# SOLUTION: 2 years ago a man's age was 3 times the square of his son's age .3 years hence his age will be 4 times his son's age .find their present ages please help me to solve this problem

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 Question 201881: 2 years ago a man's age was 3 times the square of his son's age .3 years hence his age will be 4 times his son's age .find their present ages please help me to solve this problem Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!Let x=son's present age;(x-2)=son's age 2 years ago And let y=man's present age;(y-2)=man's age two years ago so: (y-2)=3(x-2)^2 or y-2=3(x^2-4x+4) y-2=3x^2-12x+12 y=3x^2-12x+14---------------eq1 (x+3)=son's age 3 years hence (y+3)=man's age 3 years hence so: (y+3)=4(x+3) y+3=4x+12 y=4x+9------------------eq2 substitute y=4x+9 into eq1 4x+9=3x^2-12x+14 3x^2-16x+5=0 quadratic in standard form and it can be factored (3x-1)(x-5)=0 3x-1=0 x=1/3 year old---Clearly this does not work because 2 years ago he would not have been born and x-5=0 x=5 years old-----------------son's present age substitute x=5 into eq 2 y=4*5+9=29-------------------------man's present age CK Two years ago 5-2=3-----------son's age 29-2=27----------mans present age 27=3*3^2 27=27 and three years hence 5+3=8 son's age and 29+3=32 man's age 32=4*8 32=32 Hope this helps---ptaylor