+0  
 
+3
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avatar+1672 

What is the way you solve this?

 Feb 22, 2019
 #1
avatar+98060 
+3

These can be lengthy !!!

 

x + y + z = 8     (1)

x - y - z = 8        (2)

2x + y + 2z = 16    (3)

 

Add (1) and (2)    and we get that    2x = 16 ⇒  x = 8

 

Put this into (2) and (3)  for x

 

8 - y - z = 8   ⇒      -y - z = 0      (4) 

2(8) + y + 2z =  16   ⇒   16 + y + 2z = 16 ⇒   y + 2z = 0     (5)

 

Add (4) and (5)  and we get that

 

z = 0

 

Using (1)

 

8 + y + 0 = 8

 

y = 0

 

So

 

{ x, y , z }   = { 8, 0 , 0}

 

 

cool cool cool

 Feb 22, 2019
 #2
avatar+98060 
+2

Second one

x + y -  z = -1             (1) 

4x + 4y - 4z = -2     ⇒    x + y - z = -1/2    (2) 

3x + 2y + z = 0         (3)

 

Note GM....there are no solutions possible

The reason why???....look at (1) and (2)

 

x + y - z     cannot  equal  -1    and -1/2   at the same time  !!!!

 

Whatever x + y - z    is    ....it must lead to the same result every time  !!!

 

 

cool cool cool

 Feb 22, 2019
 #3
avatar+98060 
+2

Last one

 

x + y + z  = 4           (1) 

5x + 5y + z = 12      (2)

x - 4y + z = 9    ⇒   -x  + 4y - z  = -9     (3)

 

Add (1) and (3)    and we have

5y  = - 5   

y = -1

 

Put this into  (1)

x - 1 + z = 4

x + z = 5

z = 5 - x

 

Using (2)  and making the substitutions we have

 

5x + 5(-1) + 5 - x = 12

4x  = 12

x = 3

 

And

z = 5 - 3 = 2

 

So {x, y, z}   = { 3, -1, 2 }

 

 

cool cool cool

 Feb 22, 2019
edited by CPhill  Feb 22, 2019

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