SOLUTION: In 2002, capitol city had a population of 2010, and Shelbyville had a population of 1040. if capitol city grows at a rate of 150 people a year and Shelbyville grows at a rate

Question 347107: In 2002, capitol city had a population of 2010, and Shelbyville had a population of 1040. if capitol city grows at a rate of 150 people a year and Shelbyville grows at a rate of 340 people a year , when will the population of Shelbyville be greater than that of capitol city ?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In 2002, capitol city had a population of 2010, and Shelbyville had a population of 1040. if capitol city grows at a rate of 150 people a year and Shelbyville grows at a rate of 340 people a year , when will the population of Shelbyville be greater than that of capitol city ?
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Capitol Equation: C(x) = 150x+2010
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Shelby Equation: S(x) = 340x + 1040
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Question When will S(x)>C(x)
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340x+1040 > 150x+2010
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190x > 970
x > 5.1053 years
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Year: 2002 = 5.1053 = 2007.153
Rounding up Year = 2008
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Cheers,
Stan H.
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