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Solve this equation

{3x+y+2z=6

{x+y+4z=3

{2x+3y+2z=2

ManuelBautista2019  Dec 10, 2017
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3x + y + 2z  = 6    (1)

x + y +  4z = 3      (2)

2x + 3y + 2z = 2    (3)

Manuel....we can use the first two equations....write them as

y = -3x - 2z + 6

y = - x - 4z + 3          subtract the first equation from the second

0  =  2x  - 2z - 3

2x  = 2z + 3        (4)

Multiply the second equation by 2

2x + 2y + 8z  = 6      (5)

Put   (4)   into  (3)  and (5)  for 2x

(2z + 3) + 3y + 2z   = 2   ⇒     3y + 4z  = -1     (6)

(2z + 3) + 2y + 8z  =  6  ⇒       2y  + 10z =  3  (7)

Multiply   (6)  by 2  and (7)  by -3

6y + 8z =  -2

-6y - 30z  = -9        add these

-22z  =  - 11       divide both sides by -22

z  = 1/2

Using (4)     2x   =  2z + 3

2x  = 2(1/2) + 3

2x  = 4

x  = 2

And using (7)     2y + 10z  = 3

2y + 10 (1/2)  = 3

2y  + 5  = 3

2y  = -2

y = -1

So    .... { x, y, z }   =  { 1/2, -1, 2 }

CPhill  Dec 10, 2017

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