On her birthday Maya is half again ( or  times) as old as
she was 

x years ago.  Stop there. x years ago she was M-x.

So M = (M-x)

And when she was M-x, then
    
...she was half again ( or  times) as old as she was 

y years ago.  Stop there. y years ago she was M-y.

Therefore M-x = (M-y)

And when she was M-y,

...she lacked five years of being half as old as she was on her last 
birthday.

On her last birthday she was M-1.  So M-y lacks 5 being half of M-1.

So  M-y = (M-1) - 5

So we have a system of 3 equations and 3 unknowns:

M = (M-x)
M-x = (M-y)
M-y = (M-1)-5

Simplify the first equation 

 M = (M-x)
2M = 3(M-x)
2M = 3M-3x
3x = M

Simplify the second equation:

   M-x = (M-y)
2(M-x) = 3(M-y)
 2M-2x = 3M-3y
 3y-2x = M

Simplify the third equation:

M-y = (M-1)-5
2(M-y) = (M-1)-10
2M-2y = M-1-10
2M-2y = M-11
 M-2y = -11

Now the system is:


3x = M
3y-2x = M      
M-2y = -11

Solve that by substitution:

M=99, x=33, y=55

 How old is maya?

She is 99.

Checking:

On her birthday maya is 99, which is half again ( times) as old as
she was 33 years ago (when she was 66).  That checks since ·66
is 99.  Also, 33 years ago when she was 66, she was haif again (
times) as old as she was 55 years ago when she was 44.  And that was when she
lacked five years of being half as old as she was on her last birthday when she
was 98.  And indeed 44 years lacks 5 years being 49, which is half of her
age last year when she was 98.

So it checks.  Maya is the ripe old age of 99.

Edwin