SOLUTION: Fathers is three years older than mother Brother is two years older than sister Their ages added up is 73 Four years ago their ages added up to 58 Find the ages of each perso

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Question 1079856: Fathers is three years older than mother
Brother is two years older than sister
Their ages added up is 73
Four years ago their ages added up to 58
Find the ages of each person in the family
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
Fathers is three years older than mother
f = m + 3
Brother is two years older than sister
b = s + 2
Their ages added up is 73
f + m + b + s = 73
Four years ago their ages added up to 58
(f-4) + (m-4) + (b-4) + (s-4) = 58
f + m + b + s - 16 = 58
f + m + b + s = 58 + 16
f + m + b + s = 74
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Find the ages of each person in the family
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Fathers is three years older than mother
Brother is two years older than sister
Their ages added up is 73
Four years ago their ages added up to 58
Find the ages of each person in the family
If the sum of the 4 people's ages is 73, then 4 years ago, the sum would've been: 73 - 4(4), or 73 - 16 = 57, unless of course,

the youngest person was not born 4 years ago. That's what it is!



Let the father’s age, be F, and brother’s, B

Then mother’s is: F – 3, and sister’s is: B – 2

We then get: F + B + F – 3 + B – 2 = 73

2F + 2B – 5 = 73

2F + 2B = 78 ------- eq (i)



The youngest is the sister, so it appears that 4 years ago, she was not born so we get:

F – 4 + F – 3 – 4 + B – 4 = 58

2F + B – 15 = 58

2F + B = 73 ------- eq (ii)

B = 78 – 73 ------ Subtracting eq (ii) from eq (i)

B, or 



2F + 5 = 73 ------- Substituting 5 for B in eq (ii)

2F = 73 – 5

2F = 68

F, or 



 

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