A man and a woman's ages total 98.
He is twice as old as she was when he was her age now.
What are their ages?
m+w=98
m = 2(w-(m-w))
she is 42 and he is 56
He is 14 years older than she, so 14 years ago, she was 28.
Yes. The difference of their ages is the key.
To solve this problem, let's assign variables to represent the man's age and the woman's age. Let's say the man's age is M and the woman's age is W.
We are given two pieces of information:
1. The sum of their ages is 98: M + W = 98.
2. The man is twice as old as the woman was when he was her age now: M = 2(W - (M - W)).
Let's simplify the second equation:
M = 2(W - (M - W)).
M = 2(2W - M).
M = 4W - 2M.
Now, let's solve this system of equations to find the values of M and W. We can use the substitution method or the elimination method.
Using the substitution method:
From the first equation, we can solve for W:
W = 98 - M.
Substituting this value into the second equation:
M = 4(98 - M) - 2M.
M = 392 - 4M - 2M.
M + 4M + 2M = 392.
7M = 392.
M = 392 / 7.
M = 56.
Substituting the value of M back into the first equation:
56 + W = 98.
W = 98 - 56.
W = 42.
Therefore, the man is 56 years old and the woman is 42 years old.