SOLUTION: Man'S AGE IS 3TIMES OF HIS SON .TEN YEARS AGO HE WAS 5TIMES OF HIS SON .FIND AGES OF MAN AND SON.USING SIMULTANEOUS EQUATIONS TO SOLVE IT . ANS 20YEARS 60YEARS.

Question 389385: Man'S AGE IS 3TIMES OF HIS SON .TEN YEARS AGO HE WAS 5TIMES OF HIS SON .FIND AGES OF MAN AND SON.USING SIMULTANEOUS EQUATIONS TO SOLVE IT . ANS 20YEARS 60YEARS.
Answer by dudeantariksh(10) (Show Source): You can put this solution on YOUR website!
hi there!!
well, in order to solve ur problem we'll go by the track of 'LINEAR EQUATIONS IN TWO VARABLES' or 'simulataneous equations'
Thereby, let us assume that man's present age is x years and that of his son is y years.
now , as per the first given condition that' Man'S AGE IS 3TIMES his son's',
we get,
x = 3y
now we subtrat [3y] from both sides and get
x-3y = 0.............................. result no.1
now,
man's age 10 years ago was (x-10) years and that of his son will be(y-10)years.
now, as per the second given condition that' TEN YEARS AGO HE WAS 5TIMES OF HIS SON', we get,
(x-10)= 5(y-10)
thus, x-10 = 5y - 50
now we take the like terms al on one side, i.e., we take the variables[alphabets] on 1 side and the contants[numbers] on one side.
we get x-5y = -50+10
x-5y = -40............................result no.2
now we subtract equation 2 from 1 and we get,
(x-3y)-(x-5y)= 0-(-40)
now, as we have the minus sign outside d bracket in d L.HS., all the terms under it change their signs.
thus,
x - 3y -x +5y = 40/highlight x
thus, x nd -x get cancelled.
thus,
5y-3y = 40
thus,
2y = 40
now we divide by 2 throughout
and we get.(y = 20).................................result no.3/highlight y=20
now we substitute the value of y in equation 1 for obtaining x and we get,
x-3(20) = 0
therefore,
x-60 +0
therefore, (x=60)/highlight x=60
now, x = 60 years
and y = 20 years.
thus,
we find that the present age of father is 60 years and that of his sone is 20 years..
[since, the given values and obtained values are the same there is n need for verification]

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