The population \( y \) (in millions) of two states in year \( x \) can be approximated by the following, where \( x=10 \) corresponds to the year 2010. In what year did the two states have the same population? What was the population of these states at that time? State A: \( \quad 10 y-x=84 \) State B: \( 100 y-x=975 \) The two states had the same population in the year \( \square \). The population of these states at that time was million. (Type integers or decimals.)

To find the year when the populations of State A and State B were equal, we first need to rearrange the equations for each state and express the populations in terms of \( x \). For State A: \[ 10y - x = 84 \quad \text{(1)} \] Rearranging gives: \[ 10y = x + 84 \implies y = \frac{x + 84}{10} \quad \text{(2)} \] For State B: \[ 100y - x = 975 \quad \text{(3)} \] Rearranging gives: \[ 100y = x + 975 \implies y = \frac{x + 975}{100} \quad \text{(4)} \] Now, we set equations (2) and (4) equal to each other to find the year \( x \) when their populations are the same: \[ \frac{x + 84}{10} = \frac{x + 975}{100} \] To eliminate the fractions, multiply both sides by 1000: \[ 100(x + 84) = 10(x + 975) \] Expanding both sides: \[ 100x + 8400 = 10x + 9750 \] Rearranging gives: \[ 100x - 10x = 9750 - 8400 \] \[ 90x = 1350 \implies x = 15 \] Now, substituting \( x = 15 \) back into either equation (using State A's equation): \[ 10y - 15 = 84 \implies 10y = 99 \implies y = 9.9 \] Thus, the two states had the same population in the year \( 15 + 2000 = 2025 \). At that time, the population of these states was 9.9 million. The two states had the same population in the year \( 2025 \). The population of these states at that time was 9.9 million.