How do you solve the system $x + y - 2 z = 5$ $x+y-2z=5$ , $x + 2 y + z = 8$ $x+2y+z=8$ , $2 x + 3 y - z = 13$ $2x+3y-z=13$ ?

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Answer 1 · 2018-03-20T15:03:40

$x = 5 k + 2$ $x=5k+2$ , $y = 3 - 3 k$ $y=3-3k$ and $z = k$ $z=k$

Perform the Gauss Jordan elimination on the augmented matrix

$A=((1,1,-2,|,5),(1,2,1,|,8),(2,3,-1,|,13))$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R2larrR2-R1$ ; $R3larrR3-2R1$

$A=((1,1,-2,|,5),(0,1,3,|,3),(0,1,3,|,3))$

Consequently this system has infinite solutions. After choosing $z=k$ , $y$ must be $3-3k$ and $x$ must be $5k+2$

How do you solve the system x+y-2z=5x+y−2z=5x+y-2z=5, x+2y+z=8x+2y+z=8x+2y+z=8, 2x+3y-z=132x+3y−z=132x+3y-z=13?