A man and a women have the same birthday.When he was as old as she is now, the man was twice as old as the woman. When she becomes as old as he is now, the sum of their ages will be 119.How old is the man now
Call the man's age now x. Call the woman's age now y.
Then x - y is the difference in their ages.
So when the man was as old as the woman is now (y years) then the man was twice as old as the woman. So the woman was y - [x - y] years old and the man was y years. Thus
y = 2*(y - [x - y]) = 2*(2y - x) = 4y - 2x
or
3y - 2x = 0
Now, when she becomes as old as he is now (x years) the sum of their ages will be 119. The man is now x + [x - y] years old and the woman is x years. Thus
(x + [x - y]) + x = 119
or
3x - y = 119
Putting these two together gives the equations:
3y - 2x = 0
3x - y = 119
Solving the first one for y gives:
y = (2/3)x
Inserting this value of y into the second equation gives:
3x - [(2/3)x] = 119
(7/3)x = 119
x = (3/7)*119 = 51
y = (2/3)x = (2/3)*51 = 34
So the man is now 51 and the woman is 34. I leave it to you to check the ages.
-Dan