# math,algebra

posted by
**mathstudent05** on
.

Can someone help me set this up to solve or what should i do because i have no clue. thanks.

Some scientists believe there is a limit to how long humans can live.* One supporting argument is that

during the last century, life expectancy from age 65 has

increased more slowly than life expectancy from birth, so

eventually these two will be equal, at which point, according to these scientists, life expectancy should increase no further. In 1900, life expectancy at birth was 46 yr, and life

expectancy at age 65 was 76. In 2000, these figures had risen to 76.9 and 82.9, respectively. In both cases, the

increase in life expectancy has been linear. Using these assumptions and the data given, find the maximum life

expectancy for humans.

The problem is based on unfounded assumptions, but that is not the problem.

You have two linear equations, and you are looking for the common point where those lines intersect.

Line1) Expectancy0= 46 + (76.9-46)/100 * year

LIne2) Expectancy65= 76 + (82.9=76)/100 * year

where year is time in years after 1900.

solve to see where expectancy0=expectance65

Can you continue helping me. I am not seeing how to bring out this problem. this is what i had in the beginning:

letting x = # of yrs from 1900 (1900=0 and 2000 = 100)

letting y = yrs of life expected

determining the "from birth" equation

x 1 = 0 ; y 1 = 46

x2 = 100;y2=76.9

i figured out slope to be:

m= (76.9-46)/(100-0)=(30.9)/(100) = 0.309

then by using equation: y-y1=m(x-x1)

y-46=0.309 (x-0)

y=0.309x + 46

then i determine "from age 65" equation

x1=0;y1 = 76

x2= 0.309; y = 82.9

finding slope

m = (82.9-76)/(100-0)=(6.9)/(100)=0.069

then i make it into the format

y-76=0.069 (x-1)

y=0.069x+76

so then by placing both equations in standard form to solve

-0.309x+y=46

-0.069x+y=76

subtracting eliminating y

-0.24x=-30

x = (-30)/(-0.24)

x= 125 yrs (yr 2025)

Actual age find y

-0.069 (125) - y = 76

-8.625 + y = 76

y=76 +8.625

y=84.625 max age

yes, that is it.