How do you solve the following system?: $x+3y-z=1, 7x+y+z=1, 4x-2y-z=11$

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Answer 1 · 2017-10-28T19:30:20

$x=25/26$ , $y=-37/26$ and $z=-56/13$

After summing first 2 equations for removing $z$

$x+3y-z+7z+y+z=1+1$ or $8x+4y=2$

After summing second and third ones for removing $z$

$7x+y+z+4x-2y-z=1+11$ or $11x-y=12$

Now, I removed $y$ term from them. I also found $x$

$4*(11x-y)+8x+4y=4*12+2$

$52x=50$ , so $x=25/26$

Now, I found $y$ from first one.

$8*25/26+4y=2$ or $4y=-74/13$ , hence $y=-37/26$

Now, I found $z$ from third one.

$4*25/26-2*(-37/26)-z=11$

$50/13+37/13-z=11$

$z=87/13-11=-56/13$

How do you solve the following system?: x+3y-z=1, 7x+y+z=1, 4x-2y-z=11 x+3y-z=1, 7x+y+z=1, 4x-2y-z=11