Question 605210
Assume there is 1 work, 
Luke alone can finish it in L hours,
Eddie alone can finish it in E hours,
Ryan alone can finish it in R hours,
so,
Luke's speed = 1/L
Eddie's speed = 1/E
Ryan's speed = 1/R


Working alone, it take Ryan 1 hour more to complete the job than it takes Luke:
R = 1 + L


Luke works twice as fast as Eddie:
1/L = 2 * 1/E
1/L = 2/E
E = 2L


working together:
1 hour and 20 minutes = 1 20/60 hours = 1 1/3 hours = 4/3 hours
{{{1/L + 1/E + 1/R = 1/(4/3)}}}
{{{1/L + 1/(2L) + 1/(1+L) = 3/4)}}}
{{{2/(2L) + 1/(2L) + 1/(1+L) = 3/4)}}}
{{{3/(2L) + 1/(1+L) = 3/4)}}}
{{{(3*(1+L)+1*2L)/((2L)*(1+L)) = 3/4}}}
{{{(3+3L+2L)/(2L+2L^2) = 3/4}}}
{{{(3+5L)/(2L+2L^2) = 3/4}}}
{{{4*(3 + 5L) = 3*(2L+2L^2)}}}
{{{12+20L = 6L+6L^2}}}
{{{0 = 6L^2+6L-20L-12}}}
{{{0 = 6L^2-14L-12}}}
(divide all by 2)
{{{0 = 3L^2-7L-6}}}
0 = (3L + 2)*(L - 3)
3L + 2 = 0 or L - 3 = 0
3L = -2 or L = 3
L = -2/3 or L = 3
because the time can't be a negative number, we choose L = 3
R = 1 + L = 1 + 3 = 4
E = 2L = 2 * 3 = 6


so,
Luke alone takes 3 hours to complete the job
Ryan alone takes 4 hours to complete the job
Eddie alone takes 6 hours to complete the job