SOLUTION: Beth is three years older than Jay. Find their present ages if the product of Beth's age is four years from now and Jay's age three years ago is 75.Thank You!

Question 739958: Beth is three years older than Jay. Find their present ages if the product of Beth's age is four years from now and Jay's age three years ago is 75.Thank You!

Found 2 solutions by lwsshak3, josmiceli:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Beth is three years older than Jay. Find their present ages if the product of Beth's age is four years from now and Jay's age three years ago is 75
***
let x=Jay's present age
x+3=Beth's present age
x+3+4=x+7=Beth's age 4 yrs from now
x-3=Jay's age 3 yrs ago
(x+7)(x-3)=75
x^2+4x-21=75
x^2+4x-96=0
(x-8)(x+12)=0
x=-12( reject, x>0)
or
x=8
x+3=11
Jay's present age=8
Beth's present age=11
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = Beth's age now
Let = Jay's age now
= Beth's age in years
= Jay's age 3 years ago
-----------------------------
given:
(1)
(2)
--------------------------
Substitute (1) into (2)
(2)
(2)
(2)
(2)
By looking at this I can see:
(2)
( ignore negative answer )
and, since,
(1)
(1)
Beth is 11 and Jay is 8
check:
(2)
(2)
(2)
(2)
OK

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