Question 852233
Mary is younger than her husband, but their ages contain the same two digits.
let a = husband 10s digit
let b = husbands unit digit
then
10a+b = husbands age
and
10b+a = M's age
:
 The sum of their age is 11 times the difference of their ages.
(10a+b) + (10b+a) = 11[(10a+b)-(10b+a)]
11a + 11b = 11(10a - a - 10b + b)
11a + 11b = 11(9a - 9b)
simplify, divide thru by 11
a + b = 9a - 9b
b + 9b = 9a - a
10b = 8a
Simplify, divide by 2
5b = 4a
b = {{{4/5}}}a
b = .8a
the only single digit solution is when a = 5
b = .8(5)
b = 4
Husband is 54, Mary is 45
:
:
See if that checks out in the statement:
"The sum of their age is 11 times the difference of their ages."
54 + 45 = 11(54-45)
99 = 11(9)