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Linear Equations- Problems on Age
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Linear Equations- Problems on Age

If twice the son's age in years is added to the father's age, the sum is 70. But if twice the father's age is added to the son's age, the sum is 95. Find the ages of father and son.
Solution: let suppose father's age (in years) be x and that of son's be y. Then,
          x + 2y = 70
         and, 2x + y = 95
you can write the system of equations as:
         x + 2y - 70 = 0 
         2x + y - 95 = 0
By cross-multiplication, you get

x = -120 / -3 = 40 and y = -45 / -3 = 15
Hence, father's age is 40 years and the son's age is 15 years.
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Example: Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages.
Solution: Let the present ages of father and son be x years and y years respectively.
Ten years ago, Fathers's age = (x - 10) years.
         son's age = (y - 10) years
   x - 10 = 12(y - 10)  x - 12y + 110 = 0
then years later, father's age = (x + 10) years
          son's age = (y + 10) years
    x + 10 = 2(y + 10) x - 2y -10 = 0
Subtracting (ii) from (i), you get
          -10y + 120 = 0 10y = 120 y =12
Putting y = 12 in (i), you get:
           x - 144 + 110 = 0 x = 34
thus, present age of father is 34 years and the present age of son is 12 years.
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