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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 $k$?

 Jun 10, 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.

 Jun 10, 2021
 #2
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$x + y + 3z = 10$

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

$kx + z = 3$

Rearranging the 3rd equation, we have $z = 3 - kz$

Let's multiply the 1st equation by $3$: $3x + 3y + 9z = 30$

Subtract the 2nd equation from the new 1st: $7x + 4z = 27$

Plug $3-kz$ for $z:$ $7x + 4(3 - kz) = 27$

Simplifying, we have: $(7-4k)x = 15$

Note: If $7 - 4k = (- \infty, 0) \cup (0, \infty)$ then there will always be a solution for $x,$ and following, $y$ and $z$

Thus, $7-4k$ has to equal $0$ for there to be no solutions for $x,$ which will have no solutions also for $y$ and $z$

We have $7 - 4k = 0$

$4k = 7$

$k = \boxed{\frac{7}{4}}$

 Jun 10, 2021

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