Algebra 2
posted by Anonymous .
"My husband's age," remarked a lady the other day, "is represented by the figures of my own age reversed. He is my senior, and the difference between our ages is oneeleventh of their sum."
10x + y = husband's age
10y + x = wife's age
(10x+y)  (10y+x) = ((10x+y) + (10y+x))
10x + y  10y  x = (10x + y + 10y + x)
9x  9y = (11x + 11y)
9x  9y = 11(x + y)
9x  9y = x + y
9x  x = 9y + y
8x = 10y
x = y
x = 1.25y
How do you get the answer Husband is 54, wife is 45. I understand how to set it up and solve it, now how do i figure out their ages?

(10x+y)  (10y+x) = ((10x+y) + (10y+x))/11
11(9x  9y) = 11x + 11y
9x  9y = x + y
8x = 10y
4x = 5y
Both x and y are integers less than 10
Since 5 divides 5y, 5 must divide 4x. That means x = 5. Thus, y=4.