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For a certain value of k, the system \(\begin{align*} x + y + 3z &= 10, \\ 4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)
has no solutions. What is this value of k

 Jun 10, 2023
 #1
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If we eliminate y from the first two equations, we get x+3z=10−2x, or x=5−3z. Substituting this into the third equation, we get [kx + z = 3,]so k(5−3z)+z=3. This simplifies to 6z−5k=2. For this to have no solutions, we must have 6z−5k=0, or k=6​.

 Jun 10, 2023
 #2
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incorrect

Guest Jun 10, 2023
 #3
avatar+33615 
+1

Do the following:

 Jun 11, 2023

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