex]5x-2y+z=-1 \\ x-3y-z=-5 \\ -------- \\ 6x-5y=-6[/tex]
now we since we used the first and third we need to use the second to get a correct system.let us multiply the third by 2 then add the second and third
[tex]x-2y-z=-5 \\ multiply2 \\ 2x-4y-2z=-10 \\ 2x+y+2z=6 \\ -------- \\ 4x-3y=-4[/tex]
now we have two equation with the variables x and y
[tex]6x-5y=-6 \\ 4x-3y=-4[/tex]you can solve it algebraically but you can see that the only solution possible is y=0 and x=-1 we have the values for x and y let us choose one of the three main equation and substitute to get z let us pick the first equation 5x-2y+z=-1-->5(-1)-2(0)+z=-1---->-5+z=-1-------->z=4
to make sure the system works let us check by substituting into the three equations
the first one will be 5x-2y+z=-1--->5(-1)-2(0)+4=-1---->-5+4=-1--->-1=-1 first equation holds
the second equation 3x+y+2z=6---->2(-1)+0+2(4)=6--->-2+8=-6--->-6=-6 second equation holds
the third equation x-3y-z=-5----->-1-3(0)-4=-5---->-1-4=-5--->-5=-5
our third equation also holds which makes our solution correct
x=-1,y=0,z=4
