Two years ago, Dilip was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.
Check your solution.
Answers
Answer:
dilip 38 son 14 at present ages
Answer:
let the present age of Dilip be x
and present age of his son be y
by first condition
x-2=3(y-2)
x-2=3y-6
x-3y-2+6=0
x-3y+4=0
x-3y=-4-----------------equation (1)
by second condition
2x+2=5(y+2)
2x+2=5y+10
2x-5y+2-10=0
2x-5y-8=0
2x-5y=8-------------equation (2)
multiply equation(1) by 2 we get,
2x-6y=-8---------------equation (3)
subtract equation (2) and (3)
2x-5y=8
2x-6y=-8
- + +
y=16
put value of y in equation (1)
x-3y=-4
x-3(16)=-4
x-48=-4
x=-4+48
x=44
therefore the present age of Dilip is 44years and present age of his son is 16 years
checking the solution
we will take equation (1) and substitute the value of x and y that is there present ages
x-3y=-4
44-3(16)=-4
44-48=-4
-4=-4
LHS=RHS
therefore the solution is correct
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