How do you solve $5x+2y-z=-7$ , $x-2y+2z=0$ , and $3y+ z=17$ ?

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Answer 1 · 2017-08-04T14:50:54

The solution is $((x),(y),(z))=((-2),(4),(5))$

We perform the Gauss Jordan elimination with the augmented matrix

$((1,-2,2,:,0),(5,2,-1,:,-7),(0,3,1,:,17))$

$R2larrR2-5R1$

$((1,-2,2,:,0),(0,12,-11,:,-7),(0,3,1,:,17))$

$R3larr4R3-R2$

$((1,-2,2,:,0),(0,12,-11,:,-7),(0,0,15,:,75))$

$R3larrR3/15$

$((1,-2,2,:,0),(0,12,-11,:,-7),(0,0,1,:,5))$

$R2larrR2+11R3$

$((1,-2,2,:,0),(0,12,0,:,48),(0,0,1,:,5))$

$R2larrR2/12$

$((1,-2,2,:,0),(0,1,0,:,4),(0,0,1,:,5))$

$R1larrR1-2R3$

$((1,-2,0,:,-10),(0,1,0,:,4),(0,0,1,:,5))$

$R1larrR1+2R2$

$((1,0,0,:,-2),(0,1,0,:,4),(0,0,1,:,5))$

How do you solve 5x+2y-z=-75x+2y-z=-7, x-2y+2z=0x-2y+2z=0, and 3y+ z=173y+ z=17?