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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 4, 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 4, 2021
#2
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That's wrong. Someone give me the correct ansawer please.

Aug 5, 2021