Final answer:
The correct system of equations that has the solution (2,3) is Option D: Ax + By = C and (M-A)x + (N-B)y = P-C, as it is the only system that can be guaranteed to have (2,3) as a solution without knowing the specific values of A, B, C, M, N, and P. Plugging the values of x=2 and y=3 into the equations can satisfy both equations simultaneously if A, B, C, M, N, and P are chosen correctly.
Step-by-step explanation:
The student is asking about the system of equations that has the solution (2,3). To determine which system of equations has this solution, we can plug the values of x=2 and y=3 into each system and see which system is valid for these values.
If we plug in the values into the first system, option A:
(A+M)x + (B+N)y = C + P,
we get (A+M)·2 + (B+N)·3 = C + P, which does not provide enough information to assure that (2,3) is the solution without knowing the values of A, B, C, M, N, and P.
Option B, after plugging the values, gives us:
(A+M)2 + B3 = C + P.
Again, without the specific values of A, B, C, M, and P, we cannot be certain that (2,3) is the solution.
For option C, we must see if the proposed solution satisfies both equations:
Ax + By = C,
(2A + M)x + (2B + N)y = 2C + P.
Plugging in (2,3), we would have A·2 + B·3 = C for the first equation and (2A + M)·2 + (2B + N)·3 = 2C + P for the second, which is not conclusively satisfied without the actual values of A, B, C, M, N, and P.
Finally, option D:
Ax + By = C,
(M-A)x + (N-B)y = P-C.
After inserting (2,3) into both equations, we get A·2 + B·3 = C and (M-A)·2 + (N-B)·3 = P-C. For this to be true, both equations must hold for the given solution, which is possible, assuming the values of A, B, M, N, C, and P are chosen appropriately. Hence, option D is the only system that can be guaranteed to have (2,3) as a solution without knowing the specific values.
Therefore, the correct answer is option D, where the equations are Ax + By = C and (M-A)x + (N-B)y = P-C.