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For a certain value of k, the system
\(\begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)
has no solutions. What is this value of k?

 Aug 18, 2021
 #1
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The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7       --->                      -4x + 3y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z = 3        --->         z  =  3 - kx

Substituting these:        -6x - (3 - kx)  =  -15

                                         (k - 6)x  = -18

If  k - 6  =  0,  there is no value for x that will result in a product of -18

so  k = 6  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of -18.

 

So the answer is k = 6.

 Aug 18, 2021
 #2
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sorry for the bother but it isnt correct. i need some help with this

Guest Aug 18, 2021
 #3
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let me try using your way of doing to see if i can get a correct answer

Guest Aug 18, 2021
 #4
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i think you messed up when you added
 

-2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7

because -2y+3y doesnt cancel out 🙂

Guest Aug 18, 2021
 #5
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update: i got the answer it was 7/4. thanks for the help tho!!!

Guest Aug 18, 2021

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