Question

The population of two different villages is modeled by the equations shown
Suppose y represents the population and x represents the number of years
since 1980. What year(s) were the populations of both villages the same?
What was the population of both villages in the year(s) they were the same
Montville: y = x2 – 30x + 540
Havertown: y = 20x + 15

Answer 1

Answer:

Year = 1995 and population = 315

Year = 2015 and population = 715

Step-by-step explanation:

It is given that the population of two different villages is modeled by the given equations:

$y=x^2-30x+540$

$y=20x+15$

The population of both villages are same after x years after 1980 if

$x^2-30x+540=20x+15$

$x^2-30x+540-20x-15=0$

$x^2-50x+525=0$

Splitting the middle term, we get

$x^2-15x-35x+525=0$

$x(x-15)-35(x-15)=0$

$(x-15)(x-35)=0$

$x=15,35$

It means, after 15 or 35 years, the population will same.

For x=15, years is $1980+15=1995$ and population is  

$y=20(15)+15=315$

For x=35, years is $1980+35=2015$ and population is  

$y=20(35)+15=715$.

Therefore, population are equation in Year = 1995 and population = 315 or Year = 2015 and population = 715.

Answer 2

Year = 1995 and population = 315

Year = 2015 and population = 715

Calculation:

The equation is

$y = x^2 - 30x + 540\\\\y = 20x + 15$

Now

$x^2 - 30x + 540 = 20x + 15\\x^2 - 30x + 540 - 20x - 15=0\\\\x^2 - 50x + 525 = 0\\\\x^2 - 35x - 15x + 525=0\\\\x(x - 15) -35(x-15) = 0\\\\(x-15) (x-35) = 0$

x = 15,35

so,

When x is 15 , the population is = 20(15 )  +15 = 300 + 15 = 315

And, x is 35 , the population is = 20(35 )  +15 = 700 + 15 = 715

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