Let H, K and E be the amounts the persons had originally.


First equation is obvious

   H + K + E = 2190.         (1)



Hector spent 2/5 of his money; so, he left with 3/5 of his money.

Kenny spent $80;  so Kenny left with (K-80) dollars.

At the end, Kenny and Hector had the same amount, which gives us the second equation

     = K - 80

or

    3H = 5K - 400.           (2)



Hector spent half as much as Elliot, which means that Elliot spent twice as much as Hector.

Hence, Elliot spent .

It means that Elliot left with E -   dollars.


From the problem, at the end Hector had left $160 less than what Elliot had left.

It gives us the last, third equation

     + 160 = E - ,

or

    7H - 5E = -800.          (3)



Thus we have the system of 3 equations in 3 unknowns


    H +  K +  E = 2190.      (1)

   3H - 5K     =  400.       (2)

    7H    - 5E = -800.       (3)


So the problem is algebraized.


As I said at the beginning, the major step is to algebraize; solving the system is just technical part.


So, I used one of many existing online solvers www.reshish.com to save my time,

and I got this ANSWER : H= 650;  K = 470;  E = 1070.