Ages

If the product of a woman's present age (W) and wedding age (w) is subtracted from the product of her husband's present age (M) and wedding age (m) and the result of this is added to (Wm-Mw) the result is 553. Find the present ages.

I'm assuming that you're looking for an integer solution. If so, the man is presently 43 and the woman is 36. They have been married for 16 years.

Originally posted by elopawn
If the product of a woman's present age (W) and wedding age (w) is subtracted from the product of her husband's present age (M) and wedding age (m) and the result of this is added to (Wm-Mw) the result is 553. Find the present ages.

Mm - Ww + Wm - Mw
= M(m - w) + W(m - w)
= (M + W)(m - w)

But M - m = W - w
So m - w = M - W

Hence (M + W)(M - W) = 553 = 79*7

M = 43
W = 36

Originally posted by PBE6
They have been married for 16 years.

How do you get that? Seems to me that all you can get about their ages when married is the that he's 7 years older.

Originally posted by Schumi
How do you get that? Seems to me that all you can get about their ages when married is the that he's 7 years older.

The current ages must sum to 79, and differ by 7.

Originally posted by THUDandBLUNDER
Mm - Ww + Wm - Mw
= M(m - w) + W(m - w)
= (M + W)(m - w)

But M - m = W - w
So m - w = M - W

Hence (M + W)(M - W) = 553 = 79*7

M = 43
W = 36

Yep that's it. Good work.

If you still dont get it.....

We have Mm-Ww+Wm-Mw = 553
Hence M(m-w)+W(m-w) = 553
Therefore (M + W)(m - w) = 553 = 7*79

Let x be the number of years between the wedding and the present day. Then m = M-x, w = W-x and m-w = M-x-W+x = M-W

Hence (M + W)(M - W) = 7*79

Then either M+W = 553 ( this is not reasonable as the couple would be too old)

or M+W = 79

Then M-W = 7. Adding these gives 2M = 79+7 = 86 and M=43
Then W = 43 - 7 = 36

Basically what TB said. You should be able to get it now Schumi.