Money in wallet problem (haha)

[nanase]

Please confirm if I have all the answers or not?
so far I have 3 combinations of x for $5 and y for $10
(20,10)
(15,15)
(35,7)
anything else?

 

[Deleted member 4993]

oh bummer, I just realised that I got one answer only out of several possible answers.... so to find the rest, i really have to "observe and match" from a list or a table? there is really no algebraic or other mathematical methods available?

Remember two other constraints on x & y

• x & y are positive integers, and
  
  
• (x + y) < 50

 

[Dr.Peterson]

Please confirm if I have all the answers or not?
so far I have 3 combinations of x for $5 and y for $10
(20,10)
(15,15)
(35,7)
anything else?

Almost there. Keep in mind my comment that it can be y rather than x that is a multiple of 5.

 

[nanase]

I checked my table again by rearranging the equation, this time making x the subject,
I found one more set (12,30)
is that it? a total of 4 ?