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

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Answer 1 · 2017-10-28T19:30:20

$x = \frac{25}{26}$ $x=25/26$ , $y = - \frac{37}{26}$ $y=-37/26$ and $z = - \frac{56}{13}$ $z=-56/13$

After summing first 2 equations for removing $z$ $z$

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

After summing second and third ones for removing $z$ $z$

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

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

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

$52 x = 50$ $52x=50$ , so $x = \frac{25}{26}$ $x=25/26$

Now, I found $y$ $y$ from first one.

$8 \cdot \frac{25}{26} + 4 y = 2$ $8*25/26+4y=2$ or $4 y = - \frac{74}{13}$ $4y=-74/13$ , hence $y = - \frac{37}{26}$ $y=-37/26$

Now, I found $z$ $z$ from third one.

$4 \cdot \frac{25}{26} - 2 \cdot (- \frac{37}{26}) - z = 11$ $4*25/26-2*(-37/26)-z=11$

$\frac{50}{13} + \frac{37}{13} - z = 11$ $50/13+37/13-z=11$

$z = \frac{87}{13} - 11 = - \frac{56}{13}$ $z=87/13-11=-56/13$

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