For number 4)
\(x+y+3z=10\)
\(4x+2y+5z=7\)
\(kx+z=3\)
Multiply 2 on both sides of the 1st equation:
\(2x+2y+6z=20\)
Subtract it to the 2nd equation:
\(2x-z=-13\)
\(kx+z=3\)
multiply -1 to \(kx+z=3\) on both sides:
\(-kx-z=-3\)
in order for this system to have no solutions
k has to be -2 which gets us: \(2x-z=-3\)
This equation can't have 2 values since it is linear,
so this proof tells us, k = 2
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