in 1900, greenville had a population of 4000 people. Greenville's population increased by 100 people each year. Springfield had a population of 1000 people. Springfields population increased bu 4% each year. In what year were the populations about equal?

year 1948

Step-by-step explanation:

1900 is our t = 0

The population of Greenville can be written as:

G = 4000 + 100*t

the equation for Springfield can be written as:

S = 1000*(1.04)^t

we want to find the value of t that makes S(t) = G(t)

4000 + 100*t = 1000*(1.04)^t

(4000 + 100*t) = 1.04^t

4 + 0.1*t = 1.04^t

now, you can graph both of this equations (left side and right side) and see in wich value of t the graphs intersect eachother, or you also may use different values of t until the values are about the same for both sides, which is the thing i did with the equation:

4 = 1.04^t - 0.1*t

You will find that the correct value is t = 48

So we can assume that in year 1948 the populations will be about the same.

Since we want to solve for the variable x, we want to isolate x

a²x + (a - 1) = (a + 1)x ⇒ Distribute x to (a+1). Also, remove parentheses

a²x + a - 1 = ax + x ⇒ Subtract a from both sides

a²x - 1 = ax + x - a ⇒ Add 1 to both sides

a²x = ax + x - a + 1 ⇒ Subtract (ax + x) from both sides

a²x - (ax + x)= ax + x - a + 1 - (ax+x) ⇒ Simplify. Remember that multiplying positive by negative = negative

a²x - ax - x = ax + x - a + 1 - ax - x ⇒ Simplify

a²x - ax - x = -a + 1 ⇒ Factor out the x from a²x - ax - x

x(a² - a - 1) = -a + 1 ⇒ Divide both sides by (a² - a - 1)

x = (-a + 1) / (a² - a - 1)

However, we need to make sure that the denominator does not equal 0. Therefore, you set the denominator = 0 (just use the quadratic formula for this), and it gives that the denominator =0 when a = (1+√5)/2 AND (1-√5)/2

Therefore, the final answer is

x = (-a + 1) / (a² - a - 1) given that a ≠ (1+√5)/2, a ≠ (1-√5)/2