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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?

 Apr 18, 2021
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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 - (2 - 3kx)  =  -15

                                    -6x - 2 + 3kx  =  -15

                                      6x - 2 - 3kx  =  15

                                          6x - 3kx  =  1

                                          3x(2 - k)  =  17

If  2 - k  =  0,  there is no value for x that will result in a product of 10,

so  k = 2  results in no possible solution.

For every other value of k, there will be a value of x that can produce a product of 10.

 

So the answer is k = 2.

 Apr 18, 2021

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