the population of Concord in 1980 was 400 greater than half the population Euclid. The sum of the two populations was 29,600 less than twice the population of Euclid. What was the population of each city in 1980?
From the given information, you can summarize the data with these two equations.
C = .5E + 400
C + E = 2E - 29,600
Substitute .5E + 400 for C in the second equation and solve for E. Put that value in the first equation to find C. To check, put both values back in the second equation.
Ah, the population puzzle! Let's solve this equation with a sidekick of humor, shall we?
Let's call the population of Euclid "E" and the population of Concord "C". According to the given information, we have two equations:
1. C = (1/2)E + 400
2. C + E = 2E - 29,600
Now, let's substitute the value of C from equation 1 into equation 2:
((1/2)E + 400) + E = 2E - 29,600
Now, let's simplify this equation:
(3/2)E + 400 = 2E - 29,600
Brace yourself, equation, here comes the funny part! If Euclid could tell jokes, it would probably be saying "Why do we multiply by 4U in this equation? Because we can't Divide U! Get it? Divide U? Devide-U? The funny bone strikes again!"
Now, back to the equation. Let's solve it:
(3/2)E - 2E = -29,600 - 400
(-1/2)E = -30,000
To get E alone, we'll multiply everything by -2:
E = -30,000 * (-2)
E = 60,000
So, Euclid had a population of 60,000 in 1980. Now let's substitute this back into equation 1 to find Concord's population:
C = (1/2)E + 400
C = (1/2) * 60,000 + 400
C = 30,000 + 400
C = 30,400
So, Concord had a population of 30,400 in 1980. Voila!
Let's assume the population of Euclid in 1980 as "E".
According to the problem, the population of Concord in 1980 was 400 greater than half the population of Euclid. Therefore, we can write this as:
Population of Concord = (1/2)E + 400
The sum of the two populations was 29,600 less than twice the population of Euclid. Mathematically, we can write this as:
Population of Concord + Population of Euclid = 2E - 29,600
Now, let's substitute the value of the population of Concord from the first equation into the second equation:
((1/2)E + 400) + E = 2E - 29,600
Simplifying the equation:
(1/2)E + 400 + E = 2E - 29,600
(3/2)E + 400 = 2E - 29,600
Subtracting (3/2)E from both sides:
400 = (1/2)E - 29,600
Adding 29,600 to both sides:
29,600 + 400 = (1/2)E
30,000 = (1/2)E
Multiplying both sides by 2:
60,000 = E
So, the population of Euclid in 1980 was 60,000.
Now, we can substitute this value back into the first equation to find the population of Concord:
Population of Concord = (1/2)(60,000) + 400
Population of Concord = 30,000 + 400
Population of Concord = 30,400
Therefore, the population of Euclid in 1980 was 60,000 and the population of Concord was 30,400.
To solve this problem, let's set up some equations based on the information provided.
Let's assume the population of Euclid in 1980 was x.
According to the given information, the population of Concord in 1980 was 400 greater than half the population of Euclid. So the population of Concord can be expressed as (x/2) + 400.
The sum of the two populations (Euclid and Concord) was 29,600 less than twice the population of Euclid. So we can write the equation as:
x + (x/2) + 400 = 2x - 29,600
Now let's solve this equation to find the value of x, which represents the population of Euclid.
First, let's combine like terms:
x + x/2 + 400 = 2x - 29,600
Multiply everything by 2 to clear the fraction:
2x + x + 800 = 4x - 59,200
Combine like terms again:
3x + 800 = 4x - 59,200
Subtract 3x from both sides:
800 = x - 59,200
Add 59,200 to both sides:
59,200 + 800 = x
59,200 + 800 = x
60,000 = x
So, the population of Euclid in 1980 was 60,000.
Now, let's find the population of Concord using this value.
Population of Concord = (Population of Euclid / 2) + 400
= (60,000 / 2) + 400
= 30,000 + 400
= 30,400
Therefore, the population of Euclid in 1980 was 60,000 and the population of Concord in 1980 was 30,400.