Question 13

There is one pair (x, y) of integers that satisfies the equation.

The equation \((x−7)^2+(y+3)^2=0\) only holds true when both squared terms are equal to zero. This occurs at the specific point where \(x = 7\) and \(y = -3\), resulting in exactly one integer solution.

A) None

This option is incorrect because the equation does have a solution. The conditions set by the equation do allow for a specific pair of integers that satisfy it, namely (7, -3).

B) One only

This option is correct since the equation can only be satisfied by the unique solution where \(x = 7\) and \(y = -3\). This is the only pair of integers that makes both squared terms zero, confirming that there is precisely one solution.

C) Two only

This choice is incorrect as the equation does not yield two distinct pairs of integers. The nature of the equation being the sum of squares equating to zero restricts it to a single solution.

D) Three only

This option is also incorrect. The equation does not allow for three different integer solutions; it strictly defines one unique point where the conditions are met.

E) More than three

This choice is incorrect as well, as the equation does not permit more than one integer solution. The structure of the equation limits the possible pairs to just one.

Conclusion

The only valid solution to the equation \((x−7)^2+(y+3)^2=0\) is the single pair (7, -3). Therefore, the answer is definitively one pair of integers, making option B the correct choice, while all other options misinterpret the equation's constraints.