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(2198) About Me  (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