SOLUTION: the father's age 10 years ago was 35 years more than twice his son's age. After how many years from now will the father be (i) twice his son's age? (ii) thrice his son's age?

Question 201804: the father's age 10 years ago was 35 years more than twice his son's age. After how many years from now will the father be
(i) twice his son's age?
(ii) thrice his son's age?
Found 2 solutions by RAY100, MathTherapy:
Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website!
Let f = father's age 10 yrs ago, and s= son's age 10 yrs ago
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f = 2s +35,,,,,we can continue to use variable symbols, but it is much clearer to use arbitrary numbers.
Assume son is 10 at this point,,f= 2(10) +35 =55,,,all 10 yrs ago
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Adding 10 to both, today they are f=65,,,s=20
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father will be double if,,, (65+x) = 2(20+x)
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65+x = 40 +2x
25 = x,,,,,,,,,that is 25 yrs hence, father is twice age of son
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checking f= 65 + 25 =90,,,,,son = 20 +25 =45,,,90/45=2,,,ok
.
father will be tripled if ,,,(65+y) = 3 (20 +y)
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65+y =60 +3y
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5= 2y
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2.5 = y,,,,or 2.5 yrs hence father will be triple sons age
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checking 65 + 2.5 =67.5,,,20+2.5 =22.5,,,,67.5/22.5 =3,,,,ok

Answer by MathTherapy(639) (Show Source):
You can put this solution on YOUR website!
Let son’s age now be s
Then, 10 years ago, son’s age was s – 10
Therefore, father’s age, 10 years ago was 2(s – 10) + 35 = 2s – 20 +35 = 2s + 15, which means that father’s age now is 2s + 15 + 10 = 2s + 25

To find the amount of years for father’s age to be twice his son’s, and using y as those amount of years, we get: 2(s + y) = 2s + 25 + y
2s + 2y = 2s + 25 + y
y = 25

So, in , father will be twice his son’s age

To find the amount of years for father’s age to be thrice his son’s, and using y as those amount of years, we get: 3(s + y) = 2s + 25 + y
3s + 3y = 2s + 25 + y
2y = - s + 25

So, in , father will be/was thrice his son’s age, where "s" represents the son's age, today.