SOLUTION: two years ago,a man was five times as old as his son. two years later,his age be 8 more than three times the age of the son. find the present age of the man and his son

Question 878514: two years ago,a man was five times as old as his son. two years later,his age be 8 more than three times the age of the son. find the present age of the man and his son
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
M-2 = 5(S-2) [two years ago, a man was five times as old as his son.]
M-2 = 5S - 10
Add 2 to each side
M = 5S - 8
.
M+2 = 3(S+2) + 8 [two years later, his age be 8 more than three times the age of the son.] (assuming this means "two years from now")
M+2 = 3S+6+8
Subtract 2 from each side
M = 3S +12
.
Substitute the value of M from the first equation into the second.
(5S-8) = 3S+12
Add 8 to each side
5S = 3S + 20
Subtract 3S from each side
2S = 20
Divide each side by 2
S = 10
.
If the son is 10, then the father is 42 (two years ago, the son would have been 8. At that time, his father was 5x as old, or 40. Add the two years to get 42.)
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