SOLUTION: In 2005, there were 11,500 students at college A, with a projected enrollment increase of 1,200 students per year. In the same year, there were 22,700 students at college B, with a

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Question 626893: In 2005, there were 11,500 students at college A, with a projected enrollment increase of 1,200 students per year. In the same year, there were 22,700 students at college B, with a projected enrollment decline of 400 students per year. in what year will the two colleges have the same enrollment? At that time how many students be enrolled in both colleges?

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In 2005, there were 11,500 students at college A, with a projected enrollment increase of 1,200 students per year. In the same year, there were 22,700 students at college B, with a projected enrollment decline of 400 students per year. in what year will the two colleges have the same enrollment? At that time how many students be enrolled in both colleges?
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We have two straight line equations that take the standard form: y=mx+b, m=slope, b=y-intercept.
For given problem, college A:
y-intercept, b=11500
slope, m=1200 students per year increase
equation: y=1200x+11500 (number of students for given year x)
..
For given problem, college B:
y-intercept, b=22700
slope, m=-400 students per year decrease
equation: y=-400x+22700 (number of students for given year x)
..
When number of students are the same for both colleges:
1200x+11500=-400x+22700
1600x=11200
x=11200/1600
x=7
y=1200x+11500=19900.
..
The two colleges will have the same enrollment in year 2012
Number of students enrolled at both colleges at this time: 19,900
note: It would be instructive to plot both equations on an x-y coordinate system and see that the answer is the point of intersection of these two equations.