Question 974504: A man is twice twice his sons age five years older than his wife.the total of their age is 64. Find each person's age. Answer by LinnW(1048) (Show Source):
You can put this solution on YOUR website! Set M = man's age
set W = woman's age
Set S = son's age
M = 2S ( man is twice sons age )
M = W + 5 ( five years older than his wife )
M + W + S = 64
First consider M = 2S, and solve for S
divide each side by 2
M/2 = S
Now consider M = W + 5 and solve for W
add -5 to each side
M - 5 = W
Now we want to work with these 3 equations
S = M/2
W = M - 5
M + W + S = 64
In the last equation, substitute (M/2) for S, and (M - 5) for W
M + (M - 5) + (M/2) = 64
2M + M/2 - 5 = 64
add 5 to each side
5M/2 = 69
multiply each side by 2
5M = 138
divide each side by 5
M = 138/5
Since S = M/2, S = 138/10
Since W = M - 5 , W = 138/5 - 5 ,
W = 138/5 - 25/5
W = 113/5
So we have M = 138/5 , W = 113/5 and S = 138/10 or
M = 27 , W = 22 , S = 13
