Problem

According to projections through the year 2030, the population $y$ of the given state in year $x$ is approximated by
State A: $\quad-4 x+y=11,700$
State $\mathbf{B}:-140 x+y=5,000$
where $\mathrm{x}=0$ corresponds to the year 2000 and $\mathrm{y}$ is in thousands. In what year do the two states have the same population?
The two states will have the same population in the year

Answer

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Answer

So, the two states will have the same population in the year \(\boxed{2050}\).

Steps

Step 1 :We are given two equations representing the population of two states A and B. The equations are: State A: \(-4x + y = 11700\) and State B: \(-140x + y = 5000\). Here, x represents the year (with x=0 corresponding to the year 2000) and y represents the population in thousands.

Step 2 :We are asked to find the year when the two states have the same population. This means we need to find the value of x when the two equations are equal.

Step 3 :To do this, we can subtract the second equation from the first to eliminate y. This gives us a new equation: \(136x = 6700\).

Step 4 :Solving this equation for x gives us \(x = \frac{1675}{34}\).

Step 5 :Since x represents the year with x=0 corresponding to the year 2000, we add this value to 2000 to get the actual year. This gives us \(2000 + \frac{1675}{34} = 2050\).

Step 6 :So, the two states will have the same population in the year \(\boxed{2050}\).

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