Implement a procedure remove that takes in a list and returns a new list with all instances of item removed from lst

(define (filter-lst fn lst)
    'YOUR-CODE-HERE
    (if (null? lst)
        lst
        (if (fn (car lst))
            (cons (car lst)
                  (filter-lst fn (cdr lst))
                  )
            (filter-lst fn (cdr lst))
            )
        )
    )

Implement a procedure remove that takes in a list and returns a new list with all instances of item removed from lst. You may assume the list will only consist of numbers and will not have nested lists.

Hint: You might find the filter-lst procedure useful.

my code below:

(define (remove item lst)
    'YOUR-CODE-HERE
    (cond ((null? lst) '())
     (equal? item (car (lst))(remove item (cdr lst)))
            (else (cons (car lst)(remove item (cdr (lst)))))
            )
  )
;;; Tests
(remove 3 null)
; expect ()
(remove 3 '(1 3 5))
; expect (1 5)
(remove 5 '(5 3 5 5 1 4 5 4))
; expect (3 1 4 4)

My code went wrong!

application: not a procedure;
 expected a procedure that can be applied to arguments
  given: '(1 3 5)

Can some one help me fix it orz

You don't need to loop over the list in remove. As the instructions say:

Hint: You might find the filter-lst procedure useful.

So you just have to write a procedure that returns true for values that aren't the same as item, and pass this as the fn argument to filter-lst.

(define (remove item lst))
  (filter-lst (lambda (x) (not (= x item))) lst))