Assume that the suburb has a population of 698000 and is growing at a rate of 4000 per year. Assume that the city has a population of 906000 and is declining at a rate of 12000 per year.
In how many years will the populations of the suburb and the city be equal?
What will the populations be when they are equal?
i got the equations
p=698,000 + 4000y
P=906000 - 12000y
what am i doing wrong
2006-10-18 13:39:48 · 7 answers · asked by Diggler AKA The Cab Driver 1 in Science & Mathematics ➔ Mathematics
698000+4000y=906000-12000y
4000y+12000y=906000-698000
16000y=208000
y=13
2006-10-18 13:42:55 · answer #1 · answered by tranquilllity 2 · 0⤊ 0⤋
Nothing. Now just find y when the populations of the suburb and the city are equal - that is, when p=P. You get:
698,000 + 4000y = 906000 - 12000y
16,000y = 208,000
y=13
2006-10-18 13:44:30 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋
S=Population of suburb
C=Population of city
Y=Number of years
S=698,000 + 4000y
C=906,000 - 12,000y
For the first question, set the equations equal to each other
698,000 + 4000 y = 906,000 -12000 y
Add 12,000 y and subtract 698,000
16,000y = 208,000
Divide by 16,000
y =13 years
For the second question, substitue this y value into each equation
S=698,000 + 4000(13)
S=698,000 + 52,000
S=750,000
C=906,000 - 12000(13)
C=906,000 - 156,000
C=750,000
2006-10-18 13:53:30 · answer #3 · answered by Anonymous · 0⤊ 0⤋
A system of linear equations is two or more linear equations that are being solved simultaneously. In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true. In other words, it is where the two graphs intersect, what they have in common. So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system. A consistent system is a system that has at least one solution. An inconsistent system is a system that has no solution There are three ways to solve systems of linear equations in two variables: graphing substitution method elimination method
2016-05-22 00:53:06 · answer #4 · answered by Anonymous · 0⤊ 0⤋
When they are equal
698000+4000y = 906000-12000y
(12000+4000)y = 906000 - 698000
y = (906000 - 698000)/ (12000+4000) years
2006-10-18 13:45:39 · answer #5 · answered by Roadkill 6 · 0⤊ 0⤋
Just extrapolate over a certain amount of years until they're equal.
2006-10-18 13:42:33 · answer #6 · answered by Poopdragon 3 · 0⤊ 0⤋
698000+4000y=906000-12000y
16000y=906000-698000=208000
16y=208
y=13 years
2006-10-18 14:57:25 · answer #7 · answered by yupchagee 7 · 0⤊ 0⤋