For a certain value of k, the system x+y+3z=10,4x+2y+5z=7,kx+z=3 has no solutions. What is this value of k
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.
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