The population y in the year x of the cities listed is approximated by the given equations, where \( \mathrm{x}=10 \) corresponds to 2010 and \( y \) is in thousands. Solve this system of equations, and interpret the answer. \( \begin{array}{lrl}\text { City A: } & -32 x+y=1958 \\ \text { City B: } & -22 x+y=2068 \\ \text { City C: } & -73 x+2 y=3817\end{array} \) What does the answer represent? Select the correct choice and, if necessary, fill in the answer boxes to complete your choice. A. All three cities had the same population of \( \square \) in the year \( \square \) . B. Two cities had the same population of \( \square \) in the year \( \square \) . C. There will be infinitely many times when the three cities will have the same population.

Let’s first solve the system of equations. 1. From City A: \(-32x + y = 1958\) (Rearranging gives \(y = 32x + 1958\)) 2. From City B: \(-22x + y = 2068\) (Rearranging gives \(y = 22x + 2068\)) 3. From City C: \(-73x + 2y = 3817\) (Rearranging gives \(2y = 73x + 3817\), or \(y = 36.5x + 1908.5\)) Now let's set the equations for City A and City B equal to each other to find \(x\): \[ 32x + 1958 = 22x + 2068 \] \[ 10x = 110 \implies x = 11 \] Now substitute \(x = 11\) into City A’s equation to find \(y\): \[ y = 32(11) + 1958 = 352 + 1958 = 2310 \] Now substitute \(x = 11\) into City B’s equation to check: \[ y = 22(11) + 2068 = 242 + 2068 = 2310 \] Now substitute \(x = 11\) into City C's equation to check if it also yields the same population: \[ y = \frac{73(11) + 3817}{2} = \frac{803 + 3817}{2} = \frac{4620}{2} = 2310 \] Thus, all three equations yield the population of 2310 (thousands) when \(x = 11\) (which corresponds to the year 2021). So the correct answer is: A. All three cities had the same population of 2310 in the year 2021.