SOLUTION: CAN SOMEONE HELP ME WITH THIS PROBLEM? I AM A MATH DISASTER AND CANNOT FIGURE OUT HOW TO FIGURE THIS OUT? THANKS SO MUCH! Different growth rates. The combined population of Mary

Question 187098: CAN SOMEONE HELP ME WITH THIS PROBLEM? I AM A MATH DISASTER AND CANNOT FIGURE OUT HOW TO FIGURE THIS OUT? THANKS SO MUCH!
Different growth rates. The combined population of Marysville and Springfield was 25,000 in 2000. By 2005 the population of Marysville had increased by 10%, while Springfield had increased by 9%. If the total population increased by 2380 people then what was the poulation of each city in 2000?
Can anyone help me solve this problem by substitution

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The combined population of Marysville and Springfield was 25,000 in 2000.
By 2005 the population of Marysville had increased by 10%, while Springfield
had increased by 9%. If the total population increased by 2380 people then what
was the population of each city in 2000?
:
Let s = Springfield population in 2000
Let m = Marysville population in 2000
Then
.09s = Springfield increase by 2005
.1m = Marysville increase by 2005
:
"The combined population of Marysville and Springfield was 25,000 in 2000."
m + s = 25000
or
s = (25000 - m)
:
"By 2005 the population of Marysville had increased by 10%, while Springfield had increased by 9%. If the total population increased by 2380 people"
.1m + .09s = 2380
:
"then what was the population of each city in 2000?"
:
Substitute (25000-m) for s in the above equation:
.1m + .09(25000-m) = 2380
.1m + 2250 - .09m = 2380
.1m - .09m = 2380 - 2250
.01m = 130
m =
m = 13,000 is the 2000 population of Marysville
then
s = 25000 - 13000
s = 12,000 is the 2000 population of Springfield
;
;
Check solution by finding the amt of increase in 2005
.1(13000) + .09(12000) =
1300 + 1080 = 2380, confirms our solutions
;
:
did this make sense to you now? Any questions about it?