Question 567656
Let
Present age of Marilyn = m
Present age of Rosie = r



Marilyn is three years younger than her friend Rosie
m=r-3..............(1)

Five years ago
Age of Marilyn was = m-5
Age of Rosie was = r-5

In seven years
Age of Marilyn will be = m+7
Age of Rosie will be = r+7

THEN

In seven years the product of their ages will be 2 less than six times the product of their ages five years ago.

(m+7)(r+7)=6[(m-5)(r-5)]-2
mr+7m+7r+49=6(mr-5m-5r+25)-2
mr+7m+7r+49=6mr-30m-30r+150-2
mr-6mr+7m+30m+30r+7r=150-2-49
-5mr+37m+37r=150-51
-5mr+37m+37r=99..................(2)

Put the value of m from (1) to (2)
-5(r-3)r+37(r-3)+37r=99
-5r^2+15r+37r-111+37r=99
-5r^2+89r-99-111=0
-5r^2+89r-200=0
Now solve for r
*[invoke quadratic "r", -5, 89, -210 ]

r=15 or r = 2.8 (reject)

Put r=15 in (1)
m=r-3
m=15-3
m=12


Present age of Marilyn = m = 12 years old
Present age of Rosie = r = 15 years old



Check
======
Five years ago
Age of Marilyn was = 7
Age of Rosie was = 10

In seven years
Age of Marilyn will be = 19
Age of Rosie will be = 22

THEN

In seven years the product of their ages will be 2 less than six times the product of their ages five years ago.

19*22=6(7*10)-2
418=6(70)-2
418=420-2
418=418