Question 46232
OK, this is straight forward really. Just need to start correctly with a definition.


Let x be the age of Lisa NOW.
Let y be the age of Sally NOW. 
As Sally is 3 times older, then --> y = 3x


6 years ago, the ages of the 2 were:
Lisa: x-6 and 
Sally: y-6


And Sally's age then was 9 times larger than Lisa's. So
(y-6) = 9(x-6)
y-6 = 9x-54
y = 9x-48


So we have 2 equations:
y = 3x --> eqn 1 and
y = 9x-48 --> eqn2


Now we have 2 equations to solve. There are a few ways of solving them. This is one:


sub eqn1 into eqn2:
3x = 9x-48
0 = 6x-48
48 = 6x
x = 48/6
x = 8


So Lisa is 8 years old now and so Sally is 24 years old.
and 6 years ago, Lisa was 2 and Sally 18... a difference of 9 times.


jon.