Question 778562
Ten years ago, the sum of the ages of two sons was (1/3) of their father’s age. 
One son is two years older than the other 
and sum of their present ages is 14 years less than the father’s present age. Find the present ages of all
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Equations:
s1-10 + s2-10 = (1/3)f
s1 = s2+2
s1+s2 = f-14
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Simplify::
s1+s2-20 = (1/3)f
Sustitute for s1+s2
f-14 + (1/3)f
3f-42 = f
2f = 42
father's age is 21 years 
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s1+s2 = 21-14 = 7
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Substitute for s1 to get:
s2+2 + s2 = 7
2s2 = 5
s2 = 2 1/2 years
Then s1 = 4 1/2 years
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Cheers,
Stan H.
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