The population \( y \) in the year \( x \) of the cities listed is approximated by the given equations, where \( x=10 \) corresponds to 2010 and \( y \) is in thousands. Solve this system of equations, and interpret the answer. City A: \( \quad \begin{array}{r}-32 x+y=1958 \\ \text { City B: } \\ \text { City C: } \\ -22 x+y=2068 \\ -73 x+2 y=3817\end{array} \) Select the correct choice and, if necessary, fill in the answer boxes to complete your choice. A. The system has exactly one solution. The solution is ( (Type integers or fractions.) B. The system has infinitely many solutions of the form ( (Type integers or fractions.)

To solve the system of equations, we have: 1. For City A: \(-32x + y = 1958\) (Equation 1) 2. For City B: \(-22x + y = 2068\) (Equation 2) 3. For City C: \(-73x + 2y = 3817\) (Equation 3) First, let's manipulate Equations 1 and 2 to eliminate \(y\): From Equation 1, we can express \(y\): \[ y = 32x + 1958 \] Now, substitute this expression for \(y\) into Equation 2: \[ -22x + (32x + 1958) = 2068 \] \[ 10x + 1958 = 2068 \] \[ 10x = 110 \implies x = 11 \] Now plug \(x = 11\) back into the expression for \(y\): \[ y = 32(11) + 1958 = 352 + 1958 = 2310 \] So, the solution for City A and City B is \( (11, 2310) \). Next, we verify this solution with City C's equation: Substituting \(x = 11\) and \(y = 2310\) into Equation 3: \[ -73(11) + 2(2310) = -803 + 4620 = 3817 \] Thus, the equations are consistent. The system has exactly one solution. The solution is \( (11, 2310) \). So, the answer is: **A. The system has exactly one solution. The solution is (11, 2310).**