For how many pairs (x, y) of integers does(x−7)2+(y+3)2=0
There is one pair (x, y) of integers that satisfies the equation.
The equation \((x - 7)^{2} + (y + 3)^{2} = 0\) can only hold true when both squared terms equal zero. This leads to a unique solution where \(x\) must equal 7 and \(y\) must equal -3.
If there were no pairs of integers satisfying the equation, it would imply that the sum of the squares could never equal zero. However, since both squares must be equal to zero simultaneously, this option is incorrect as there is indeed one valid solution.
The only solution arises when both \((x - 7)^{2} = 0\) and \((y + 3)^{2} = 0\). Solving these gives \(x = 7\) and \(y = -3\), indicating that there is precisely one pair of integers \((7, -3)\) that satisfies the equation.
For there to be two pairs of integers, the equation would need to allow for multiple combinations of \(x\) and \(y\) that yield zero. However, since both squared terms must independently equal zero, it is impossible to have more than one solution in this scenario.
Similar to the reasoning for option C, three pairs would suggest that the equation permits multiple distinct solutions. The nature of the equation, requiring both terms to equal zero, restricts the solution to a single pair, making this option incorrect.
This option would imply an even greater number of integer pairs satisfying the equation. Given that we can only satisfy both conditions with one unique pair, this choice is also incorrect.
The equation \((x - 7)^{2} + (y + 3)^{2} = 0\) allows for only one solution: the integer pair \((7, -3)\). All other options incorrectly suggest the existence of additional pairs, which is impossible given the requirements of the equation. Thus, the only valid count of integer pairs is one.
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