let

x = be the time Myline needs to do the work, x>0

y = be the time Jeana needs to do the work, y>0

it takes Myline twice as long as Jeana to do a certain piece of work

x = 2y

working together, they can finish the work in 6 hours

1/x + 1/y = 1/6 multiply both sides by 6xy

6y + 6x = xy

6y + 6*2y = 2y*y

2y^2 - 18y = 0 divide both sides by 2

y^2 - 18y = 0

y(y - 18) = 0

the quadratic equation has 2 solutions

y1 = 0

y2 = 18

since y>0 the solution the our problem is y = 18 hours.

Jeana needs 18 hours to do the work alone.