Question 607031
The sum of the ages of a man and his wife is 80 years. 10 years ago, the ages of the wife was exactly two-third the age of the man. Find the present ages of the man and his wife.


Let:
man's age = x
wife's age = y


x+y=80 <--sum of their age
y-10=2/3x <--10 years ago, the age of the woman was 2/3 of the man's current age


y=2/3x+10 <--solve the 2nd equation for y
x+2/3x+10=80 <--plug above into 1st equation
5/3x=70 <--simplify and solve
x=42 <--plug back into original equation to find y
42+y=80
y=80-42
y=38


Since x=42 and y=38
The man's age is 42 and the wife's age is 38