SOLUTION: In 9 years time a mother will be twice as old as her son. Three years ago she was four times as old as her son. Find their present age.

Let M be the mother's present age, and S be the son's present age.

Then in 9 years  

M + 9 = 2*(S+9).   (1)


Three years ago 

M - 3 = 4*(S-3).   (2)


So, you have these two equations

M + 9 = 2S + 18    (3)
M - 3 = 4S - 12.   (4)


Subtract equation (3) from equation (4) (both sides). You will get

-3 - 9 = 2S -12 - 18,   or

-12 = 2S - 30,   or

2S = 30 - 12 = 18,   which implies   S =  = 9.


Thus the son's present age is 9 years. 
Then that of the mother is 2*(S+9) - 9 = 2(9+9) - 9 = 2*18-9 = 27.  (from the equation (1)).


Check.  You need to check the equation (2).
        M - 3 = 27 - 3 = 24.   S - 3 = 9 - 3 = 6.   24 = 4*6.   Correct.