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For a certain value of k, the system

x + y + 3z = 10, 
-4x + 2y + 5z = 7, 

kx + z &= 3
has no solutions. What is this value of k?

 Oct 29, 2019
 #1
avatar+109202 
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x + y + 3z = 10    ⇒ -2x - 2y - 6z  = -20      (1)
-4x + 2y + 5z = 7        (2)

kx + z = 3       (3)

 

Add (1)  and (2)   and we get that

 

-6x - z  =  -13      Multiply through by  -1

6x + z  = 13       (4)

 

Note that, in (3) , if  k = 6 we have that

 

6x + z  =  3      (5)

 

But this is impossible because  if 6x + z  = 13 in (4), it cannot also = 3   in (5)

 

So....k = 6   gives no solutions

 

 

cool cool cool

 Oct 30, 2019

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