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For coincident lines, the second equation must be twice the first. Therefore k = 10.
The second equation must be exactly twice the first. Therefore k = 14.
The second equation is twice the first, so a = 2.
4/8 = 3/6 = 1/2, but 12/25 ≠ 1/2. Hence the lines are parallel and there is no solution.
Substituting x = 3 and y = 2 gives 3k + 8 = 17, so k = 3.
From 2x + y = 7, y = 7 - 2x. Substituting in x + 2y = 8 gives x = 2 and y = 3.
Substituting (4, -1) in 2x + y gives 2(4) - 1 = 7, so the equation is satisfied.
The second equation is twice the first, so both equations represent the same line.
Adding the equations gives 3x = 15, so x = 5. Therefore y = 5.
Coincident lines are the same line, so they have infinitely many common points.
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